10. Parametric Surfaces and Surface Integrals

Exercises

a. Introduction to Parametric Surfaces

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1. Parametrize the paraboloid \(z=x^2+y^2\) in terms of \(r\) and \(\theta\).

   

   Start with cylindrical coordinates.

   \(\vec R(r,\theta)=\left\langle r\cos\theta,r\sin\theta,r^2\right\rangle\)

   We start with the cylindrical coordinate system: \[ \vec R(r,\theta,z)=\left\langle r\cos\theta,r\sin\theta,z\right\rangle \] We write the equation of the paraboloid in cylindrical coordinates: \[ z=x^2+y^2=r^2\cos^2\theta+r^2\sin^2\theta=r^2 \] Thus the parametrization of the paraboloid becomes: \[ \vec R(r,\theta)=\left\langle r\cos\theta,r\sin\theta,r^2\right\rangle \]

   dm 

2. Parametrize the cone \(z=5r\) in terms of \(r\) and \(\theta\) and find the point \((x,y,z)\) where \(r=3\) and \(\theta=\pi\)

   

   Start with cylindrical coordinates.

   \(\vec R(r,\theta)=\left\langle r\cos\theta,r\sin\theta,5r\right\rangle\)
   \(\vec R(3,\pi)=\left\langle -3,0,15\right\rangle\)

   Again we start with the cylindrical coordinate system: \[\vec R(r,\theta,z)=\left\langle r\cos\theta,r\sin\theta,z\right\rangle\] This time our constraint is \(z=5r\) which we can simply plug in: Thus the parametrization of the paraboloid becomes: \[ \vec R(r,\theta)=\left\langle r\cos\theta,r\sin\theta,5r\right\rangle \] To find where \(r=3\) and \(\theta=\pi\), we plug in: \[\begin{aligned} x&=r\cos\theta=3\cos\pi=-3 \\ y&=r\sin\theta=3\sin\pi=0 \\ z&=5r=15 \end{aligned}\] Or \(\vec R(3,\pi)=\left\langle -3,0,15\right\rangle\).

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3. Parametrize the cylinder \(y^2+z^2=25\) (centered on the \(x\)-axis) in terms of \(x\) and \(\theta\). Be sure \(\theta=0\) is on the positive \(y\) axis and \(\theta=\dfrac{\pi}{2}\) is on the positive \(z\) axis. Then find the rectangular coordinates of the point where \(x=1\) and \(\theta=\dfrac{\pi}{4}\).

   

   Start with cylindrical coordinates centered around the \(x\) axis.

   \(\vec R(x,\theta) =\left\langle x,5\cos\theta,5\sin\theta\right\rangle\)
   \(\vec R\left(1,\dfrac{\pi}{4}\right) =\left\langle 1,5\dfrac{\sqrt{2}}{2},5\dfrac{\sqrt{2}}{2}\right\rangle\)

   This time we start with the cylindrical coordinate system oriented along the x axis: \[ \vec R=\left\langle x,r\cos\theta,r\sin\theta\right\rangle \] This time our constraint is \(y^2+z^2=25\) which we use to eliminate \(r\): \[ y^2+z^2=r^2\cos^2\theta+r^2\sin^2\theta=r^2=25 \quad \rightarrow \quad r=5 \] Therefore \[ \vec R(x,\theta)=\left\langle x,5\cos\theta,5\sin\theta\right\rangle \] To find where \(x=1\) and \(\theta=\dfrac{\pi}{4}\), we plug in: \[\begin{aligned} x&=1 \\ y&=5\cos\pi/4=5\dfrac{\sqrt{2}}{2} \\ z&=5\sin\pi/4=5\dfrac{\sqrt{2}}{2} \end{aligned}\] Or \(\vec R\left(1,\dfrac{\pi}{4}\right) =\left\langle 1,5\dfrac{\sqrt{2}}{2},5\dfrac{\sqrt{2}}{2}\right\rangle\).

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4. Parametrize the apple \(\rho=1-\cos\phi\) in terms of \(\theta\) and \(\phi\)

   

   Begin with spherical coordinates and then eliminate one of the variables by using the constraint equation given.

   \(\vec R(\theta,\phi) =\left\langle (1-\cos\phi)\sin\phi\cos\theta, (1-\cos\phi)\sin\phi\sin\theta,(1-\cos\phi)\cos\phi\right\rangle\)

   This time we start with the spherical coordinates: \[ \vec R =\left\langle \rho\sin\phi\cos\theta,\rho\sin\phi\sin\theta,\rho\cos\phi\right\rangle \] This time our constraint is \(\rho=1-\cos\phi\) which we plug in for \(\rho\): \[ \vec R(\theta,\phi) =\left\langle (1-\cos\phi)\sin\phi\cos\theta, (1-\cos\phi)\sin\phi\sin\theta,(1-\cos\phi)\cos\phi\right\rangle \]

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b. Tangent and Normal Vectors

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5. Consider the cone parameterized as \(\vec R(r,\theta) =\left\langle r\cos\theta,r\sin\theta,5r\right\rangle\). Find the coordinate curves for \(r=2\) and for \(\theta=\dfrac{\pi}{3}\). Then find tangent and normal vectors to the surface of the cone and the length of the normal.

   

   Differentiate the parametrization with respect to the parameters. Then take their cross product.

   \(\vec R(2,\theta)=\left\langle 2\cos\theta,2\sin\theta,10\right\rangle\) This is a circle of radius \(2\) at height \(z=10\).
   \(\vec R(r,\pi/3)=\left\langle \dfrac{r}{2},\dfrac{\sqrt{3}r}{2},2r\right\rangle\) This is a line through the origin in the direction \(\vec v=\left\langle \dfrac{1}{2},\dfrac{\sqrt{3}}{2},2\right\rangle\).
   \(\begin{aligned} \vec{e}_r&=\left\langle \cos\theta,\sin\theta,5\right\rangle \\ \vec{e}_\theta&=\left\langle -r\sin\theta,r\cos\theta,0\right\rangle \\ \vec{N}&=\left\langle -5r\cos\theta,-5r\sin\theta,r\right\rangle \\ \left|\vec{N}\right|&=\sqrt{25r^2\cos^2\theta+25r^2\sin^2\theta+r^2}=\sqrt{26}r \end{aligned}\)

   To find the coordinate curve for \(r=2\) we plug this value into the parametrization \[ \vec R(2,\theta)=\left\langle 2\cos\theta,2\sin\theta,10\right\rangle \] This is a circle of radius \(2\) at height \(z=10\). Similarly for \(\theta=\pi/3\) we get \[ \vec R(r,\pi/3) =\left\langle \dfrac{r}{2},\dfrac{\sqrt{3}r}{2},2r\right\rangle. \] which is a line through the origin in the direction \(\vec v=\left\langle \dfrac{1}{2},\dfrac{\sqrt{3}}{2},2\right\rangle\). To find the tangent vectors, we differentiate the parametrization: \[ \vec R(r,\theta) =\left\langle r\cos\theta,r\sin\theta,5r\right\rangle \] to get: \[\begin{aligned} \vec{e}_r&=\dfrac{\partial\vec R}{\partial r} =\left\langle \cos\theta,\sin\theta,5\right\rangle \\ \vec{e}_\theta&=\dfrac{\partial\vec R}{\partial\theta} =\left\langle -r\sin\theta,r\cos\theta,0\right\rangle \end{aligned}\] To find the normal vector, we compute the cross product of the tangent vectors: \[\begin{aligned} \vec{N}&=\vec{e}_\phi\times\vec{e}_\theta =\left| \begin{array}{ccc} \hat{\imath} & \hat{\jmath} & \hat{k} \\ \cos\theta & \sin\theta & 5\\ -r\sin\theta & r\cos\theta& \quad0\quad \end{array} \right| \\ &=\hat{\imath}(-5r\cos\theta)-\hat{\jmath}(5r\sin\theta) +\hat{k}(r\cos^2\theta+r\sin^2\theta) \\ &=\left\langle -5r\cos\theta,-5r\sin\theta,r\right\rangle \end{aligned}\] The length of the normal vector is: \[\left|\vec{N}\right|=\sqrt{25r^2\cos^2\theta+25r^2\sin^2\theta+r^2} =\sqrt{26}\,r\]

   dm,lk 

6. Consider the cylinder parameterized as \(\vec R(x,\theta) =\left\langle x,5\cos\theta,5\sin\theta\right\rangle\). Find the tangent and normal vectors to the surface of the cylinder.

   

   Differentiate the parametrization with respect to the parameters. Then take their cross product.

   \(\begin{aligned} \vec{e}_x&=\left\langle 1,0,0\right\rangle \\ \vec{e}_\theta&=\left\langle 0,-5\sin\theta,5\cos\theta\right\rangle \\ \vec{N}&=\left\langle 0,-5\cos\theta,-5\sin\theta\right\rangle \end{aligned}\)

   To find the tangent vectors, we differentiate the parametrization: \[ \vec R(x,\theta) =\left\langle x,5\cos\theta,5\sin\theta\right\rangle \] to get: \[ \begin{aligned} \vec{e}_x&=\dfrac{\partial\vec R}{\partial x} =\left\langle 1,0,0\right\rangle \\ \vec{e}_\theta&=\dfrac{\partial\vec R}{\partial\theta} =\left\langle 0,-5\sin\theta,5\cos\theta\right\rangle \end{aligned} \] To find the normal vector, we compute the cross product of the tangent vectors: \[\begin{aligned} \vec{N}&=\vec{e}_x\times\vec{e}_\theta =\left| \begin{array}{ccc} \hat{\imath} & \hat{\jmath} & \hat{k} \\ \left\langle 1\right. & 0 & \left.0\right\rangle\\ \left\langle 0\right. & -5\sin\theta& \left.5\cos\theta\right\rangle \end{array} \right| \\ &=\hat{\imath}(0)-\hat{\jmath}(5\cos\theta)+\hat{k}(-5\sin\theta\theta) \\ &=\left\langle 0,-5\cos\theta,-5\sin\theta\right\rangle \end{aligned}\]

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7. Consider the hyperbolic paraboloid \(z=x^2-y^2\)

   

   1. Parametrize the surface starting from cylindrical coordinates.

      What is \(z=x^2-y^2\) in cylindrical coordinates?

      \(\vec R(r,\theta) =\left\langle r\cos\theta,r\sin\theta,r^2(\cos^2\theta-\sin^2\theta)\right\rangle \)

      Cylindrical coordinates are: \[ \vec R(r,\theta,z) =\langle r\cos\theta,r\sin\theta,z\rangle \] but \[ z=x^2-y^2=r^2\cos^2\theta-r^2\sin^2\theta \] So the hyperbolic paraboloid can be parametrized as: \[\vec R(r,\theta) =\left\langle r\cos\theta,r\sin\theta,r^2(\cos^2\theta-\sin^2\theta)\right\rangle \]

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   2. Find the tangent vectors to the surface.

      Compute the \(r\) and \(\theta\) derivatives of the parametrization.

      \(\begin{aligned} \vec{e}_r&=(\cos\theta,\sin\theta,2r(\cos^2\theta-\sin^2\theta)) \\ \vec{e}_\theta&=(-r\sin\theta,r\cos\theta,-4r^2\sin\theta\cos\theta) \end{aligned}\)

      The hyperbolic paraboloid is parametrized as: \[ \vec R(r,\theta) =\left\langle r\cos\theta,r\sin\theta,r^2(\cos^2\theta-\sin^2\theta)\right\rangle \] Now we find \(\vec{e}_r\) and \(\vec{e}_\theta\) by taking partial derivatives: \[\begin{aligned} \vec{e}_r=\dfrac{\partial\vec R}{\partial r} &=(\cos\theta,\sin\theta,2r(\cos^2\theta-\sin^2\theta)) \\ \vec{e}_\theta=\dfrac{\partial\vec R}{\partial\theta} &=\left\langle -r\sin\theta,r\cos\theta, r^2(-2\cos\theta\sin\theta-2\sin\theta\cos\theta\right\rangle \\ &=\left\langle -r\sin\theta,r\cos\theta, -4r^2\sin\theta\cos\theta\right\rangle \\ \end{aligned}\]

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   3. Find the normal vector to the surface and its length.

