27. Gauss' Theorem

a. The Theorem

What happens to Gauss' Theorem if the region has holes? Here we do not mean holes which go all the way through like a donut. Rather, we mean holes inside, like in Swiss cheese. The theorem still holds, but the boundary of the region is two or more closed surfaces and there are some special directional considerations.

2. Solids with Holes

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 \[ \iiint\limits_V \vec\nabla\cdot\vec F\,dV =\iint\limits_{\partial V} \vec F\cdot d\vec S \] 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.

Technical details:   The boundary \(\partial V\) of the volume \(V,\) is usually a single closed surface surrounding the solid. However, if the solid has holes, (like swiss cheese, as in the figure at the right) then its boundary consists of a single outer closed surface and \(1\) or more inner closed surfaces. Each piece of the boundary must be oriented with the normal pointing out of the solid. In particular, the normal to outer surface points outward, while the normal(s) to the inner surface(s) point inward.

We now look at Gauss' Theorem for some regions with holes.

Compute \(\displaystyle \iint_{\partial C} \vec F\cdot d\vec S\) for the vector field \(\vec F=\langle xz^2,yz^2,z(x^2+y^2)\rangle\) over the complete surface of the region between the cylinder \(x^2+y^2=9\) for \(0 \le z \le 4\) oriented outward and the cylinder \(x^2+y^2=1\) for \(1 \le z \le 2\) oriented inward. Compute the integral without and with Gauss' Theorem.

Surface Integral:

On the previous page, we already computed the integrals for the outer cylinder and disks. We now compute them for the inner cylinder and disks and combine the answers.

Inner Cylinder:  We parameterize the cylinder of radius \(1\) as: \[\begin{aligned} &\vec R(\theta,z)=(\cos\theta,\sin\theta,z) \\ \text{for} &\quad 0 \le \theta \le 2\pi \quad \text{and} \quad 1 \le z \le 2 \end{aligned}\] We compute the normal vector \[\begin{aligned} \vec N &=\vec e_\theta\times\vec e_z= \begin{vmatrix} \hat{\imath} & \hat{\jmath} & \hat{k} \\ -\sin\theta & \cos\theta & 0 \\ 0 & 0 & 1 \end{vmatrix} \\ &=\hat{\imath}(\cos\theta) -\hat{\jmath}(-\sin\theta) +\hat{k}(0) \\ &=\langle \cos\theta,\sin\theta,0\rangle \end{aligned}\] The normal is horizontal and points outward. So we reverse it: \[ \vec N=\langle -\cos\theta,-\sin\theta,0\rangle \] On this cylinder the vector field, \(\vec F=\langle xz^2,yz^2,z(x^2+y^2)\rangle\), is \[ \vec F =\langle \cos\theta z^2,\sin\theta z^2,z\rangle \] So the dot product with the normal is \[ \vec F\cdot\vec N =-\cos^2\theta z^2-\sin^2\theta z^2 =-z^2 \] We are ready to do the integral: \[\begin{aligned} \iint\limits_\text{inner cyl} \vec F\cdot d\vec S =\int_1^2\int_0^{2\pi} -z^2\,d\theta\,dz =-2\pi\left[\dfrac{z^3}{3}\right]_1^2 =-\,\dfrac{14\pi}{3} \end{aligned}\] Inner Upper Disk:  We parameterize the inner upper disk as: \[\begin{aligned} &\vec R(r,\theta)=(r\cos\theta,r\sin\theta,2) \\ \text{for} &\quad 0 \le r \le 1 \quad \text{and} \quad 0 \le \theta \le 2\pi \end{aligned}\] 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 points upward. So we reverse it: \[ \vec N=\langle 0,0,-r\rangle \] On the disk the vector field, \(\vec F=\langle xz^2,yz^2,z(x^2+y^2)\rangle\), is \[ \vec F =\langle 4r\cos\theta,4r\sin\theta,2r^2\rangle \] So the dot product with the normal is just \[ \vec F\cdot\vec N=-2r^3 \] We do the integral: \[ \iint\limits_\text{inner upper} \vec F\cdot d\vec S =\int_0^{2\pi}\int_0^1 -2r^3\,dr\,d\theta =-\pi\left[\rule{0pt}{10pt}r^4\right]_0^1 =-\pi \] Inner Lower Disk:  We parameterize the inner lower disk as: \[\begin{aligned} &\vec R(r,\theta)=(r\cos\theta,r\sin\theta,1) \\ \text{for} &\quad 0 \le r \le 1 \quad \text{and} \quad 0 \le \theta \le 2\pi \end{aligned}\] 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, \(\vec F=\langle xz^2,yz^2,z(x^2+y^2)\rangle\), is \[ \vec F =\langle r\cos\theta,r\sin\theta,r^2\rangle \] So the dot product with the normal is just \[ \vec F\cdot\vec N=r^3 \] We do the integral: \[ \iint\limits_\text{inner lower} \vec F\cdot d\vec S =\int_0^{2\pi}\int_0^1 r^3\,dr\,d\theta =\pi\left[\dfrac{r^4}{2}\right]_0^1 =\dfrac{\pi}{2} \] Total:  We complete the computation of the surface integral by adding together the surface integrals from the six pieces of the boundary: (The first three terms are from the previous page.) \[\begin{aligned} \iint\limits_{\partial C} \vec F\cdot d\vec S &=\iint\limits_\text{outer cyl} \vec F\cdot d\vec S +\iint\limits_\text{outer upper} \vec F\cdot d\vec S +\iint\limits_\text{outer lower} \vec F\cdot d\vec S \\ &\quad+\iint\limits_\text{inner cyl} \vec F\cdot d\vec S +\iint\limits_\text{inner upper} \vec F\cdot d\vec S +\iint\limits_\text{inner lower} \vec F\cdot d\vec S \\ &=384\pi+162\pi+0-\dfrac{14\pi}{3}-\pi+\dfrac{\pi}{2} =\dfrac{3245\pi}{6} \end{aligned}\]

