27. Gauss' Theorem

Let \(V\) be a nice solid region in \(\mathbb{R}^3\) with a nice properly oriented boundary, \(\partial V\), and let \(\vec F\) be a nice vector field on \(V\). Then \[ \iiint_V \vec\nabla\cdot\vec F\,dV =\iint_{\partial V} \vec F\cdot d\vec S \] The outer boundary must be oriented outward while any inner boundaries must be oriented inward. This means that all pieces of the boundary are oriented away from the solid.

b. Verification

To verify a theorem means to check it works for specific examples. (It is not a proof.) On previous pages, we have already verified Gauss' Theorem for several functions and solids, with \(1\) or more pieces to the boundary. So we here just give a few more exercises for practice.

1. LHS: Volume Integral:

   Always compute a divergence or curl in rectangular coordinates! It can be done in curvilinear coordinates, but it is much more complicated. When you are ready to do the integral, then you convert the divergence or curl into the coordinates needed for the integral.

   \(\displaystyle \iiint_V \vec\nabla\cdot\vec F\,dV=5120\pi\)

   We want to compute the solid integral in cylindrical coordinates: \[ \vec R(r,\theta,z)=(r\cos\theta,r\sin\theta,z) \] But, we first need to compute the divergence in rectangular coordinates: \[ \vec\nabla\cdot\vec F=\partial_x(xy^2) +\partial_y(x^2y)+\partial_z(z^3) =y^2+x^2+3z^2 \] In cylindrical coordinates, the divergence is: \[ \vec\nabla\cdot\vec F=r^2+3z^2 \] The volume differential is \(dV=r\,dr\,d\theta\,dz\) and the bounding surfaces are \(z=2r\) and \(z=8\).

   The limits can be taken as \(0 \le z \le 8\), \(0 \le \theta \le 2\pi\) and \(0 \le r \le \dfrac{z}{2}\). So the integral is \[\begin{aligned} \iiint_V &\vec\nabla\cdot\vec F\,dV =\int_0^8\int_0^{2\pi}\int_0^{z/2} (r^2+3z^2) r\,dr\,d\theta\,dz \\ &=2\pi\int_0^8 \left[\dfrac{r^4}{4}+3z^2\dfrac{r^2}{2}\right]_{r=0}^{z/2}\,dz =2\pi\int_0^8 \dfrac{z^4}{64}+\dfrac{3z^4}{8}\,dz \\ &=2\pi\int_0^8 \dfrac{25}{64}z^4\,dz =\left.\dfrac{5\pi}{32}z^5\right|_0^8 =\dfrac{5\pi}{32}8^5=5120\pi \end{aligned}\]

   Alternatively, the limits can be taken as \(0 \le r \le 4\), \(0 \le \theta \le 2\pi\) and \(2r \le z \le 8\). In this case, the integral is \[\begin{aligned} \iiint_V &\vec\nabla\cdot\vec F\,dV =\int_0^4\int_0^{2\pi}\int_{2r}^8 (r^2+3z^2)r\,dz\,d\theta\,dr \\ &=2\pi\int_0^4 \left[\rule{0pt}{10pt}r^3z+rz^3\right]_{z=2r}^8\,dr =2\pi\int_0^4 (8r^3+8^3r-10r^4)\,dr \\ &=2\pi\left[ 2r^4+8^3\dfrac{r^2 }{2}-2r^5\right]_0^4 =2\pi\left(2\cdot4^4+8^3\dfrac{4^2 }{2}-2\cdot4^5\right) =5120\pi \end{aligned}\] Either choice of limits is acceptable. There was no way to know in advance which might be easier.

2. RHS: Surface Integral(s):

   \(\displaystyle \iint_{\partial V} \vec F\cdot d\vec S=5120\pi\)

   There are \(2\) pieces to the boundary: the cone on the bottom and the disk on the top. The cone must be oriented down while the disk must be oriented up.