      Take the cross product of the tangent vectors. Don't forget to include the surface differential in the integral.

      \(\vec{N} =\langle -2r^2\cos\theta,2r^2\sin\theta,r\rangle\)
      \(|\vec{N}|=\sqrt{4r^4+r^2}=r\sqrt{4r^2+1}\)

      We take their cross product of the tangent vectors to get the normal vector: \[\begin{aligned} \vec{N}&=\vec{e}_r\times\vec{e}_\theta =\begin{vmatrix} \hat\imath & \hat \jmath & \hat k \\ \cos\theta & \sin\theta & 2r(\cos^2\theta-\sin^2\theta) \\ -r\sin\theta & r\cos\theta & -4r^2\sin\theta\cos\theta \end{vmatrix} \\[2pt] &=\quad\hat{\imath}(-4r^2\sin^2\theta\cos\theta -2r^2\cos\theta(\cos^2\theta-\sin^2\theta)) \\ &\quad-\hat{\jmath}(-4r^2\sin\theta\cos^2\theta +2r^2\sin\theta(\cos^2\theta-\sin^2\theta)) \\ &\quad+\hat{k}(r\cos^2\theta+r\sin^2\theta) \\[2pt] &=\langle -2r^2\cos\theta,2r^2\sin\theta,r\rangle \end{aligned}\] The length of the normal vector is \[|\vec{N}|=\sqrt{4r^4\cos^2\theta+4r^4\sin^2\theta+r^2}=r\sqrt{4r^2+1}\]

      dm,lk 

8. Consider the heart-shaped surface given in spherical coordinates by \(\rho=\phi\).

   

   1. Parametrize the surface starting from spherical coordinates.

      Start with the spherical coordinates and replace \(\rho\).

      \(\vec R(\theta,\phi) =\left\langle \phi\sin\phi\cos\theta,\phi\sin\phi\sin\theta, \phi\cos\phi\right\rangle\)

      We start with the spherical coordinates: \[ \vec R =\left\langle \rho\sin\phi\cos\theta,\rho\sin\phi\sin\theta, \rho\cos\phi\right\rangle \] We use the constraint \(\rho=\phi\) to eliminate \(\rho\): \[ \vec R(\theta,\phi) =\left\langle \phi\sin\phi\cos\theta,\phi\sin\phi\sin\theta,\phi\cos\phi\right\rangle \]

      dm 

   2. Find the tangent vectors to the surface.

      The tangent vectors are the partial derivatives of the parametrization with respect to the parameters.

      \(\begin{aligned} \vec{e}_\theta &=(-\phi\sin\phi\sin\theta,\phi\sin\phi\cos\theta,0) \\ \vec{e}_\phi &=(\sin\phi\cos\theta+\phi\cos\phi\cos\theta, \sin\phi\sin\theta+\phi\cos\phi\sin\theta,\cos\phi-\phi\sin\phi) \end{aligned}\)

      The tangent vectors are: \[\begin{aligned} \vec{e}_\theta &=(-\phi\sin\phi\sin\theta,\phi\sin\phi\cos\theta,0) \\ \vec{e}_\phi &=(\sin\phi\cos\theta+\phi\cos\phi\cos\theta, \sin\phi\sin\theta+\phi\cos\phi\sin\theta,\cos\phi-\phi\sin\phi) \end{aligned}\]

      dm 

   3. Find the normal vector to the surface.

      The normal is the cross product of the tangent vectors.

      \(\begin{aligned} \vec{N} &=\left\langle \phi\sin\phi\cos\theta(\cos\phi-\phi\sin\phi),\right. \\ &\qquad\quad\phi\sin\phi\sin\theta(\cos\phi-\phi\sin\phi), \\ &\qquad\qquad\left.-\phi\sin^2\phi-\phi^2\sin\phi\cos\phi\right\rangle \end{aligned}\)

      The normal is the cross product of the tangent vectors: \[\begin{aligned} \vec{N}&=\vec{e}_\theta\times\vec{e}_\phi \\ &=\begin{vmatrix} i & j & k \\ -\phi\sin\phi\sin\theta & \phi\sin\phi\cos\theta & 0 \\ \sin\phi\cos\theta+\phi\cos\phi\cos\theta & \sin\phi\sin\theta +\phi\cos\phi\sin\theta & \cos\phi-\phi\sin\phi \end{vmatrix} \\ &=\quad\hat{\imath}(\phi\sin\phi\cos\phi\cos\theta-\phi^2\sin^2\phi\cos\theta) \\ &\quad-\hat{\jmath}(-\phi\sin\phi\cos\phi\sin\theta+\phi^2\sin^2\phi\sin\theta) \\ &\quad+\hat{k}(-\phi\sin^2\phi\sin^2\theta-\phi^2\sin\phi\cos\phi\sin^2\theta \\ &\qquad-\phi\sin^2\phi\cos^2\theta-\phi^2\sin\phi\cos\phi\cos^2\theta) \\ &=\left\langle \phi\sin\phi\cos\theta(\cos\phi-\phi\sin\phi),\right. \\ &\qquad\quad\phi\sin\phi\sin\theta(\cos\phi-\phi\sin\phi), \\ &\qquad\qquad\left.-\phi\sin^2\phi-\phi^2\sin\phi\cos\phi\right\rangle \end{aligned}\]

      dm 

9. Consider the cylinder from problem 3, \(y^2+z^2=25\). Find the tangent vectors and the magnitude of the normal vector to the surface.

   The tangent vectors are:
   \(\vec{e}_x=\dfrac{\partial R}{\partial x} =\left\langle 1,0,0\right\rangle\)
   \(\vec{e}_\theta=\dfrac{\partial R}{\partial\theta} =\left\langle 0,-5\sin\theta,5\cos\theta\right\rangle\)
   The magnitude of the normal vector is
   \(|\vec{N}|=5\)

   We already know that the surface is parametrized by \(\vec R(x,\theta) =\left\langle x,5\cos\theta,5\sin\theta\right\rangle\)
   The tangent vectors are: \[\begin{aligned} \vec{e}_x&=\dfrac{\partial R}{\partial x} =\left\langle 1,0,0\right\rangle \\ \vec{e}_\theta&=\dfrac{\partial R}{\partial\theta} =\left\langle 0,-5\sin\theta,5\cos\theta\right\rangle \end{aligned}\]
   The normal vector is found from the cross product: \[\begin{aligned} \vec{N}&=\vec{e}_x\times\vec{e}_\theta =\begin{vmatrix} \hat{\imath} & \hat{\jmath} & \hat{k} \\ 1 & 0 & 0 \\ 0 &-5\sin\theta & 5\cos\theta \end{vmatrix} \\ &=\hat{\imath}(0)-\hat{\jmath}(5\cos\theta)+\hat{k}(-5\sin\theta) \\ &=\left\langle 0,-5\cos\theta,-5\sin\theta\right\rangle \end{aligned}\] and its length is: \[ |\vec{N}|=\sqrt{25\cos^2\theta+25\sin^2\theta} =5 \]

   dm 

c. Scalar Surface Integrals

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10. Find the normal vector and the scalar differential of surface area of the sphere \(x^2+y^2+z^2=36\)

    

    The scalar differential of surface area is \(dS=|\vec N|\,dr\,d\theta\).

    \(\vec{N}=\langle-36\sin^2\phi\cos\theta,-36\sin^2\phi\sin\theta, -36\sin\phi\cos\phi\rangle\)
    \(dS=36\sin\phi\,d\phi\,d\theta\)

    First, parametrize the sphere by \[\vec R(\theta,\phi) =\left\langle 6\sin\phi\cos\theta,6\sin\phi\sin\theta,6\cos\phi\right\rangle\] Now find the tangent vectors \(\vec{e}_\theta\) and \(\vec{e}_\phi\): \[\vec{e}_\theta=\left\langle-6\sin\phi\sin\theta,6\sin\phi\cos\theta,0\right\rangle\] \[\vec{e}_\phi=\left\langle6\cos\phi\cos\theta,6\cos\phi\sin\theta,-6\sin\phi\right\rangle\] Now find the normal vector \[\begin{aligned} \vec{N} &=\vec{e}_\theta\times\vec{e}_\phi =\begin{vmatrix} \hat{\imath} & \hat{\jmath} & \hat{k} \\ -6\sin\phi\sin\theta & 6\sin\phi\cos\theta & 0 \\ 6\cos\phi\cos\theta & 6\cos\phi\sin\theta & -6\sin\phi \end{vmatrix} \\ &=\hat\imath(-36\sin^2\phi\cos\theta) -\hat\jmath(36\sin^2\phi\sin\theta) \\ &\quad+\hat k(-36\sin\phi\cos\phi\sin^2\theta- 36\sin\phi\cos\phi\cos^2\theta) \\ &=\langle-36\sin^2\phi\cos\theta,-36\sin^2\phi\sin\theta, -36\sin\phi\cos\phi\rangle \end{aligned}\] and now we can find the magnitude of the normal vector \[\begin{aligned} |\vec{N}| &=36\sqrt{\sin^4\phi\cos^2\theta+\sin^4\phi\sin^2\theta+\sin^2\phi\cos^2\phi} \\ &=36\sqrt{\sin^2\phi(\sin^2\phi\cos^2\theta+\sin^2\phi\sin^2\theta+\cos^2\phi)} \\ &=36\sqrt{\sin^2\phi(\sin^2\phi+\cos^2\phi)} =36\sin\phi. \end{aligned}\] Recalling the general definition of the scalar differential for surface area is \(dS = |\vec N|\,d\phi\,d\theta\), we find that \[ dS=|\vec N|\,d\phi\,d\theta =36\sin\phi\,d\phi\,d\theta. \]

    dm,lk 

11. Find the normal vector and the scalar differential of surface area of the cylinder \(x^2+y^2=36\).

    \(\vec{N}=\left\langle 6\cos\theta,6\sin\theta,0\right\rangle\)
    \(dS=6\,d\theta\,dz\)

    First, we parametrize the cylinder by: \[ \vec R(\theta,z)=\left\langle 6\cos\theta,6\sin\theta,z\right\rangle \] Now we find the tangent vectors \(\vec{e}_\theta\) and \(\vec{e}_z\): \[\begin{aligned} \vec{e}_\theta&=(-6\sin\theta,6\cos\theta,0) \\ \vec{e}_z&=(0,0,1) \end{aligned}\] and the normal vector: \[\begin{aligned} \vec{N}&=\vec{e}_\theta\times\vec{e}_z =\begin{vmatrix} \hat{\imath} & \hat{\jmath} & \hat{k} \\ -6\sin\theta & 6\cos\theta & 0 \\ 0 & 0 & 1 \end{vmatrix} \\ &=\hat{\imath}(6\cos\theta)-\hat{\jmath}(-6\sin\theta)+\hat{k}(0) =\left\langle 6\cos\theta,6\sin\theta,0\right\rangle \end{aligned}\] and its magnitude: \[|\vec{N}|=\sqrt{N_x^2+N_y^2+N_z^2} =\sqrt{36\cos^2\theta+36\sin^2\theta+0}=6\] So the scalar differential is \[dS=|\vec{N}|\,d\theta\,dz=6\,d\theta\,dz\]

    dm 

12. Parametrize the plane \(2x-3y+z=-1\) and find the normal vector. Is there more than one way to find the normal vector? Then find the scalar surface differential of the plane.

    \[\vec{N}=\left\langle 2,-3,1\right\rangle\] You can also find the normal vector as the coefficients in the plane equation or as the gradient of the function \(f(x,y,z)=2x-3y+z\). \[dS=|\vec{N}|\,dx\,dy=\sqrt{14}\,dx\,dy\]

    To parametrize, let's start with the rectangular coordinates: \[\vec R=\left\langle x,y,z\right\rangle\] The equation of the plane relates these three variables, so we can eliminate one of them. Let's eliminate \(z\), since \(z=-2x+3y-1\) we see that \[\vec R=\left\langle x,y,-2x+3y-1\right\rangle\] Now to find the tangent vectors \(\vec{e}_x\) and \(\vec{e}_y\): \[\vec{e}_x=(1,0,-2)\] \[\vec{e}_y=(0,1,3)\] Now find the normal vector \[\begin{aligned} \vec{N}&=\vec{e}_x\times\vec{e}_y= \begin{vmatrix} \hat{\imath} & \hat{\jmath} & \hat{k} \\ 1 & 0 & -2 \\ 0 & 1 & 3 \end{vmatrix} \\ &=\hat{\imath}(2)-\hat{\jmath}(3)+\hat{k}( 1)=\left\langle 2,-3,1\right\rangle \end{aligned}\] The scalar surface differential \(dS\) is \[dS=|\vec{N}|=\sqrt{2^2+(-3)^2+1^2}\,dx\,dy =\sqrt{14}\,dx\,dy\] Note that since the equation of a plane is \(\vec N \cdot X=\vec N \cdot P\), the components of \(\vec N\) can be read off as the coefficients in the plane equation. Also note that the plane is a level set of the function \(f(x,y,z)=2x-3y+z\). So the normal is it gradient, \(\vec \nabla f = \langle 2,-3,1\rangle\).

    dm,lk 

    Reading off the coefficients in the equation of a plane (or computing the gradient of the equation defining the surface) may not give the correct area of the surface. Only the normal found as the cross product of the tangent vectors gives the correct area.

