22. Parametric Surfaces and Surface Integrals

b. Tangent and Normal Vectors

3. Normal Vector

For a parametric curve, the important vector is the tangent vector, v\vec{v}.

For a parametric surface, the important vector is the normal vector, N\vec{N}.

Since the tangent vectors eu\vec{e}_u and ev\vec{e}_v are tangent to a parametric surface, the normal vector N\vec{N} can be computed as the cross product of eu\vec{e}_u and ev\vec{e}_v.

Normal Vector
The normal vector, N\vec{N}, to a parametric surface is: N=eu×ev=ı^ȷ^k^xuyuzuxvyvzv=(yuzvzuyv)ı^(xuzvzuxv)ȷ^+(xuyvyuxv)k^=(y,z)(u,v),(x,z)(u,v),(x,y)(u,v)\begin{aligned} \vec{N}&=\vec{e}_u\times\vec{e}_v =\left| \begin{array}{ccc} \hat{\imath} & \hat{\jmath} & \hat{k} \\[3pt] \dfrac{\partial x}{\partial u} & \dfrac{\partial y}{\partial u} & \dfrac{\partial z}{\partial u} \\[8pt] \dfrac{\partial x}{\partial v} & \dfrac{\partial y}{\partial v} & \dfrac{\partial z}{\partial v} \end{array} \right| \\[5pt] &=\left(\dfrac{\partial y}{\partial u}\dfrac{\partial z}{\partial v}- \dfrac{\partial z}{\partial u}\dfrac{\partial y}{\partial v}\right) \hat{\imath} -\left(\dfrac{\partial x}{\partial u}\dfrac{\partial z}{\partial v}- \dfrac{\partial z}{\partial u}\dfrac{\partial x}{\partial v}\right) \hat{\jmath} +\left(\dfrac{\partial x}{\partial u}\dfrac{\partial y}{\partial v}- \dfrac{\partial y}{\partial u}\dfrac{\partial x}{\partial v}\right) \hat{k} \\[5pt] &=\left\langle \dfrac{\partial(y,z)}{\partial(u,v)}, -\,\dfrac{\partial(x,z)}{\partial(u,v)}, \dfrac{\partial(x,y)}{\partial(u,v)}\right\rangle \end{aligned}

Notice that the components of the normal vector are 22-dimensional Jacobian determinants. For the xx component the variables are yy and zz. For the yy component the variables are xx and zz. For the zz component the variables are xx and yy. Also notice that in each component the variables x,y,zx,y,z occur in alphabetical order, yzyz, xzxz, xyxy, but there is a minus on the yy component. However, back up a step and distribute the minus sign in the yy component. Then:

N=(yuzvzuyv)ı^+(zuxvxuzv)ȷ^+(xuyvyuxv)k^=(y,z)(u,v),(z,x)(u,v),(x,y)(u,v)\begin{aligned} \vec{N} &=\left(\dfrac{\partial y}{\partial u}\dfrac{\partial z}{\partial v}- \dfrac{\partial z}{\partial u}\dfrac{\partial y}{\partial v}\right) \hat{\imath} +\left(\dfrac{\partial z}{\partial u}\dfrac{\partial x}{\partial v}- \dfrac{\partial x}{\partial u}\dfrac{\partial z}{\partial v}\right) \hat{\jmath} +\left(\dfrac{\partial x}{\partial u}\dfrac{\partial y}{\partial v}- \dfrac{\partial y}{\partial u}\dfrac{\partial x}{\partial v}\right) \hat{k} \\[5pt] &=\left\langle \dfrac{\partial(y,z)}{\partial(u,v)}, \dfrac{\partial(z,x)}{\partial(u,v)}, \dfrac{\partial(x,y)}{\partial(u,v)}\right\rangle \end{aligned}

In this version, there is no minus sign but the variables are now in cyclic order, xyzxyzxyzxyz\cdots. You can remember N\vec{N} in either version. However, in computing N\vec{N} we rarely use the Jacobian determinants, but rather, go back to the cross product formula.

Caution: We previously saw Jacobian determinants in the computation of the differentials of area and volume. In those cases, the sign of the determinant did not matter because we were taking the absolute value to form the Jacobian factor. However, in computing the components of the normal vector, the sign of each Jacobian determinant is crucial since if you make a mistake, the vector will no longer be normal! This is particularly true for the sign of the yy-component.