   Cone:  We parameterize the cone as: \[ \vec R(r,\theta)=(r\cos\theta,r\sin\theta,2r) \quad \text{for} \quad 0 \le r \le 4 \quad \text{and} \quad 0 \le \theta \le 2\pi \] We compute the normal vector \[\begin{aligned} \vec N &=\vec e_r\times\vec e_\theta= \begin{vmatrix} \hat{\imath} & \hat{\jmath} & \hat{k} \\ \cos\theta & \sin\theta & 2 \\ -r\sin\theta & r\cos\theta & 0 \end{vmatrix} \\ &=\hat{\imath}(-2r\cos\theta) -\hat{\jmath}(2r\sin\theta) +\hat{k}(r\cos^2\theta+r\sin^2\theta) \\ &=\langle -2r\cos\theta,-2r\sin\theta,r\rangle \end{aligned}\] The normal vector needs to point down and out. So we reverse it: \[ \vec N=\langle 2r\cos\theta,2r\sin\theta,-r\rangle \] On the cone the vector field, \(\vec F=\langle xy^2,x^2y,z^3\rangle\), is \[ \vec F =\langle r^3\sin^2\theta\cos\theta,r^3\sin\theta\cos^2\theta,8r^3\rangle \] So the dot product with the normal is \[\begin{aligned} \vec F\cdot\vec N &=2r^4\sin^2\theta\cos^2\theta+2r^4\sin^2\theta\cos^2\theta-8r^4 \\ &=r^4(4\sin^2\theta\cos^2\theta-8) \end{aligned}\] We are ready to do the integral: \[ \iint\limits_\text{cone} \vec F\cdot d\vec S =\int_0^{2\pi}\int_0^4 r^4(4\sin^2\theta\cos^2\theta-8)\,dr\,d\theta \] The integral factors. For the \(\theta\) integral, we could use the two identities: \[ \sin^2A=\dfrac{1-\cos2A}{2} \quad \text{and} \quad \cos^2A=\dfrac{1+\cos2A}{2} \] but it turns out to be easier to first use the identity: \[ \sin2A=2\sin A\cos A \] Thus \[\begin{aligned} \iint\limits_\text{cone} \vec F\cdot d\vec S &=\left[\dfrac{r^5 }{5}\right]_0^4\int_0^{2\pi} \left(\sin^2\left(2\theta\right)-8\right)\,d\theta \\ &=\dfrac{1024}{5}\int_0^{2\pi} \left(\dfrac{1-\cos(4\theta)}{2}-8\right)\,d\theta \\ &=\dfrac{1024}{5}\left[\dfrac{1}{2} \left(\theta-\dfrac{\sin(4\theta)}{4}\right)-8\theta\right]_0^{2\pi} \\ &=\dfrac{1024}{5}(\pi-16\pi)=-3072\pi \end{aligned}\] Disk:  We parameterize the disk as: \[ \vec R(r,\theta)=(r\cos\theta,r\sin\theta,8) \quad \text{for} \quad 0 \le r \le 4 \quad \text{and} \quad 0 \le \theta \le 2\pi \] We compute the normal vector: \[ \vec N=\vec e_r\times\vec e_\theta= \begin{vmatrix} \hat{\imath} & \hat{\jmath} & \hat{k} \\ \cos\theta & \sin\theta & 0 \\ -r\sin\theta & r\cos\theta & 0 \end{vmatrix} =\langle 0,0,r\rangle \] which correctly points upward. On the disk the vector field \(\vec F=\langle xy^2,x^2y,z^3\rangle\) is \[ \vec F =\langle r^3\sin^2\theta\cos\theta,r^3\sin\theta\cos^2\theta,8^3\rangle \] So the dot product with the normal is just \[ \vec F\cdot\vec N=8^3r \] We are ready to do the integral: \[ \displaystyle \iint\limits_\text{disk} \vec F\cdot d\vec S =\int_0^{2\pi}\int_0^4 8^3r\,dr\,d\theta =2\pi8^3\left[\dfrac{r^2 }{2}\right]_0^4 =8^34^2\pi=8192\pi \] Total:  We complete the computation of the right hand side by adding together the surface integrals from the two pieces of the boundary: \[ \iint\limits_{\partial S} \vec F\cdot d\vec s =\iint\limits_\text{cone} \vec F\cdot d\vec s +\iint\limits_\text{disk} \vec F\cdot d\vec s =-3072\pi+8192\pi=5120\pi \]

   Comparison:   The left and the right sides of the equation gave the same answer. Thus Gauss' Theorem is verified for this example.

1. LHS: Volume Integral:

   Always compute a divergence or curl in rectangular coordinates! It can be done in curvilinear coordinates, but it is much more complicated. When you are ready to do the integral, then you convert the divergence or curl into the coordinates needed for the integral.

   \(\displaystyle \iiint_V \vec\nabla\cdot\vec F\,dV=124\pi\)

   We want to compute the integral in spherical coordinates: \[ \vec R(\rho,\theta,z)=(\rho\sin\phi\cos\theta,\rho\sin\phi\sin\theta,\rho\cos\phi) \] We compute the divergence in rectangular coordinates and then convert to spherical: \[ \vec F=\langle x^3+xy^2+xz^2,x^2y+y^3+yz^2,x^2z+y^2z+z^3\rangle \] \[\begin{aligned} \vec\nabla\cdot\vec F &=\partial_x(x^3+xy^2+xz^2) +\partial_y(x^2y+y^3+yz^2) \\ &\quad+\partial_z(x^2z+y^2z+z^3) \\ &=(3x^2+y^2+z^2)+(x^2+3y^2+z^2) \\ &\quad+(x^2+y^2+3z^2) \\ &=5(x^2+y^2+z^2)=5\rho^2 \end{aligned}\] In spherical coordinates, the volume differential is \(dV=\rho^2\sin\phi\,d\rho\,d\phi\,d\theta\). So the integral is: (Notice the limits on the \(\rho\) integral.) \[\begin{aligned} \iiint_V &\vec\nabla\cdot\vec F\,dV =\int_0^{2\pi}\int_0^\pi\int_1^2 (5\rho^2)\rho^2\sin\phi\,d\rho\,d\phi\,d\theta \\ &=2\pi\left[\rule{0pt}{10pt}\rho^5\right]_1^2\left[\rule{0pt}{10pt}-\cos\theta\right]_0^\pi =2\pi(2^5-1)(2)=124\pi \end{aligned}\]