13. Find the normal vector and the scalar differential of surface area of the paraboloid \(z=x^2+y^2\).

    \[\vec N=\left\langle -2r^2\cos\theta,-2r^2\sin\theta,r\right\rangle\] \[dS=\sqrt{4r^4+r^2}\,dr\,d\theta\]

    Again, we must first parametrize the surface. We start with cylindrical coordinates and substitute \(z=x^2+y^2=r^2\): \[ \vec R(r,\theta) =\left\langle r\cos\theta,r\sin\theta,r^2\right\rangle \] Now we find the tangent vectors: \[\begin{aligned} \vec{e}_r&=\left\langle \cos\theta,\sin\theta,2r\right\rangle \\ \vec{e}_\theta&=\left\langle -r\sin\theta,r\cos\theta,0\right\rangle \end{aligned}\] And now the normal vector: \[\begin{aligned} \vec{N}&= \begin{vmatrix} \hat{\imath} & \hat{\jmath} & \hat{k} \\ \cos\theta & \sin\theta & 2r \\ -r\sin\theta & r\cos\theta & 0 \end{vmatrix} \\ &=\hat{\imath}(-2r^2\cos\theta)-\hat{\jmath}(2r^2\sin\theta) \\ &\quad+\hat{k}(r\cos^2\theta+r\sin^2\theta) \\ &=\left\langle -2r^2\cos\theta,-2r^2\sin\theta,r\right\rangle \end{aligned}\] Then the length of the normal is: \[\begin{aligned} |\vec N|&=\sqrt{4r^4\cos^2\theta+4r^4\sin^2\theta+r^2} \\ &=\sqrt{4r^4+r^2}=r\sqrt{4r^2+1} \end{aligned}\] And the scalar differential of surface area is: \[ dS=|\vec N|\,dr\,d\theta =r\sqrt{4r^2+1}\,dr\,d\theta \]

    dm 

14. Find the surface area of the sphere, \(x^2+y^2+z^2=36\), by using the scalar surface differential \(dS\) found in a previous problem. Check your answer with the formula for sphere surface area, \(A=4\pi r^2\).

    \(A=144\pi\)

    Recall from that we alredy know from a previous problem the scalar surface differential \[ dS = 36\sin(\phi)\,d\phi\,d\theta \] Notice that the ranges are \(0\le \theta \le 2\pi\) and \(0\le \phi\le \pi\). Then the area is the surface integral \[\begin{aligned} A&=\iint_{\vec R}1\,dS=\int_{0}^{2\pi}\int_{0}^{\pi} 36\sin(\phi)\,d\phi\,d\theta\\ &=-36\int_{0}^{2\pi}\left[\rule{0pt}{10pt}\cos(\phi)\right]_{0}^{\pi}d\,\theta =72\int_{0}^{2\pi}1\,d\theta \\ &=72\left[\rule{0pt}{10pt}\theta\right]_{0}^{2\pi}=144\pi\text{.} \end{aligned}\]

    Now, recall the area of a sphere is \(A=4\pi r^2\). With \(r=6\), this is \(A=4\pi (6)^2=144\pi\), and we have verified our answer.

15. Find the area of the bowl shaped surface \(z=x^2+y^2\) for \(z \le 4\).

    \[ A=\dfrac{\pi}{6}(17^{3/2}-1) \]

    To parametrize the bowl, we start with cylindrical coordinates and plug in \(z=x^2+y^2=r^2\): \[ \vec R(r,\theta)=\left\langle r\cos\theta,r\sin\theta,r^2\right\rangle \] We find the tangent vectors \[ \vec{e}_r=\left\langle \cos\theta,\sin\theta,2r\right\rangle \] \[ \vec{e}_\theta=\left\langle -r\sin\theta,r\cos\theta,0\right\rangle \] The normal vector and its length are: \[\begin{aligned} \vec{N}&=\vec{e}_r\times\vec{e}_\theta=\det \begin{vmatrix} \hat{\imath} & \hat{\jmath} & \hat{k} \\ \cos\theta & \sin\theta & 2r \\ -r\sin\theta & r\cos\theta & 0 \end{vmatrix} \\ &=\hat{\imath}(-2r^2\cos\theta) -\hat{\jmath}(2r^2\sin\theta) +\hat{k}(r\cos^2\theta+r\sin^2\theta)\\ &=\left\langle -2r^2\cos\theta,-2r^2\sin\theta,r\right\rangle \\ |\vec N|&=\sqrt{4r^4\cos^2\theta+4r^4\sin^2\theta+r^2} \\ &=\sqrt{4r^4+r^2} =r\sqrt{4r+1} \end{aligned}\] So the scalar surface differential is: \[ dS=|\vec{N}|\,dr\,d\theta =r\sqrt{4r^2+1}\,dr\,d\theta . \] To find the surface area we compute \(\displaystyle A=\iint dS\), but first we need to find the bounds of \(r\) and \(\theta\). The bounds are \(0\le\theta \le 2\pi\) since the bowl goes all the way around. The minimum value of \(r\) is \(0\) and the maximum value occurs at \(z=4=r^2\) or \(r=2\). Now, we can calculate the area: \[\begin{aligned} A&=\iint dS =\int_0^{2\pi}\int_0^2 r\sqrt{4r^2+1}\,dr\,d\theta \\ &=\int_0^{2\pi} \,d\theta\int_0^2 r\sqrt{4r^2+1}\,dr=2\pi \int_0^2 r\sqrt{4r^2+1}\,dr . \end{aligned}\] Now, we make the substitution \(u=4r^2+1\) and \(du=8r\,dr\). \[\begin{aligned} A&=2\pi\int_0^2 r\sqrt{4r^2+1}\,dr\,d\theta =2\pi\int_1^{17} \dfrac{1}{8}\sqrt{u}\,du \\ &=\dfrac{\pi}{4}\left[\dfrac{2u^{3/2}}{3}\right]_1^{17} =\dfrac{\pi}{6}(17^{3/2}-1) \end{aligned}\]

    dm 

16. Compute \(\displaystyle \iint_S 15xy\,dS\) over the piece of the elliptic paraboloid \(z=x^2+y^2\) above the rectangle \([0,2]\times[0,1]\).

    \(\displaystyle \iint_S 15xy\,dS =\dfrac{1}{16}(21^{5/2}-5^{5/2}-17^{5/2}+1)\)

    In a previous example, we parametrized the surface as \[ \vec R(u,v)=\left\langle u,v,u^2+v^2\right\rangle \qquad \text{where} \qquad 0 \le u \le 2 \quad \text{and} \quad 0 \le v \le 1 \] Then the tangent and normal vectors are: \[ \begin{array}{c} \\[8pt] \vec{e}_u=\dfrac{\partial R}{\partial u}=\, \\[8pt] \vec{e}_v=\dfrac{\partial R}{\partial v}=\, \end{array} \left| \begin{array}{ccc} \hat{\imath} & \hat{\jmath} & \hat{k} \\[8pt] \left\langle 1 \right. & 0 & \left. 2u\right\rangle \\[8pt] \left\langle 0 \right. & 1 & \left. 2v\right\rangle \end{array} \right|=\vec N \] \[\begin{aligned} \vec N &=\vec{e}_u\times\vec{e}_v \\ &=\hat{\imath}(0-2u)-\hat{\jmath}(2v-0)+\hat{k}(1-0) =\left\langle -2u,-2v,1\right\rangle \end{aligned}\] The length of the normal vector is \[ |\vec N| =\sqrt{4u^2+4v^2+1} \] The integrand is \(15xy=15uv\). So the integral is \[\begin{aligned} \iint_S 15xy\,dS &=\int_0^1\int_0^2 15uv|\vec N|\,du\,dv \\ &=\int_0^1\int_0^2 15uv\sqrt{4u^2+4v^2+1}\,du\,dv \\ &=\int_0^1 \left[\dfrac{5}{4}v(4u^2+4v^2+1)^{3/2}\right]_{u=0}^2\,dv \\ &=\int_0^1 \dfrac{5}{4}v(4v^2+17)^{3/2} -\dfrac{5}{4}v(4v^2+1)^{3/2}\,dv \\ &=\left[\dfrac{1}{16}(4v^2+17)^{5/2} -\,\dfrac{1}{16}(4v^2+1)^{5/2}\right]_0^1 \\ &=\dfrac{1}{16}(21^{5/2}-5^{5/2}-17^{5/2}+1) \\ \end{aligned}\]

17. Consider the cone \(z=2r\) from \(z=0\) to \(z=10\).

    1. Find the lateral surface area of the cone.

       The lateral surface is the sides of the cone. Start with cylindrical coordinates and set \(z=2r\).

       \(A_\text{lateral}=25\sqrt{5}\,\pi\)

       First we need to parametrize the lateral surface of the cone. Starting with cylindrical coordinates, we plug in \(z=2r\): \[ \vec R(r,\theta)=\left\langle r\cos\theta,r\sin\theta,2r\right\rangle \] The tangent vectors and the normal vector and its length are: \[\begin{aligned} \vec{e}_r&=\left\langle \cos\theta,\sin\theta,2\right\rangle \\ \vec{e}_\theta&=\left\langle -r\sin\theta,r\cos\theta,0\right\rangle \end{aligned}\] \[\begin{aligned} \vec{N}&=\vec{e}_r\times\vec{e}_\theta= \begin{vmatrix} \hat{\imath} & \hat{\jmath} & \hat{k} \\ \cos\theta & \sin\theta & 2 \\ -r\sin\theta & r\cos\theta & 0 \end{vmatrix} \\ &=\hat{\imath}(-2r\cos\theta) -\hat{\jmath}(2r\sin\theta) +\hat{k}(r\cos^2\theta+r\sin^2\theta) \\ &=\left\langle -2r\cos\theta,-2r\sin\theta,r\right\rangle \end{aligned}\] \[\begin{aligned} |\vec{N}|&=\sqrt{4r^2\cos^2\theta+4r^2\sin^2\theta+r^2} \\ &=\sqrt{5r^2} =\sqrt{5}\,r \end{aligned}\] To compute the integral, we need to know the bounds. The maximum height is \(z=10\). So the maximum radius is \(r=5\). Then the lateral area is: \[\begin{aligned} A_\text{lateral}&=\int_0^{2\pi}\int_0^5 \sqrt{5}\,r\,dr\,d\theta =2\pi\int_0^5\sqrt{5}\,r\,dr \\ &=2\pi\left[\dfrac{\sqrt{5}}{2}r^2\right]_0^5 =25\sqrt{5}\,\pi \end{aligned}\]

       dm 

    2. Find the surface area of the top of the cone.

       Just find the area of a circle, no integral.