In dealing with parametric surfaces, you will have to compute the normal many times and the length of the formulas will keep growing. If you try to copy the components of eu\vec{e}_u and ev\vec{e}_v to take the cross product, you will invariably make a copying mistake at some time. Don't re-copy. It is a waste of time and leads to mistakes. If you carefully calculate eu\vec{e}_u and ev\vec{e}_v and keep things lined up, you just need to put the row ı^\hat{\imath}, ȷ^\hat{\jmath}, k^\hat{k} on top of the values of eu\vec{e}_u and ev\vec{e}_v and take the determinant of the resulting matrix. On paper, your work would look like this. First, compute the two tangent vectors: eu=Ru=ev=Rv=ı^ȷ^k^xuyuzuxvyvzv=N \begin{array}{c} \\ \vec{e}_u=\dfrac{\partial R}{\partial u}=\, \\[10pt] \vec{e}_v=\dfrac{\partial R}{\partial v}=\, \end{array} \left| \begin{array}{ccc} \hat{\imath} & \hat{\jmath} & \hat{k} \\[3pt] \left\langle \dfrac{\partial x}{\partial u}\right. & \dfrac{\partial y}{\partial u} & \left. \dfrac{\partial z}{\partial u}\right\rangle\\[10pt] \left\langle \dfrac{\partial x}{\partial v}\right. & \dfrac{\partial y}{\partial v} & \left. \dfrac{\partial z}{\partial v}\right\rangle \end{array} \right|=\vec N Now add an ı^\hat{\imath}, ȷ^\hat{\jmath} and k^\hat{k} and some vertical bars and compute the determinant. Click Here to delete the determinant.

On the next page, we will see that it is also useful to know the length of the normal vector. So these examples and exercises also ask for the length.

Find the normal vector to the sphere of radius ρ=2\rho=2, parametrized by R(ϕ,θ)=2sinϕcosθ,2sinϕsinθ,2cosϕ \vec R(\phi,\theta) =\left\langle 2\sin\phi\cos\theta,2\sin\phi\sin\theta,2\cos\phi\right\rangle Then find the magnitude of the normal.

Sphere

We previously found the tangent vectors: eϕ=Rϕ=2cosϕcosθ,2cosϕsinθ,2sinϕeθ=Rθ=2sinϕsinθ,2sinϕcosθ,0\begin{aligned} \vec{e}_\phi&=\dfrac{\partial\vec R}{\partial\phi} =\left\langle 2\cos\phi\cos\theta,2\cos\phi\sin\theta,-2\sin\phi\right\rangle \\ \vec{e}_\theta&=\dfrac{\partial\vec R}{\partial\theta} =\left\langle -2\sin\phi\sin\theta,2\sin\phi\cos\theta,\quad 0\quad \right\rangle \end{aligned} Now add the ı^\hat{\imath}, ȷ^\hat{\jmath}, k^\hat{k} and some vertical bars and complete the computation of the cross product: N=eϕ×eθ=ı^ȷ^k^2cosϕcosθ2cosϕsinθ2sinϕ2sinϕsinθ2sinϕcosθ0=ı^(0+4sin2ϕcosθ)ȷ^(04sin2ϕsinθ)+k^(4sinϕcosϕcos2θ+4sinϕcosϕsin2θ)=4sin2ϕcosθ,4sin2ϕsinθ,4sinϕcosϕ\begin{aligned} \vec{N}&=\vec{e}_\phi\times\vec{e}_\theta =\left| \begin{array}{ccc} \hat{\imath} & \hat{\jmath} & \hat{k} \\ \left\langle 2\cos\phi\cos\theta\right. & 2\cos\phi\sin\theta& \left. -2\sin\phi\right\rangle\\ \left\langle -2\sin\phi\sin\theta\right. & 2\sin\phi\cos\theta& \left. \quad 0\quad\right\rangle \end{array} \right| \\ &=\hat{\imath}(0+4\sin^2\phi\cos\theta) -\hat{\jmath}(0-4\sin^2\phi\sin\theta) \\ & \qquad+\hat{k}(4\sin\phi\cos\phi\cos^2\theta+4\sin\phi\cos\phi\sin^2\theta) \\ &=\left\langle 4\sin^2\phi\cos\theta, 4\sin^2\phi\sin\theta, 4\sin\phi\cos\phi\right\rangle \end{aligned}

Notice that N\vec{N} points in the same direction as the position vector since N=2sinϕR(ϕ,θ)\vec{N}=2\sin\phi\,\vec R(\phi,\theta). This should be expected since the position vector from the origin to the sphere is perpendicular to the surface.