2. RHS: Surface Integral(s):

   \(\displaystyle \iint_{\partial V} \vec F\cdot d\vec S=124\pi\)

   There are \(2\) pieces to the boundary: the inner and outer spheres. They may both be parametrized by \[ \vec R_a(\phi,\theta) =\langle a\sin\phi\cos\theta,a\sin\phi\sin\theta,a\cos\phi\rangle \] where \(a=1\) for the inner sphere and \(a=2\) for the outer sphere. We compute the normal vector \[\begin{aligned} \vec N_a &=\vec e_\phi\times\vec e_\theta= \begin{vmatrix} \hat{\imath} & \hat{\jmath} & \hat{k} \\ a\cos\phi\cos\theta & a\cos\phi\sin\theta & -a\sin\phi\\ -a\sin\phi\sin\theta & a\sin\phi\cos\theta & 0 \end{vmatrix} \\ &=\hat{\imath}(a^2\sin^2\phi\cos\theta) -\hat{\jmath}(-a^2\sin^2\phi\sin\theta) \\ &\quad+\hat{k}(a^2\sin\phi\cos\phi\cos^2\theta+a^2\sin\phi\cos\phi\sin^2\theta) \\ &=\langle a^2\sin^2\phi\cos\theta,a^2\sin^2\phi\sin\theta,a^2\sin\phi\cos\phi\rangle \end{aligned}\] Notice that the normal is a multiple of the position since \[ \vec N_a =a\sin\phi\,\langle a\sin\phi\cos\theta,a\sin\phi\sin\theta,a\cos\phi\rangle =a\sin\phi\,\langle x,y,z\rangle \] This makes sense since the normal to a sphere is parallel to the radius vector.

   Outer Sphere:  For the outer sphere, the normal is: \[ \vec N_2 =2\sin\phi\,\langle x,y,z\rangle \] This points outward which is correct for the outer sphere.

   Notice, that on the outer sphere, \(x^2+y^2+z^2=\rho^2=4\). So rather than evaluating \(\vec F=\langle x(x^2+y^2+z^2),y(x^2+y^2+z^2),z(x^2+y^2+z^2)\rangle\) on the surface, we write is as: \[\vec F=\langle 4x,4y,4z\rangle\] Then, when evaluation its dot product with the normal, we again use \(x^2+y^2+z^2=4\): \[\begin{aligned} \vec F\cdot\vec N_2 &=\langle 4x,4y,4z\rangle\cdot2\sin\phi\,\langle x,y,z\rangle \\ &=8\sin\phi(x^2+y^2+z^2) \\ &=8\sin\phi\cdot4 =32\sin\phi \end{aligned}\] So the outer surface integral is \[\begin{aligned} \iint\limits_\text{outer} &\vec F\cdot d\vec S =\int_0^{2\pi}\int_0^\pi \vec F\cdot\vec N_2\,d\phi\,d\theta \\ &=\int_0^{2\pi}\int_0^\pi 32\sin\phi\,d\phi\,d\theta \\ &=64\pi\left[\rule{0pt}{10pt}-\cos\phi\right]_0^\pi =128\pi \end{aligned}\] Inner Sphere:  For the inner sphere, the normal is: \[ \vec N_1 =\sin\phi\,\langle x,y,z\rangle \] This points outward which is incorrect for the inner sphere. So we reverse it: \[ \vec N_1 =-\sin\phi\,\langle x,y,z\rangle \] On the inner surface \(x^2+y^2+z^2=\rho^2=1\). So: \[\begin{aligned} \vec F&=\langle x(x^2+y^2+z^2),y(x^2+y^2+z^2),z(x^2+y^2+z^2)\rangle \\ &=\langle x,y,z\rangle \end{aligned}\] and its dot product with the normal is: \[ \vec F\cdot\vec N_1 =-\langle x,y,z\rangle\cdot\sin\phi\,\langle x,y,z\rangle =-\sin\phi \] So the inner surface integral is \[\begin{aligned} \iint\limits_\text{inner} &\vec F\cdot d\vec S =\int_0^{2\pi}\int_0^\pi \vec F\cdot\vec N_1\,d\phi\,d\theta \\ &=\int_0^{2\pi}\int_0^\pi (-\sin\phi)\,d\phi\,d\theta =2\pi\left[\rule{0pt}{10pt}\cos\phi\right]_0^\pi =-4\pi \end{aligned}\] Total:  We complete the computation of the right hand side by adding together the surface integrals from the two pieces of the boundary: \[ \iint\limits_{\partial V} \vec F\cdot d\vec S =\iint\limits_\text{outer} \vec F\cdot d\vec S +\iint\limits_\text{inner} \vec F\cdot d\vec S =128\pi-4\pi=124\pi \]

   Comparison:   The left and the right sides of the equation gave the same answer. Thus Gauss' Theorem is verified for this example.