       \(A_\text{top}=25\pi\)

       The top of the cone is at \(z=10\). So \(r=5\). A circle with radius \(r=5\) has area \[ A_\text{top}=\pi r^2=25\pi \]

       dm 

    3. What is the total surface area of the cone?

       Just add.

       \(A_\text{total}=25\pi(\sqrt{5}+1)\)

       We simply add the lateral and top surface areas to find the total surfacea area of the cone:
       \[\begin{aligned} A_\text{total}&=A_\text{lateral}+A_\text{top} \\ &=25\sqrt{5}\pi+25\pi\\ &=25\pi(\sqrt{5}+1) \end{aligned}\]

       dm 

d. Applications of Scalar Surface Integrals

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18. Consider the cylindrical surface \(x^2+y^2=9\) between \(z=0\) and \(z=1\) with surface density \(\delta=z(2x^2+2y^2)\).

    1. Find the mass of the cylinder.

       Parametrize the cylinder starting from cylindrical coordinates.

       \(\displaystyle M=\iint \delta\,dS=54\pi \)

       In terms of cylindrical coordinates, we can parametrize the surface as \[ \vec R=\left\langle 3\cos\theta,3\sin\theta,z\right\rangle \] and our density function becomes \(\delta=2zr^2=18z\). We find the tangent vectors \[\begin{aligned} e_\theta&=\langle -3\sin\theta,3\cos\theta,0\rangle \\ e_z&=\langle 0,0,1\rangle \end{aligned}\] and the normal vector \[\begin{aligned} \vec{N}&= \begin{vmatrix} \hat{\imath} & \hat{\jmath} & \hat{k} \\ -3\sin\theta & 3\cos\theta & 0 \\ 0 & 0 & 1 \end{vmatrix} \\ &=\hat{\imath}(3\cos\theta)-\hat{\jmath}(-3\sin\theta)+\hat{k}(0) \\ &=\left\langle 3\cos\theta,3\sin\theta,0\right\rangle \end{aligned}\] Thus, the surface differential is \[ dS=|\vec{N}|d\theta\,dz=\sqrt{9\cos^2\theta+9\sin^2\theta}\,d\theta\,dz =3\,d\theta\,dz \] To find the mass, we integrate the density function over the surface, remembering to include the scalar surface area differential: \[\begin{aligned} M&=\iint \delta\,dS =\int_0^1\int_0^{2\pi} (18z)3\,d\theta\,dz \\ &=54\int_0^1 z\,dz\int_0^{2\pi} \,d\theta =54\left[\dfrac{1}{2}z^2\right]_0^1 2\pi \\ &=54\pi \end{aligned}\]

       dm,lk 

    2. Find the center of mass of the cylinder.

       Find the \(x\) and \(y\) components by symmetry.

       \(\displaystyle (\bar x,\bar y,\bar z)=\left(0,0,\dfrac{2}{3}\right) \)

       By symmetry, the center of mass must be on the \(z\)-axis. Therefore \(\bar{x}=\bar{y}=0\). The density is \(\delta=18z\) and the magintude of the normal is \(|\vec N|=3\). So, the \(z\)-moment of mass is \[\begin{aligned} M_z&=\iint z\delta\,dS=\iint z\delta\,|\vec N|\,d\theta\,dz \\ &=54\int_0^1\int_0^{2\pi}z^2\,d\theta\,dz =36\pi\left[\rule{0pt}{10pt}z^3\right]_0^1=36\pi \end{aligned}\] and thus the \(z\)-component of center of mass is \(\bar{z}=\dfrac{36\pi}{54\pi}=\dfrac{2}{3}\).

       lk 

19. An ice cream cone with ice cream on top consists of a cone at an angle of \(\phi=\dfrac{\pi}{6}\) measured from the north pole, and the piece of the sphere \(\rho=3\) above the cone. We emphasize that we are discussing the surface of the ice cream cone and not discussing the volume inside.

    1. Find the total surface area of the cone and the ice cream.

       The sides of the ice cream cone is the cone \(\phi=\dfrac{\pi}{6}\). The top is the piece of the sphere \(\rho=3\) above the angle \(\phi=\dfrac{\pi}{6}\). You need to separately parametrize both surfaces. For the cone, start from cylindrical coordinates. For the sphere, start from spherical coordinates.

       \(A_\text{tot}=\dfrac{45}{2}\pi-9\pi\sqrt{3}\approx 21.7\)

       The ice cream on top of the cone is the piece of the sphere with radius \(\rho=3\) for \(0<\phi<\dfrac{\pi}{6}\). To find its area, we parametrize the sphere as \[ \vec R(\phi,\theta) =\left\langle 3\sin\phi\cos\theta,3\sin\phi\sin\theta,3\cos\phi\right\rangle \] Then the tangent vectors are \[\begin{aligned} \vec{e}_\phi &=\left\langle 3\cos\phi\cos\theta,3\cos\phi\sin\theta,-3\sin\phi\right\rangle \\ \vec{e}_\theta &=\left\langle -3\sin\phi\sin\theta,3\sin\phi\cos\theta,0\right\rangle \end{aligned}\] and the normal vector is \[\begin{aligned} \vec{N}&=\vec{e}_\phi\times\vec{e}_\theta =\begin{vmatrix} \hat{\imath} & \hat{\jmath} & \hat{k} \\ 3\cos\phi\cos\theta & 3\cos\phi\sin\theta & -3\sin\phi\\ -3\sin\phi\sin\theta & 3\sin\phi\cos\theta & 0 \end{vmatrix} \\ &=\hat{\imath}(9\sin^2\phi\cos\theta) -\hat{\jmath}(-9\sin^2\phi\sin\theta)\\ &\qquad +\hat{k}(9\sin\phi\cos\phi\cos^2\theta+9\sin\phi\cos\phi\sin^2\theta) \\ &=\left\langle 9\sin^2\phi\cos\theta,9\sin^2\phi\sin\theta, 9\sin\phi\cos\phi\right\rangle \end{aligned}\] The length of the normal is: \[\begin{aligned} |\vec{N}| &=\sqrt{81\sin^4\phi\cos^2\theta+81\sin^4\phi\sin^2\theta+81\sin^2\phi\cos^2\phi} \\ &=9\sqrt{\sin^2\phi(\sin^2\phi(\cos^2\theta+\sin^2\theta)+\cos^2\phi)} =9\sin\phi \end{aligned}\] So the surface area is \[\begin{aligned} A_\text{top}&=\iint \,dS =\int_0^{2\pi}\int_0^{\pi/6} 9\sin\phi\,d\phi\,d\theta \\ &=18\pi\left[-\cos\phi\dfrac{}{}\right]_0^{\pi/6} \\ &=18\pi\left(-\,\dfrac{\sqrt{3}}{2}+1\right) =9\pi(2-\sqrt{3}) \end{aligned}\] To find the area of the cone, we need to parametrize the cone at \(\phi=\pi /6\). We start with cylindrical coordinates \[ (x,y,z)=(r\cos\theta,r\sin\theta,z) \] We need to relate \(r\) and \(z\) to eliminate one of them. Using the relation between cylindrical and spherical coordinates \[ \dfrac{r}{z}=\tan\phi=\tan\dfrac{\pi}{6}=\dfrac{1}{\sqrt{3}} \] we have \(z=\sqrt{3}r\). So the parametrization is \[ \vec R(r,\theta) =\left\langle r\cos\theta,r\sin\theta,\sqrt{3}r\right\rangle \] The tangent vectors are: \[\begin{aligned} \vec{e}_r&=\left\langle \cos\theta,\sin\theta,\sqrt{3}\right\rangle \\ \vec{e}_\theta&=\left\langle -r\sin\theta,r\cos\theta,0\right\rangle \end{aligned}\] with the normal vector being: \[\begin{aligned} \vec{N}&=\vec{e}_r\times\vec{e}_\theta =\begin{vmatrix} \hat{\imath} & \hat{\jmath} & \hat{k} \\ \cos\theta & \sin\theta & \sqrt{3} \\ -r\sin\theta & r\cos\theta & 0 \end{vmatrix} \\ &=\hat{\imath}(-\sqrt{3}r\cos\theta) -\hat{\jmath}(\sqrt{3}r\sin\theta) +\hat{k}(r\cos^2\theta+r\sin^2\theta) \\ &=\left\langle -\sqrt{3}r\cos\theta,-\sqrt{3}r\sin\theta,r\right\rangle \end{aligned}\] Then the length of the normal vector is \[ |\vec{N}|=\sqrt{3r^2\cos^2\theta+3r^2\sin^2\theta+r^2}=\sqrt{4r^2}=2r \] Finally, since \(\rho \le 3\), we have \[ r=\rho\sin\phi=\rho\sin\dfrac{\pi}{6}=\dfrac{\rho}{2} \le \dfrac{3}{2} \] So the surface area of the cone is: (Don't forget the magnitude of the normal.) \[\begin{aligned} A_\text{cone}&=\iint \,dS=\int_0^{2\pi}\int_0^{3/2} 2r\,dr\,d\theta \\ &=2\pi\left[\rule{0pt}{10pt}r^2\right]_0^{3/2} =\dfrac{9}{2}\pi \end{aligned}\] Finally we add the two areas together to find the total surface area of the ice cream cone: \[\begin{aligned} A_\text{tot}&=A_\text{top}+A_\text{cone} =9\pi(2-\sqrt{3})+\dfrac{9}{2}\pi \\ &=\dfrac{45}{2}\pi-9\pi\sqrt{3} \approx 21.7 \end{aligned}\]

       dm,lk 

    2. Find the mass of the cone (without the ice cream) if the density of the cone is given by \(\delta(r,\theta)=6-z\)

       \(\displaystyle M=\iint \delta\,dS=27\pi-\,\dfrac{9\pi}{2}\sqrt{3}\approx 60.3\)

       We continue using the parametrization of the cone given by: \[ \vec R(r,\theta) =\left\langle r\cos\theta,r\sin\theta,\sqrt{3}r\right\rangle \]and the differential \(dS=2r\,dr\,d\theta\). Since \(z=\sqrt{3}r\), the density is \(\delta=6-z=6-\sqrt{3}r\). We find the mass by integrating the density over the surface of the cone: \[\begin{aligned} M&=\iint \delta\,dS =\int_0^{2\pi}\int_0^{3/2} (6-\sqrt{3}r)2r\,dr\,d\theta \\ &=2\pi\left[6r^2-2\sqrt{3}\dfrac{r^3}{3}\right]_0^{3/2} \\ &=2\pi\left(6\left(\dfrac{3}{2}\right)^2 -\,\dfrac{2\sqrt{3}}{3}\left(\dfrac{3}{2}\right)^3\right) \\ &=27\pi-\,\dfrac{9\pi}{2}\sqrt{3}\approx 60.3 \end{aligned}\]

       dm 

    3. Find the center of mass of the cone (without the ice cream) in rectangular coordinates.

       By symmetry, what are \(\bar{x}\) and \(\bar{y}\)?

       \((\bar{x},\bar{y},\bar{z})=(0,0,1.644)\)

       From symmetry, we know that \(\bar{x}=\bar{y}=0\), thus we just need to find the center of mass along the \(z\) axis. We recall the mass is: \[ M=27\pi-\,\dfrac{9\pi}{2}\sqrt{3}\approx 60.337 \] We compute the \(z\)-moment of the mass, we recall \(z=\sqrt{3}r\) and \(|\vec{N}|=2r\) \[\begin{aligned} M_z&=\iint z\delta\,dS =\int_0^{2\pi}\int_0^{3/2} \sqrt{3}r(6-\sqrt{3}r)2r\,dr\,d\theta\\ &=2\pi\left[4\sqrt{3}r^3-\,\dfrac{3}{2}r^4\right]_0^{3/2} \\ &=2\pi\left(4\sqrt{3}\left(\dfrac{3}{2}\right)^3 -\,\dfrac{3}{2}\left(\dfrac{3}{2}\right)^4\right) \\ &=27\sqrt{3}\pi-\,\dfrac{243}{16}\pi\approx 99.2 \end{aligned}\] Finally, we divide the moment by the mass to get center of mass along the \(z\) axis: \[ \bar{z}=\dfrac{M_z}{M}\approx \dfrac{99.2}{60.3}=1.644 \] Putting everything together we find the center of mass to be \[(\bar x,\bar y, \bar z)=(0,0,1.644)\]

       dm,lk 

20. Compute the average of the temperature, \(T=300(1+16x^2+16y^2)\), on the paraboloid \(z=2x^2+2y^2\) below \(z=\dfrac{2}{9}\).