This result is the normal vector. Its magnitude will also be useful in doing integrals. So we compute: N=(4sin2ϕcosθ)2+(4sin2ϕsinθ)2+(4sinϕcosϕ)2=16sin4ϕcos2θ+16sin4ϕsin2θ+16sin2ϕcos2ϕ=16sin4ϕ+16sin2ϕcos2ϕ=16sin2ϕ(sin2ϕ+cos2ϕ)=16sin2ϕ=4sinϕ\begin{aligned} |\vec{N}| &=\sqrt{(4\sin^2\phi\cos\theta)^2+(4\sin^2\phi\sin\theta)^2+(4\sin\phi\cos\phi)^2} \\ &=\sqrt{16\sin^4\phi\cos^2\theta+16\sin^4\phi\sin^2\theta+16\sin^2\phi\cos^2\phi} \\ &=\sqrt{16\sin^4\phi+16\sin^2\phi\cos^2\phi} =\sqrt{16\sin^2\phi(\sin^2\phi+\cos^2\phi)} \\ &=\sqrt{16\sin^2\phi}=4\sin\phi \end{aligned}

Notice this result resembles the Jacobian factor for spherical coordinates: J=ρ2sinϕJ=\rho^2\sin\phi. This is coincidence for the spherical geometry.

Find the tangent vectors, normal vector, and length of the normal vector for the hyperbolic paraboloid z=x2y2z=x^2-y^2 discussed in a previous exercise and parametrized by: R(r,θ)=rcosθ,rsinθ,r2(cos2θsin2θ)\vec R(r,\theta) =\left\langle r\cos\theta,r\sin\theta,r^2(\cos^2\theta-\sin^2\theta)\right\rangle

Hyperbolic_paraboloid

Answer

The tangent vectors are:
er=cosθ,sinθ,2r(cos2θsin2θ)eθ=rsinθ,rcosθ,r2(4sinθcosθ)\begin{aligned} \vec{e}_r &=\left\langle \cos\theta,\sin\theta,2r(\cos^2\theta-\sin^2\theta)\right\rangle \\ \vec{e}_\theta &=\left\langle -r\sin\theta,r\cos\theta,r^2(-4\sin\theta\cos\theta)\right\rangle \end{aligned}
The normal vector is:
N=2r2cosθ,2r2sinθ,r\vec{N}=\left\langle -2r^2\cos\theta,2r^2\sin\theta,r\right\rangle
Its length is:
N=4r4+r2|\vec{N}|=\sqrt{4r^4+r^2}

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Solution

To find the tangent vectors, we take the derivative of the parametrization with respect to the two parameters: er=Rr=cosθ,sinθ,2r(cos2θsin2θ)eθ=Rθ=rsinθ,rcosθ,r2(4sinθcosθ)\begin{aligned} \vec{e}_r &=\dfrac{\partial\vec R}{\partial r} =\left\langle \cos\theta,\sin\theta,2r(\cos^2\theta-\sin^2\theta)\right\rangle \\ \vec{e}_\theta &=\dfrac{\partial\vec R}{\partial\theta} =\left\langle -r\sin\theta,r\cos\theta,r^2(-4\sin\theta\cos\theta)\right\rangle \end{aligned} To find the normal vector we compute the cross product of the two tangent vectors: (On paper, don't recopy the tangent vectors; just add ı^\hat{\imath}, ȷ^\hat{\jmath} and k^\hat{k} and vertical bars.) N=er×eθ=ı^ȷ^k^cosθsinθ2r(cos2θsin2θ)rsinθrcosθr2(4sinθcosθ)=ı^(4r2sin2θcosθ2r2cos3θ+2r2cosθsin2θ)ȷ^(4r2sinθcos2θ+2r2sinθcos2θ2r2sin3θ)+k^(rcos2θ+rsin2θ)=ı^(2r2sin2θcosθ2r2cos3θ)ȷ^(2r2sinθcos2θ2r2sin3θ)+k^(r)=ı^(2r2cosθ)ȷ^(2r2sinθ)+k^(r)=2r2cosθ,2r2sinθ,r\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 & r^2(-4\sin\theta\cos\theta) \end{vmatrix} \\ &=\hat{\imath}(-4r^2\sin^2\theta\cos\theta-2r^2\cos^3\theta+2r^2\cos\theta\sin^2\theta) \\ &\qquad-\hat{\jmath}(-4r^2\sin\theta\cos^2\theta+2r^2\sin\theta\cos^2\theta-2r^2\sin^3\theta) \\ &\qquad+\hat{k}(r\cos^2\theta+r\sin^2\theta) \\ &=\hat{\imath}(-2r^2\sin^2\theta\cos\theta-2r^2\cos^3\theta) \\ &\qquad-\hat{\jmath}(-2r^2\sin\theta\cos^2\theta-2r^2\sin^3\theta)+\hat{k}(r) \\ &=\hat{\imath}(-2r^2\cos\theta)-\hat{\jmath}(-2r^2\sin\theta)+\hat{k}(r) \\ &=\left\langle -2r^2\cos\theta,2r^2\sin\theta,r\right\rangle \end{aligned} Finally, its length is: N=(2r2cosθ)2+(2r2sinθ)2+r2=4r4+r2 |\vec{N}| =\sqrt{(2r^2\cos\theta)^2+(2r^2\sin\theta)^2+r^2} =\sqrt{4r^4+r^2}