    Recall that \(\displaystyle T_{\text{ave}}=\dfrac{1}{A}\iint_P T\,dS\).

    \(T_{\text{ave}}=\dfrac{28820}{49}\approx 588.2\)

    First, notice that we can parametrize the elliptic parabaloid by \(\vec{R}(r,\theta)=\langle r\cos\theta,r\sin\theta,2r^2\rangle\). Since \(z=2r^2\) and \(z \le \dfrac{2}{9}\), we have \(0\leq r\leq \dfrac{1}{3}\) and \(0\leq \theta\leq 2\pi\). The coordinate tangent vectors, the normal and its length are: \[\begin{aligned} \vec{e}_r&=\langle \cos\theta,\sin\theta,4r\rangle \\ \vec{e}_\theta&=\langle -r\sin\theta,r\cos\theta,0\rangle \\ \vec{N}&=\vec{e}_r\times\vec{e}_\theta =\langle -4r^2\cos\theta,-4r^2\sin\theta,r\rangle \\ |\vec{N}|&=\sqrt{16r^4\cos^2\theta+16r^4\sin^2\theta+r^2} =r\sqrt{16r^2+1} \end{aligned}\] Then, the area of the parabaloid is: \[\begin{aligned} A&=\iint_P\,dS=\iint_P\,|\vec{N}|\,dr\,d\theta \\ &=\int_0^{2\pi}\int_0^{1/3} r\sqrt{16r^2+1}\,dr\,d\theta =2\pi\int_0^{1/3} r\sqrt{16r^2+1}\,dr \end{aligned}\] Using \(u=16r^2+1\) and \(du=32r\,dr\) we see that \[\begin{aligned} A&=2\pi\int_1^{25/9}\dfrac{1}{32}\sqrt{u}\,du =\dfrac{\pi}{16}\left[\dfrac{2}{3}u^{3/2}\right]_1^{25/9} \\ &=\dfrac{\pi}{24}\left(\left[\dfrac{25}{9}\right]^{3/2}-1\right) =\dfrac{\pi}{24}\left(\dfrac{125}{27}-1\right) \\ &=\dfrac{\pi}{24}\dfrac{98}{27} =\dfrac{49\pi}{324} \end{aligned}\] To find the integral of the temperature, \(T=300(1+16x^2+16y^2)\), we will use the same substitution \(u=16r^2+1\). Then \(T=300(1+16r^2)=300u\) and its integral is: \[\begin{aligned} \iint_P T\,dS&=\iint_P T|\vec{N}|\,dr\,d\theta \\ &=300\int_0^{2\pi}\int_0^{1/3} (1+16r^2)r\sqrt{16r^2+1}\,dr\,d\theta \\ &=\dfrac{300\pi}{16}\int_0^{25/9}u^{3/2}\,du =\dfrac{75\pi}{4}\left[\dfrac{2}{5}u^{5/2}\right]_1^{25/9}\\ &=\dfrac{15\pi}{2}\left(\left[\dfrac{25}{9}\right]^{5/2}-1\right) =\dfrac{15\pi}{2}\left(\dfrac{3125}{243}-1\right) \\ &=\dfrac{15\pi}{2}\dfrac{2882}{243} =\dfrac{7205\pi}{81} \\ \end{aligned}\] Finally, the average temperature is: \[\begin{aligned} T_{\text{ave}}&=\dfrac{1}{A}\displaystyle\iint_P T\,dS \\ &=\dfrac{7205\pi}{81}\dfrac{324}{49\pi} \\ &=\dfrac{28820}{49}\approx 588.2 \end{aligned}\]

    lk 

e. Vector Surface Integrals

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21. Consider the surface which is an eighth of a sphere with radius \(\rho=3\) in the first octant and the vector field \(\vec{F}=\langle xz^2,0,y^2z\rangle\).

    1. Parameterize the sphere and express the vector field \(\vec F\) on the surface.

       Start from spherical coordinates.

       \(\vec R(\phi,\theta) =\left\langle 3\sin\phi\cos\theta,3\sin\phi\sin\theta,3\cos\phi\right\rangle\)
       \(\vec{F}(\vec R(\phi,\theta)) =\langle 27\sin\phi\cos\theta\cos^2\phi,0,27\sin^2\phi\sin^2\theta\cos\phi\rangle\)

       The spherical to rectangular conversion is \[\begin{aligned} x&=\rho\sin\phi\cos\theta \\ y&=\rho\sin\phi\sin\theta \\ z&=\rho\cos\phi \end{aligned}\] We substitute \(\rho=3\) to get the parameterization: \[ \vec R(\phi,\theta) =\left\langle 3\sin\phi\cos\theta,3\sin\phi\sin\theta,3\cos\phi\right\rangle \] We evaluate the vector \(\vec{F}=\langle xz^2,0,y^2z\rangle\) on the surface to get: \[ \vec{F}(\vec R(\phi,\theta)) =\langle 27\sin\phi\cos\theta\cos^2\phi,0,27\sin^2\phi\sin^2\theta\cos\phi\rangle \]

       dm,lk 

    2. Find the tangent and normal vectors of the sphere. Orient the normal vector to point radially outwards.

       \(\vec{e}_\phi=\left\langle 3\cos\phi\cos\theta,3\cos\phi\sin\theta, -3\sin\phi\right\rangle\)
       \(\vec{e}_\theta =\left\langle -3\sin\phi\sin\theta,3\sin\phi\cos\theta,0\right\rangle\)
       \(\vec{N}=\left\langle 9\sin^2\phi\cos\theta,9\sin^2\phi\sin\theta, 9\sin\phi\cos\phi\right\rangle\)

       Recall that the parametrization of the surface is: \[ \vec R(\phi,\theta) =\left\langle 3\sin\phi\cos\theta,3\sin\phi\sin\theta,3\cos\phi\right\rangle \] The tangent vectors are: \[\begin{aligned} \vec{e}_\phi&=\dfrac{\partial\vec R}{\partial\phi} =\left\langle 3\cos\phi\cos\theta,3\cos\phi\sin\theta,-3\sin\phi\right\rangle \\ \vec{e}_\theta&=\dfrac{\partial\vec R}{\partial\theta} =\left\langle -3\sin\phi\sin\theta,3\sin\phi\cos\theta,0\right\rangle \end{aligned}\] So the normal vector is \[\begin{aligned} \vec{N}&=\vec{e}_\phi\times\vec{e}_\theta= \begin{vmatrix} \hat{\imath} & \hat{\jmath} & \hat{k} \\ 3\cos\phi\cos\theta & 3\cos\phi\sin\theta & -3\sin\phi \\ -3\sin\phi\sin\theta & 3\sin\phi\cos\theta & 0 \end{vmatrix} \\ &=\hat{\imath}(9\sin^2\phi\cos\theta) -\hat{\jmath}(-9\sin^2\phi\sin\theta) \\ &\quad+\hat{k}(9\sin\phi\cos\phi\cos^2\theta+9\sin\phi\cos\phi\sin^2\theta) \\ &=\left\langle 9\sin^2\phi\cos\theta,9\sin^2\phi\sin\theta, 9\sin\phi\cos\phi\right\rangle \end{aligned}\] We look at how this normal vector is oriented. All components are positive in the \(1^\text{st}\) octant, which means that the normal vector is pointing outward as desired.

       dm,lk 

    3. Compute \(\displaystyle \iint \vec{F}\cdot\,d\vec{S}\) over the eighth of the sphere oriented outward.

       \[ \iint_S \vec{F}\cdot\,d\vec{S} =\iint_S \vec{F}(\vec R(u,v))\cdot\vec{N}\,du\,dv \] When integrating, be careful with the bounds on \(\theta\) and \(\phi\)!

       \(\displaystyle \iint_S \vec{F}\cdot\,d\vec{S}=\dfrac{81}{5}\pi \)

       Recall: \[\begin{aligned} \left.\vec{F}\right|_{\vec R} &=\langle 27\sin\phi\cos\theta\cos^2\phi,0, 27\sin^2\phi\sin^2\theta\cos\phi\rangle \\ \vec{N} &=\left\langle 9\sin^2\phi\cos\theta,9\sin^2\phi\sin\theta, 9\sin\phi\cos\phi\right\rangle \end{aligned}\] So their dot product is: \[\begin{aligned} \left.\vec{F}\right|_{\vec R}\cdot\vec{N} &=243\sin^3\phi\cos^2\theta\cos^2\phi +243\sin^3\phi\sin^2\theta\cos^2\phi \\ &=243\sin^3\phi\cos^2\phi \end{aligned}\] Then the integral is: \[\begin{aligned} \iint_S &\vec{F}\cdot\,d\vec{S} =\int_0^{\pi/2}\int_0^{\pi/2} 243\sin^3\phi\cos^2\phi\,d\phi\,d\theta \\ &=243\dfrac{\pi}{2}\int_0^{\pi/2} (1-\cos^2\phi)\cos^2\phi\sin\phi\,d\phi \\ \end{aligned}\] To complete the integral, we substitute \(u=\cos\phi\) and change the limits: \[\begin{aligned} \iint_S &\vec{F}\cdot\,d\vec{S} =-243\dfrac{\pi}{2}\int_1^0 (1-u^2)u^2\,du \\ &=-243\dfrac{\pi}{2}\left[\dfrac{u^3}{3}-\dfrac{u^5}{5}\right]_1^0 =-243\dfrac{\pi}{2}\left(0-\dfrac{1}{3}+\dfrac{1}{5}\right) \\ &=243\dfrac{\pi}{2}\left(\dfrac{1}{3}-\dfrac{1}{5}\right) =243\dfrac{\pi}{2}\left(\dfrac{2}{15}\right) =\dfrac{81}{5}\pi \end{aligned}\]

       dm,lk 

22. Consider the paraboloid \(z=4x^2+4y^2\) oriented down and out and the vector field \(\vec{G}=\left\langle z,3y,xy\right\rangle\).

    1. Parameterize the paraboloid and evaluate \(\vec{G}\) on the surface.

       Start from cylindrical coordinates.

       \( \vec R(r,\theta) =\left\langle r\cos\theta,r\sin\theta,4r^2\right\rangle \)
       \( \vec{G}(\vec R(r,\theta)) =\left\langle 4r^2,3r\sin\theta,r^2\sin\theta\cos\theta\right\rangle \)

       We can paramterize the paraboloid starting from cylindrical coordinates, by replacing \(z\) by \(z=4x^2+4y^2=4r^2\): \[ \vec R(r,\theta)=\langle r\cos\theta,r\sin\theta,4r^2\rangle \] Now, we substitute into the vector \(\vec{G}=\langle z,3y,xy\rangle\) and find: \[ \vec G(\vec R(r,\theta))=\langle 4r^2,3r\sin\theta,r^2\sin\theta\cos\theta\rangle \]

       dm,lk 

    2. Find the normal vector to the paraboloid and check its orientation.

       \(\vec{N}=\vec{e}_r\times\vec{e}_\theta\)

       \[ \vec{N}=\left\langle 8r^2\cos\theta,8r^2\sin\theta,-r\right\rangle \]

       For the surface, \(\vec R(r,\theta)=\langle r\cos\theta,r\sin\theta,4r^2\rangle\) we first calculate the tangent vectors: \[\vec e_r=\langle \cos\theta,\sin\theta,8r\rangle\] \[\vec e_\theta=\langle -r\sin\theta,r\cos\theta,0\rangle\] Then we calculate the normal vector: \[\begin{aligned} \vec{N}&=\vec{e}_r\times\vec{e}_\theta= \begin{vmatrix} \hat{\imath} & \hat{\jmath} & \hat{k} \\ \cos\theta & \sin\theta & 8r\\ -r\sin\theta & r\cos\theta & 0 \end{vmatrix} \\ &=\hat{\imath}(-8r^2\cos\theta) -\hat{\jmath}(8r^2\sin\theta)+\hat{k}(r^2) \\ &=\langle-8r^2\cos\theta,-8r^2\sin\theta,r\rangle \end{aligned}\] This is up (since \(\vec N_3\gt0\)) and in (since \(\vec N_1\lt0\) and \(\vec N_2\lt0\)). We want down and out. So we reverse \(\vec N\): \[ \vec{N}=\langle8r^2\cos\theta,8r^2\sin\theta,-r\rangle \]

       dm 

    3. Compute \(\displaystyle \iint_S\vec{G}\cdot\,d\vec{S}\) over the part of the paraboloid with \(0 < r < 3\) and \(0 < \theta < \pi\).

       \(\displaystyle \iint_S \vec{G}\cdot\,d\vec{S} =\int_0^\pi\int_0^3 \vec G\cdot\vec N\,dr\,d\theta\)

       \(\displaystyle \iint_S \vec{G}\cdot\,d\vec{S}=243\pi\)

       From part (a), \[ \vec{G}(\vec R(r,\theta)) =\left\langle 4r^2,3r\sin\theta,r^2\cos\theta\sin\theta\right\rangle \] From part (b), \[ \vec{N} =\left\langle 8r^2\cos\theta,8r^2\sin\theta,-r\right\rangle \] Their dot product is: \[ \vec{G}\cdot\vec{N} =32r^4\cos\theta+24r^3\sin^2\theta-r^3\sin\theta\cos\theta \] We do the integral by factoring the integrals in each term: \[\begin{aligned} \iint_S &\vec{G}\cdot\,d\vec{S} =\int_0^\pi\int_0^3 \vec G\cdot\vec N\,dr\,d\theta \\ &=\int_0^\pi\int_0^3 (32r^4\cos(\theta)+24r^3\sin^2(\theta)-r^3\sin(\theta)\cos(\theta))\,dr\,d\theta \\ &=\int_0^\pi \cos\theta\,d\theta\int_0^3 32r^4\,dr +\int_0^\pi \dfrac{1-\cos2\theta}{2}\,d\theta\int_0^3 24r^3\,dr\\ &\qquad-\int_0^\pi \cos\theta\sin\theta\,d\theta\int_0^3 r^3\,dr \\ &=\left[\rule{0pt}{10pt}\sin\theta\right]_0^\pi\left[\dfrac{32}{5}r^5\right]_0^3 +\left[\dfrac{\theta}{2}-\,\dfrac{\sin2\theta}{4}\right]_0^\pi \left[\rule{0pt}{10pt}6r^4\right]_0^3 -\left[\dfrac{\sin^2\theta}{2}\right]_0^\pi\left[\dfrac{r^4}{4}\right]_0^3 \\ &=0+\dfrac{\pi}{2}\cdot6\cdot3^4+0=243\pi \end{aligned}\]

       dm,lk 

23. Compute \(\displaystyle\iint_S \vec{F}\cdot d\vec{S}\) inward over the upper hemisphere \(z=\sqrt{16-x^2-y^2}\) with the vector field \(\vec{F}=\langle xz,yz,z^2\rangle\).

    \(\displaystyle \iint_S \vec{F}\cdot\,d\vec{S}=-256\pi\)

    The parametrization of a sphere of radius \(\rho=4\) is \[ \vec{R}(\phi,\theta)=\langle 4\sin\phi\cos\theta,4\sin\phi\sin\theta, 4\cos\phi\rangle \] and plugging this into the vector field we get \[ \vec{F}(\vec{R}(\phi,\theta)) =\langle 16\sin\phi\cos\phi\cos\theta, 16\sin\phi\cos\phi\sin\theta, 16\cos^2\phi\rangle \] The tangent vectors are: \[\begin{aligned} \vec e_\phi&=\langle 4\cos\phi\cos\theta, 4\cos\phi\sin\theta, -4\sin\phi\rangle \\ \vec e_\theta&=\langle -4\sin\phi\sin\theta,4\sin\phi\cos\theta, 0\rangle \end{aligned}\] Then we calculate the normal vector: \[\begin{aligned} \vec{N}&=\vec{e}_\phi\times\vec{e}_\theta= \begin{vmatrix} \hat{\imath} & \hat{\jmath} & \hat{k} \\ 4\cos\phi\cos\theta & 4\cos\phi\sin\theta & -4\sin\phi\\ -4\sin\phi\sin\theta & 4\sin\phi\cos\theta & 0 \end{vmatrix} \\ &=\hat{\imath}(16\sin^2\phi\cos\theta) -\hat{\jmath}(-16\sin^2\phi\sin\theta)+\hat{k}(16\sin\phi\cos\phi) \\ &=\langle 16\sin^2\phi\cos\theta,16\sin^2\phi\sin\theta,16\sin\phi\cos\phi\rangle \end{aligned}\] Since this normal is outward, we need to reverse it: \[ \vec{N}=\langle -16\sin^2\phi\cos\theta,-16\sin^2\phi\sin\theta,-16\sin\phi\cos\phi\rangle \] Now, we calculate the dot product of the vector field with the normal vector: \[\begin{aligned} \vec{F}\cdot \vec{N}&= -16^2\sin^3\phi\cos\phi\cos^2\theta - 16^2\sin^3\phi\cos\phi\sin^2\theta -16^2\sin\phi\cos^3\phi\\ &=-16^2\sin^3\phi\cos\phi-16^2\sin\phi\cos^3\phi=-16^2\sin\phi\cos\phi \end{aligned}\] Finally, setting the bounds to be \(0\leq \phi\leq \pi/2\) and \(0\leq\theta\leq 2\pi\) we can calculate the integral \[\begin{aligned} \iint_S \vec{F}\cdot\,d\vec{S} &=\int_0^{2\pi}\int_0^{\pi/2} \vec F\cdot\vec N\,d\phi\,d\theta \\ &=\int_0^{2\pi}\int_0^{\pi/2} -16^2\sin\phi\cos\phi\,d\phi\,d\theta \\ &=-16^2(2\pi)\left[\dfrac{1}{2}\sin^2\phi\right]_0^{\pi/2} \\ &=-256\pi \end{aligned}\]

    lk 

f. Applications of Vector Surface Integrals

1

24. Find the flux of the vector field \(\vec{F}=\left\langle x,y,z\right\rangle\) through the piece of the plane \(2x-3y+6z=6\) for \(0 \le x \le 5\) and \(0 \le y \le3\), oriented upward.

    \(\displaystyle \text{Flux}=\iint_R \vec{F}\cdot d\vec S\)

    \(\displaystyle \text{Flux}=\iint_R \vec{F}\cdot d\vec S=15 \)

    We parameterize the plane using \(x\) and \(y\) as the coordinates. To do this, we solve the plane for \(z\): \[ z=\dfrac{6-2x+3y}{6}=1-\,\dfrac{x}{3}+\dfrac{y}{2} \] \[ \vec R(x,y) =\left\langle x,y,1-\,\dfrac{x}{3}+\dfrac{y}{2}\right\rangle \] Next we restrict the vector field to the plane: \[ \vec{F}(\vec R(x,y)) =\left\langle x,y,1-\,\dfrac{x}{3}+\dfrac{y}{2}\right\rangle \] Now we find the normal vector from the parametrization: \[ \vec{N} =\vec e_x\times\vec e_y =\left| \begin{array}{ccc} \hat{\imath} & \hat{\jmath} & \hat{k} \\ 1 & 0 & -\,\dfrac{1}{3} \\ 0 & 1 & \dfrac{1}{2} \end{array} \right|=\left\langle \dfrac{1}{3},-\,\dfrac{1}{2},1\right\rangle \] Since \(N_3=1\gt0\), \(\vec N\) is oriented upward. Next, we compute their dot product: \[ \vec{F}\cdot\vec{N}=\dfrac{x}{3}-\,\dfrac{y}{2}+1-\,\dfrac{x}{3}+\dfrac{y}{2}=1 \] Finally, we can integrate over the surface \[\begin{aligned} \text{Flux}&=\iint_R \vec{F}(\vec R(u,v))\cdot\vec{N}\,du\,dv =\int_0^3\int_0^5 1\,dx\,dy \\ &=(3-0)(5-0)=15 \end{aligned}\]

    dm 

25. Compute the expansion of the vector function \(\vec{G}=\langle xy^2,yx^2,z(x^2+y^2)\rangle\) outward thorugh the complete surface of the solid above the paraboloid \(z=x^2+y^2\) and below the disk in the plane \(z=4\).

    Recall the trig identities: \(\sin 2\theta=2\sin\theta\cos\theta\) and \(\sin^2\theta=\dfrac{1}{2}-\dfrac{1}{2}\cos2\theta\).

    \(\text{Expansion}=\dfrac{2^6}{3}\pi\)

    The expansion is the sum of the flux integrals \[ \text{Expansion} =\iint \vec{G}\cdot\,d\vec{S} =\iint_R \vec{G}(\vec R(r,\theta))\cdot\vec{N}\,dr\,d\theta \] through the paraboloid and the disk.

    We parametrize the paraboloid in cylindrical coordinates with \(z=r^2\) and evaluate the vector field on the surface: \[\begin{aligned} \vec{R}(r,\theta)&=\langle r\cos\theta,r\sin\theta,r^2\rangle \\ \vec{G}\left(\vec{R}(r,\theta)\right) &=\left\langle r^3\cos\theta\sin^2\theta, r^3\sin\theta\cos^2\theta,r^4\right\rangle \end{aligned}\] The normal to the paraboloid is: \[\begin{aligned} \vec{N} &=\vec e_r\times\vec e_\theta =\left| \begin{array}{ccc} \hat{\imath} & \hat{\jmath} & \hat{k} \\ \cos\theta & \sin\theta & 2r \\ -r\sin\theta & r\cos\theta & 0 \end{array} \right| \\ &=\left\langle -2r^2\cos\theta,-2r^2\sin\theta,r\right\rangle \end{aligned}\] Since this is up and in, and we want down and out, we reverse the normal: \[ \vec{N} =\left\langle 2r^2\cos\theta,2r^2\sin\theta,-r\right\rangle \] Then the dot product of \(\vec{G}\) and \(\vec{N}\) is: \[\begin{aligned} \vec{G}\cdot\vec{N} &=2r^5\cos^2\theta\sin^2\theta+2r^5\sin^2\theta\cos^2\theta-r^5 \\ &=4r^5\sin^2\theta\cos^2\theta-r^5 =r^5[(2\sin\theta\cos\theta)^2-1] \\ &=r^5[\sin^2(2\theta)-1] =-r^5\cos^2\theta \end{aligned}\] The upper edge of the paraboloid is at \(z=r^2=4\) or \(r=2\). So the flux is: \[\begin{aligned} \iint_P \vec{G}\cdot\,d\vec{S} &=\int_0^{2\pi}\int_0^2 (-r^5\cos^2\theta)\,dr\,d\theta \\ &=-\int_0^2 r^5\,dr\int_0^{2\pi}\cos^2\theta\,d\theta \\ &=-\left[\dfrac{r^6}{6}\right]_0^2\int_0^{2\pi}\dfrac{1}{2}[1+\cos(2\theta)]\,d\theta \\ &=-\,\dfrac{2^6}{6}\dfrac{1}{2}\left[\theta+\dfrac{\sin(2\theta)}{2}\right]_0^{2\pi} =-\,\dfrac{2^5}{3}\pi \end{aligned}\]