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Check

We can check the normal vector by computing its dot product with each tangent vector: Ner=2r2cos2θ+2r2sin2θ+2r2(cos2θsin2θ)=0Neθ=2r3cosθsinθ+2r3sinθcosθ+r3(4sinθcosθ)=0\begin{aligned} \vec{N}\cdot\vec{e}_r &=-2r^2\cos^2\theta+2r^2\sin^2\theta+2r^2(\cos^2\theta-\sin^2\theta) \\ &=0 \\ \vec{N}\cdot\vec{e}_\theta &=2r^3\cos\theta\sin\theta+2r^3\sin\theta\cos\theta+r^3(-4\sin\theta\cos\theta) \\ &=0 \end{aligned}

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Find the normal vector, and length of the normal vector for the cylinder x2+z2=4x^2+z^2=4 discussed in two previous exercises and parametrized by R(θ,y)=2cosθ,y,2sinθ\vec R(\theta,y)=\left\langle 2\cos\theta,y,2\sin\theta\right\rangle.

Cylinder

Answer

The normal vector is:
N=2cosθ,0,2sinθ\vec{N}=\left\langle -2\cos\theta,0,-2\sin\theta\right\rangle
Its length is:
N=22|\vec{N}|=2\sqrt{2}

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Solution

We previously found the tangent vectors: eθ=Rθ=2sinθ,0,2cosθey=Ry=0,1,0\begin{aligned} \vec{e}_\theta &=\dfrac{\partial\vec R}{\partial\theta} =\left\langle -2\sin\theta,0,2\cos\theta\right\rangle \\ \vec{e}_y &=\dfrac{\partial\vec R}{\partial y} =\left\langle 0,1,0\right\rangle \end{aligned} Again, to find the normal vector we compute the cross product of the two tangent vectors: N=eθ×ey=ı^ȷ^k^2sinθ02cosθ010=ı^(02cosθ)ȷ^(00)+k^(2sinθ)=2cosθ,0,2sinθ\begin{aligned} \vec{N} &=\vec{e}_\theta\times\vec{e}_y =\begin{vmatrix} \hat{\imath} & \hat{\jmath} & \hat{k} \\ -2\sin\theta & 0 & 2\cos\theta\\ 0 & 1 & 0 \end{vmatrix} \\ &=\hat{\imath}(0-2\cos\theta) -\hat{\jmath}(0-0) +\hat{k}(-2\sin\theta) \\ &=\left\langle -2\cos\theta,0,-2\sin\theta\right\rangle \end{aligned} Finally, its length is: N=(2cosθ)2+0+(2sinθ)2=8=22 |\vec{N}|=\sqrt{(2\cos\theta)^2+0+(-2\sin\theta)^2}=\sqrt{8}=2\sqrt{2}

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Check

We can check the normal vector by computing its dot product with each tangent vector: Neθ=4cosθsinθ4sinθcosθ=0Ney=2cosθ0+012sinθ0=0\begin{aligned} \vec{N}\cdot\vec{e}_\theta &=4\cos\theta\sin\theta-4\sin\theta\cos\theta \\ &=0 \\ \vec{N}\cdot\vec{e}_y &=-2\cos\theta\cdot0+0\cdot1-2\sin\theta\cdot0 \\ &=0 \end{aligned}

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