    We parametrize the disk in cylindrical coordinates with \(z=4\) and evaluate the vector field, \(\vec{G}=\langle xy^2,yx^2,z(x^2+y^2)\rangle\), on the surface: \[\begin{aligned} \vec R&=\langle r\cos\theta,r\sin\theta,4\rangle \\ \vec{G}&=\left\langle r^3\cos\theta\sin^2\theta, r^3\cos^2\theta\sin\theta,4r^2\right\rangle \end{aligned}\] The normal to the disk is: \[\begin{aligned} \vec{N} =\vec e_r\times\vec e_\theta =\left| \begin{array}{ccc} \hat{\imath} & \hat{\jmath} & \hat{k} \\ \cos\theta & \sin\theta & 0 \\ -r\sin\theta & r\cos\theta & 0 \end{array} \right|=\left\langle 0,0,r\right\rangle \end{aligned}\] and since the normal vector is upward, it is oriented correctly. The dot product between \(\vec{G}\) and \(\vec{N}\) is \[\vec{G}\cdot\vec{N}=4r^3\] So the flux is: \[\begin{aligned} \iint_D \vec{G}\cdot\,d\vec{S} &=\int_0^{2\pi} \int_{0}^2 4 r^3\,dr\,d\theta \\ &=2\pi\left[\rule{0pt}{10pt}r^4\right]_0^2 =2^5\pi \end{aligned}\]

    To find the Expansion we add these up: \[\begin{aligned} \text{Expansion} &=\iint_P \vec{G}\cdot\,d\vec{S}+\iint_D \vec{G}\cdot\,d\vec{S} \\ &=-\,\dfrac{2^5}{3}\pi+2^5\pi=\dfrac{2^6}{3}\pi \end{aligned}\]

    dm,lk 

26. What is the expansion of the vector function \(\vec{H} =\left\langle x,y,z\right\rangle\) outward through the hemisphere \(x^2+y^2+z^2=16\) for \(z>0\). outward.

    \[ \iint\limits_\text{total} \vec{H}(\vec R(u,v))\cdot\vec{N}\,du\,dv=128\pi \]

    The expansion is the flux: \[ \text{Expansion}=\iint \vec{H}\cdot\,d\vec{S} \] which is the sum of the flux integrals through the upper hemisphere and the disk at the bottom.

    The hemisphere is parameterized as: \[ \vec R(\theta,\phi) =\left\langle 4\sin\phi\cos\theta,4\sin\phi\sin\theta,4\cos\phi\right\rangle \] On the hemisphere, the vector field is: \[ \vec{H}(\vec R(\theta,\phi)) =\left\langle 4\sin\phi\cos\theta,4\sin\phi\sin\theta,4\cos\phi\right\rangle \] The normal to the hemisphere is: \[\begin{aligned} \vec{N} &=\vec e_\theta\times\vec e_\phi =\left| \begin{array}{ccc} \hat{\imath} & \hat{\jmath} & \hat{k} \\ -4\sin\phi\sin\theta & 4\sin\phi\cos\theta & 0 \\ 4\cos\phi\cos\theta & 4\cos\phi\sin\theta & -4\sin\phi \end{array} \right| \\ &=\left\langle -16\sin^2\phi\cos\theta,-16\sin^2\phi\sin\theta, -16\sin\phi\cos\phi\right\rangle \end{aligned}\] Since this is down and in, and we want up and out, we reverse the normal \[\vec{N}=\left\langle 16\sin^2\phi\cos\theta,16\sin^2\phi\sin\theta, 16\sin\phi\cos\phi\right\rangle\] Then the dot product of \(\vec{H}\) and \(\vec{N}\) is: \[\begin{aligned} \vec{H}\cdot\vec{N} &=64\sin^3\phi\cos^2\theta+64\sin^3\phi\sin^2\theta+64\sin\phi\cos^2\phi\\ &=64\sin^3\phi+64\sin\phi\cos^2\phi =64\sin\phi(\sin^2\phi+\cos^2\phi) \\ &=64\sin\phi \end{aligned}\] Since it is half a sphere, the bounds are \(0\leq\theta\leq 2\pi\) and \(0\leq\phi\leq \dfrac{\pi}{2}\). So the flux is: \[\begin{aligned} \iint_\text{hemi} \vec{H}\cdot\,d\vec{S} &=\iint_\text{hemi} \vec{H}(\vec R(\theta,\phi))\cdot\vec{N}\,d\phi\,d\theta \\ &=64\int_0^{2\pi}\int_0^{\pi/2} \sin\phi\,d\phi\,d\theta =128\pi \end{aligned}\]

    The disk is parameterized as \[ R=\left\langle r\cos\theta,r\sin\theta,0\right\rangle \] On the disk, the vector field is: \[ \vec{H}=\left\langle r\cos\theta,r\sin\theta,0\right\rangle \] Since the normal to the disk is \(\vec{N}=\left\langle 0,0,r\right\rangle\), its dot product with \(\vec{H}\) is \(\vec{H}\cdot\vec{N}=0\). So the flux is: \[ \iint_\text{disk} \vec{H}\cdot\,d\vec{S}=0 \]

    The total expansion is their sum: \[\begin{aligned} \text{Expansion} &=\iint_{\text{hemi}} \vec{H}\cdot\,d\vec{S} +\iint_{\text{disk}} \vec{H}\cdot\,d\vec{S} \\ &=128\pi+0=128\pi \end{aligned}\]

    dm,lk 

27. The electric field of a positive point charge is given by: \[\begin{aligned} \vec E&(x,y,z)=\dfrac{\hat r}{r^2}=\dfrac{\vec r}{r^3} \\ &=\left\langle \dfrac{x}{\left(x^2+y^2+z^2\right)^{3/2}}, \dfrac{y}{\left(x^2+y^2+z^2\right)^{3/2}}, \dfrac{z}{\left(x^2+y^2+z^2\right)^{3/2}}\right\rangle \end{aligned}\]

    1. Find the expansion of the electric field out of the sphere, \(S\), given by \(x^2+y^2+z^2 \le 4\).

       To find the parametrization, start with spherical coordinates and set \(\rho=2\).

       \(\displaystyle \text{Expansion}=\iint_S \vec E\cdot d\vec S=4\pi\)

       The surface of the sphere is parametrized in spherical coordinates by setting \(\rho=2\): \[ \vec R(\phi,\theta) =\left\langle 2\sin\phi\cos\theta,2\sin\phi\sin\theta,2\cos\phi\right\rangle \] The tangent and normal vectors are: \[\begin{aligned} \vec e_\phi&=\left\langle 2\cos\phi\cos\theta,2\cos\phi\sin\theta,-2\sin\phi\right\rangle \\ \vec e_\theta&=\left\langle -2\sin\phi\sin\theta,2\sin\phi\cos\theta,0\right\rangle \\ \vec N&=\left\langle 4\sin^2\phi\cos\theta,4\sin^2\phi\sin\theta,4\sin\phi\cos\phi\right\rangle \end{aligned}\] As should be expected, the normal is parallel to the position vector, and it points outward, as required.

       On the surface, the electric field becomes: \[ \vec E(\theta,z) =\left\langle \dfrac{\sin\phi\cos\theta}{4}, \dfrac{\sin\phi\sin\theta}{4}, \dfrac{\cos\phi}{4}\right\rangle \] and its dot product with the normal is: \[\begin{aligned} \vec E\cdot\vec N &=\dfrac{4\sin^3\phi\cos^2\theta}{4} +\dfrac{4\sin^3\phi\sin^2\theta}{4} +\dfrac{4\sin\phi\cos^2\phi}{4} \\ &=\sin^3\phi+\sin\phi\cos^2\phi =\sin\phi \end{aligned}\] So the flux is: \[\begin{aligned} \text{Flux}_S &=\iint_S \vec E\cdot d\vec S =\iint_S \vec E\cdot \vec N\,d\phi\,d\theta \\ &=\int_0^{2\pi}\int_0^\pi \sin\phi\,d\phi\,d\theta =2\pi\left[-\cos\phi\dfrac{}{}\right]_0^\pi =4\pi \end{aligned}\]

    2. Find the expansion of the electric field out of the sphere, \(S\), given by \(x^2+y^2+z^2 \le R^2\).

       \(\displaystyle \text{Expansion}=\iint_S \vec E\cdot d\vec S=4\pi\)

       The surface of the sphere is parametrized in spherical coordinates by setting \(\rho=R\): \[ \vec R(\phi,\theta) =\left\langle R\sin\phi\cos\theta,R\sin\phi\sin\theta,R\cos\phi\right\rangle \] The tangent and normal vectors are: \[\begin{aligned} \vec e_\phi&=\left\langle R\cos\phi\cos\theta,R\cos\phi\sin\theta,-R\sin\phi\right\rangle \\ \vec e_\theta&=\left\langle -R\sin\phi\sin\theta,R\sin\phi\cos\theta,0\right\rangle \\ \vec N&=\left\langle R^2\sin^2\phi\cos\theta,R^2\sin^2\phi\sin\theta,R^2\sin\phi\cos\phi\right\rangle \end{aligned}\] As should be expected, the normal is parallel to the position vector, and it points outward, as required.

       On the surface, the electric field becomes: \[\begin{aligned} \vec E(\theta,z) &=\left\langle \dfrac{R\sin\phi\cos\theta}{R^3}, \dfrac{R\sin\phi\sin\theta}{R^3}, \dfrac{R\cos\phi}{R^3}\right\rangle \\ &=\left\langle \dfrac{\sin\phi\cos\theta}{R^2}, \dfrac{\sin\phi\sin\theta}{R^2}, \dfrac{\cos\phi}{R^2}\right\rangle \end{aligned}\] and its dot product with the normal is: \[\begin{aligned} \vec E\cdot\vec N &=\dfrac{R^2\sin^3\phi\cos^2\theta}{R^2} +\dfrac{R^2\sin^3\phi\sin^2\theta}{R^2} +\dfrac{R^2\sin\phi\cos^2\phi}{R^2} \\ &=\sin^3\phi+\sin\phi\cos^2\phi =\sin\phi \end{aligned}\] So the flux is: \[\begin{aligned} \text{Flux}_S &=\iint_S \vec E\cdot d\vec S =\iint_S \vec E\cdot \vec N\,d\phi\,d\theta \\ &=\int_0^{2\pi}\int_0^\pi \sin\phi\,d\phi\,d\theta =2\pi\left[-\cos\phi\dfrac{}{}\right]_0^\pi =4\pi \end{aligned}\]

    3. Compare the expansion for various values of \(R\). Does the expansion get larger or smaller as \(R\) gets bigger.

       The expansion is independent of \(R\).

       The expansion, \(\displaystyle \iint_S \vec E\cdot d\vec S=4\pi\), is independent of \(R\).

       In a later chapter, we will learn that Gauss' Theorem says this integral is surface independent. Consequently, changing the value of \(R\) does not change the value of the integral.

Review Exercises

28. Find the mass of the piece of the hyperbolic paraboloid \(z=x^2-y^2\) within the cylinder \(x^2+y^2 \le 4\) if the density is \(\delta=z^2\). The hyperbolic paraboloid was considered in a previous exercise.

    \(\displaystyle M=\dfrac{\pi}{840}\left(7769\sqrt{17}-1\right)\)

    The hyperbolic paraboloid is parametrized by: \[\begin{aligned} \vec R(r,\theta) &=\left\langle r\cos\theta,r\sin\theta,r^2(\cos^2\theta-\sin^2\theta)\right\rangle \\ &=\left\langle r\cos\theta,r\sin\theta,r^2\cos(2\theta)\right\rangle \end{aligned}\] In a previous exercise the length of the normal was found to be: \[ |\vec N|=r\sqrt{4r^2+1} \] We evaluate the density on the surface: \[ \delta=z^2=r^4\cos^2(2\theta) \] So the mass is: \[\begin{aligned} M=\iint_S \delta\,dS &=\int_0^{2\pi}\int_0^2 r^4\cos^2(2\theta)r\sqrt{4r^2+1}\,dr\,d\theta \\ &=\int_0^{2\pi} \cos^2(2\theta)\,d\theta\int_0^2 r^5\sqrt{4r^2+1}\,dr \end{aligned}\] The first factor is \[\begin{aligned} \int_0^{2\pi} \cos^2(2\theta)\,d\theta &=\int_0^{2\pi} \dfrac{1+\cos(4\theta)}{2}\,d\theta \\ &=\dfrac{1}{2}\left[\theta+\dfrac{\sin(4\theta)}{4}\right]_0^{2\pi} =\pi \end{aligned}\] In the second factor, we make the substitution \(u=4r^2+1\) with \(du=8r\,dr\). Then \(r^2=\dfrac{u-1}{4}\) and \(r^4=\dfrac{u^2-2u+1}{16}\). \[\begin{aligned} \int_0^2 r^5\sqrt{4r^2+1}\,dr &=\dfrac{1}{8}\int_1^{17} \dfrac{u^2-2u+1}{16}\sqrt{u}\,du \\ &=\dfrac{1}{128}\int_1^{17} (u^{5/2}-2u^{3/2}+u^{1/2})\,du \\ &=\dfrac{1}{128}\left[\dfrac{2u^{7/2}}{7}-\dfrac{4u^{5/2}}{5} +\dfrac{2u^{3/2}}{3}\right]_1^{17} \\ &=\dfrac{1}{128}\left(\dfrac{2\cdot17^{7/2}}{7}-\dfrac{4\cdot17^{5/2}}{5} +\dfrac{2\cdot17^{3/2}}{3}-\dfrac{2}{7}+\dfrac{4}{5}-\dfrac{2}{3}\right) \\ &=\dfrac{1}{128}\left(\left[\dfrac{2\cdot17^3}{7}-\dfrac{4\cdot17^2}{5} +\dfrac{2\cdot17}{3}\right]\sqrt{17}-\dfrac{2}{7}+\dfrac{4}{5}-\dfrac{2}{3}\right) \\ &=\dfrac{1}{840}\left(7769\sqrt{17}-1\right) \\ \end{aligned}\] So the mass is: \[\begin{aligned} M&=\int_0^{2\pi} \cos^2(2\theta)\,d\theta\int_0^2 r^5\sqrt{4r^2+1}\,dr \\ &=\dfrac{\pi}{840}\left(7769\sqrt{17}-1\right) \end{aligned}\]

29. The vorticity of fluid is the curl of its fluid velocioty \(\vec V\). The total vorticity through a surface, \(S\), is the flux of the vorticity through that surface: \[ \text{Total Vorticity}=\iint_S \vec\nabla\times\vec V\,d\vec S \] We will see in the chapter on Stokes' Theorem, that the total vorticity is equal to the circulation of the fluid velocity around the surface's boundary curve, \(C\). \[ \text{Circulation}=\oint_C \vec V\cdot d\vec s \] Find the total vorticity of the atmosphere counterclockwise over the first octant of the Earth's surface, oriented outward, (i.e. \(0\leq\theta\leq \dfrac{\pi}{2}\) and \(0\leq\phi\leq \dfrac{\pi}{2}\)) at an altitude of \(22\,\text{km}\) above the surface. Assume the wind velocity is given by the vector field \(\vec{V}=\langle x,x^2+y^2,yz \rangle\) where distances are in kilometers.
    Note: the radius of the Earth is \(R_E=6,378\,\text{km}\) and the first octant is the quarter of the northern hemisphere containing Europe.
    Use these steps:

    1. Find the vorticity of the wind, i.e. curl of \(\vec{V}\).

       Recall that \[\vec{\nabla}\times\vec{V}= \begin{vmatrix} \hat\imath & \hat\jmath & \hat k \\ \partial_x & \partial_y & \partial_z \\ V_1 & V_2 & V_3 \end{vmatrix}\]

       \(\vec{\nabla}\times\vec{V}=\left\langle z,0,2x\right\rangle\)

       We compute the vorticity which is the curl: \[\begin{aligned} \vec{\nabla}\times\vec{V}&= \begin{vmatrix} \hat\imath & \hat\jmath & \hat k \\ \partial_x & \partial_y & \partial_z \\ x & x^2+y^2 & yz \end{vmatrix}\\ &=\left\langle z-0,0,2x-0\right\rangle =\left\langle z,0,2x\right\rangle. \end{aligned}\]

       lk 

    2. Compute the total vorticity, i.e. the flux of the curl over the first octant of the Earth with outward normal at the total radius \(R=6,400\,\text{km}\).
       (Don't put in the radius until the end.)

       \(\displaystyle \text{Total Vorticity}=\iint_S \vec\nabla\times\vec V\,d\vec S\)

       \(\displaystyle \iint_\text{first oct} \vec{\nabla}\times\vec{V}\cdot\,d\vec{S} =R^3=6400^3\)

       The piece of the sphere may be parametrized as \[ \vec R(\theta,\phi) =\langle R\sin\phi\cos\theta,R\sin\phi\sin\theta,R\cos\phi\rangle \] Plugging this into the curl of the vector field we get: \[ \left[\vec{\nabla}\times\vec{V}\right]_{\vec{R}(\theta,\phi)} =\left\langle z,0,2x\right\rangle_{\vec{R}(\theta,\phi)} =\left\langle R\cos\phi,0,2R\sin\phi\cos\theta\right\rangle \] The normal to the northern hemisphere is: \[\begin{aligned} \vec{N} &=\vec e_\theta\times\vec e_\phi =\left| \begin{array}{ccc} \hat{\imath} & \hat{\jmath} & \hat{k} \\ -R\sin\phi\sin\theta & R\sin\phi\cos\theta & 0 \\ R\cos\phi\cos\theta & R\cos\phi\sin\theta & R\sin\phi \end{array} \right| \\ &=\left\langle -R^2\sin^2\phi\cos\theta,-R^2\sin^2\phi\sin\theta, -R^2\sin\phi\cos\phi\right\rangle \end{aligned}\] Since this is down and in and we want up and out, we reverse the normal to get: \[ \vec{N}=\left\langle R^2\sin^2\phi\cos\theta,R^2\sin^2\phi\sin\theta, R^2\sin\phi\cos\phi\right\rangle \] Then the dot product of \(\vec{\nabla}\times\vec{V}\) and \(\vec{N}\) is: \[\begin{aligned} \vec{\nabla}\times\vec{V}\cdot \vec{N}&= R^3\sin^2\phi\cos\phi\cos\theta+0+2R^3\sin^2\phi\cos\phi\cos\theta \\ &=3R^3\sin^2\phi\cos\phi\cos\theta \end{aligned}\] For the first octant, the flux (and therefore the total vorticity) is: \[\begin{aligned} \iint_\text{first oct} &\vec{\nabla}\times\vec{V}\cdot\,d\vec{S} =\int_0^{\pi/2}\int_0^{\pi/2} \vec{\nabla}\times\vec{V}\cdot\vec{N}\,d\phi\,d\theta \\ &=R^3\int_0^{\pi/2}\int_0^{\pi/2} 3\sin^2\phi\cos\phi\cos\theta\,d\phi\,d\theta \\ &=R^3\int_0^{\pi/2}\cos\theta\,d\theta \int_0^{\pi/2}3\sin^2\phi\cos\phi\,d\phi \\ &=R^3\left[\rule{0pt}{10pt}\sin\theta\right]_0^{\pi/2} \left[\rule{0pt}{10pt}\sin^3\phi\right]_0^{\pi/2} =R^3=6400^3 \end{aligned}\]

       lk 

    3. To check your work, compute the circulation of the fluid velocity, \(\vec V\), counterclockwise around the boundary of the first octant. (This consists of \(3\) quarter circles.)
       (Don't put in the radius until the end.)

       Review line integrals and in particular: \[ \text{Circulation}=\oint_C \vec V\cdot d\vec s \]

       \(\displaystyle \oint_C \vec V\cdot d\vec s =R^3=6400^3\)

       The \(3\) quarter circles are:
       \(C_1\): North Pole to the Equator at \(\theta=0\)
       \(C_2\): Along the Equator from \(\theta=0\) to \(\theta=\dfrac{\pi}{2}\)
       \(C_3\): From the Equator to the North Pole along \(\theta=\dfrac{\pi}{2}\)
       We compute these one at a time and then add them up.
       Recall the surface is parametrized by: \[ \vec R(\theta,\phi) =\langle R\sin\phi\cos\theta,R\sin\phi\sin\theta,R\cos\phi\rangle \]

       Curve \(C_1\) may be parametrized by setting \(\theta=0\) on the surface: \[ \vec r(\phi)=\langle R\sin\phi,0,R\cos\phi\rangle \] The curve tangent vector and the fluid velocity on the surface are: \[\begin{aligned} \vec v&=\langle R\cos\phi,0,-R\sin\phi\rangle \\ \vec V&=\langle x,x^2+y^2,yz \rangle=\langle R\sin\phi,R^2\sin^2\phi,0\rangle \end{aligned}\] Their dot product is: \[ \vec V\cdot\vec v=R^2\sin\phi\cos\phi \] So the line integral is: \[\begin{aligned} \int_{C_1} \vec V\cdot d\vec s &=\int_0^{\pi/2} R^2\sin\phi\cos\phi\,d\phi \\ &=R^2\left[\dfrac{\sin^2\phi}{2}\right]_0^{\pi/2} =\dfrac{R^2}{2} \end{aligned}\]

       Curve \(C_2\) may be parametrized by setting \(\phi=\dfrac{\pi}{2}\) on the surface: \[ \vec r(\phi)=\langle R\cos\theta,R\sin\theta,0\rangle \] The curve tangent vector and the fluid velocity on the surface are: \[\begin{aligned} \vec v&=\langle -R\sin\theta,R\cos\theta,0\rangle \\ \vec V&=\langle x,x^2+y^2,yz \rangle=\langle R\cos\theta,R^2,0\rangle \end{aligned}\] Their dot product is: \[ \vec V\cdot\vec v=-R^2\sin\theta\cos\theta+R^3\cos\theta \] So the line integral is: \[\begin{aligned} \int_{C_2} \vec V\cdot d\vec s &=\int_0^{\pi/2} -R^2\sin\theta\cos\theta+R^3\cos\theta\,d\theta \\ &=\left[-R^2\dfrac{\sin^2\theta}{2}+R^3\sin\theta\right]_0^{\pi/2} =-\,\dfrac{R^2}{2}+R^3 \end{aligned}\]

       Curve \(C_3\) may be parametrized by setting \(\theta=\dfrac{\pi}{2}\) on the surface: \[ \vec r(\phi)=\langle 0,R\sin\phi,R\cos\phi\rangle \] The curve tangent vector and the fluid velocity on the surface are: \[\begin{aligned} \vec v&=\langle 0,R\cos\phi,-R\sin\phi\rangle \\ \vec V&=\langle x,x^2+y^2,yz \rangle=\langle 0,R^2\sin^2\phi,R^2\sin\phi\cos\phi\rangle \end{aligned}\] The tangent vector points down and we need it to point up. So we reverse it: \[ \vec v=\langle 0,-R\cos\phi,R\sin\phi\rangle \] Their dot product is: \[ \vec V\cdot\vec v=-R^3\sin^2\phi\cos\phi+R^3\sin^2\phi\cos\phi=0 \] So the line integral is: \[ \int_{C_3} \vec V\cdot d\vec s=0 \] All totaled, the circulation is: \[\begin{aligned} \oint_C \vec V\cdot d\vec s &=\int_{C_1} \vec V\cdot d\vec s +\int_{C_2} \vec V\cdot d\vec s +\int_{C_3} \vec V\cdot d\vec s \\ &=\left(\dfrac{R^2}{2}\right)+\left(-\,\dfrac{R^2}{2}+R^3\right)+(0) \\ &=R^3=6400^3 \end{aligned}\] This agrees with the total vorticity!

       Note that we will discuss vorticity and circulation further when we discuss applications of Stokes' Theorem, and it will become clear where the connection comes from. It is also worth noting that vorticity and circulation are powerful tools in the study of atmospheric and oceanic fluid dynamics.

PY: All Checked