Volume Integral:

We now compute the integral over the inner solid cylinder and subtract it from that for the outer solid cylinder. Everything is the same except for the limits of integration: \[\begin{aligned} \iiint\limits_\text{inner solid} &\vec\nabla\cdot\vec F\,dV =\int_1^2\int_0^{2\pi}\int_0^1 (r^2+2z^2)r\,dr\,d\theta\,dz \\ &=2\pi\left(\left[\dfrac{r^4}{4}\right]_0^1\left[\rule{0pt}{10pt}z\right]_1^2 +2\left[\dfrac{r^2}{2}\right]_0^1\left[\dfrac{z^3}{3}\right]_1^2\right) \\ &=2\pi\left(\dfrac{1}{4}+\dfrac{7}{3}\right) =\dfrac{31\pi}{6} \end{aligned}\] The net integral is: (The first term is from the previous page.) \[\begin{aligned} \iiint\limits_C \vec\nabla\cdot\vec F\,dV &=\iiint\limits_\text{outer solid} \vec\nabla\cdot\vec F\,dV -\iiint\limits_\text{inner solid} \vec\nabla\cdot\vec F\,dV \\ &=546\pi-\dfrac{31\pi}{6}=\dfrac{3245\pi}{6} \end{aligned}\]

Both methods gave the same answer. This checks both computations and also verifies Gauss' Theorem for this vector field and solid. Notice how much easier the volume integral was. This may not always be the case.

Compute \(\displaystyle \iint_{\partial S} \vec F\cdot d\vec S\) for the vector field \(\vec F=\langle x z^2, y z^2, z^3\rangle\) over the two spheres \(x^2+y^2+z^2=4\) oriented inward and \(x^2+y^2+z^2=16\) oriented outward.

  1. Surface Integral:  First solve it directly as a surface integral. Be sure to add the integrals over both spheres, correctly oriented.

\(\displaystyle \iint\limits_{\partial S} \vec F\cdot d\vec S =\dfrac{3968\pi}{3}\)

On the previous page, we already computed the integral for the outer sphere. We now compute it for the inner sphere and combine the answers.

We parameterize the sphere of radius \(2\) as: \[\begin{aligned} \vec R(\phi,&\theta)=(2\sin\phi\cos\theta,2\sin\phi\sin\theta,2\cos\phi) \\ &\text{for} \quad 0 \le \phi \le \pi \quad \text{and} \quad 0 \le \theta \le 2\pi \end{aligned}\] The normal vector is: \[ \vec N =\langle 4\sin^2\phi\cos\theta,4\sin^2\phi\sin\theta,4\sin\phi\cos\phi\rangle \] which is outward. So we reverse it: \[ \vec N =\langle -4\sin^2\phi\cos\theta,-4\sin^2\phi\sin\theta,-4\sin\phi\cos\phi\rangle \] On the surface, the vector field, \(\vec F=\langle x z^2, y z^2, z^3\rangle\) becomes: \[ \vec F=\langle 8\sin\phi\cos^2\phi\cos\theta, 8\sin\phi\cos^2\phi\sin\theta, 8\cos^3\phi\rangle \] And its dot product with the normal is: \[ \vec F\cdot\vec N=-32\sin\phi\cos^2\phi \] So the integral is: \[\begin{aligned} \iint_{\partial V} &\vec F\cdot d\vec S =\int_0^{2\pi}\int_0^{\pi} \vec F\cdot\vec N\,d\phi\,d\theta \\ &=\int_0^{2\pi}\int_0^{\pi} -32\sin\phi\cos^2\phi\,d\phi\,d\theta \\ &=-64\pi\left[-\,\dfrac{\cos^3\phi}{3}\right]_0^{\pi} =-\,\dfrac{128\pi}{3} \end{aligned}\] We complete the computation of the surface integral by adding together the surface integrals for the inner and outer spheres: \[\begin{aligned} \iint\limits_{\partial S} \vec F\cdot d\vec S &=\iint\limits_\text{outer sph} \vec F\cdot d\vec S +\iint\limits_\text{inner sph} \vec F\cdot d\vec S \\ &=\dfrac{4096\pi}{3}-\,\dfrac{128\pi}{3} =\dfrac{3968\pi}{3} \end{aligned}\]

  1. Volume Integral:  Next use Gauss' Theorem to convert the surface integral into a volume integral over the region between the spheres.

\(\displaystyle \iiint_S \vec\nabla\cdot\vec F\,dV =\dfrac{3968\pi}{3}\)

The divergence of \(\vec F\) is still: \[ \vec\nabla \cdot\vec F=5\rho^2\cos^2\phi \] So the integral is: (The only change is the limits on \(\rho\).) \[\begin{aligned} \iiint_S &\vec\nabla\cdot\vec F\,dV =\int_0^{2\pi}\int_0^\pi\int_2^4 5\rho^2\cos^2\phi\,\rho^2\sin\phi\,d\rho\,d\phi\,d\theta \\ &=2\pi\left[\rho^5\right]_2^4\left[-\,\dfrac{\cos^3\phi}{3}\right]_0^\pi =2\pi(4^5-2^5)\cdot\dfrac{2}{3} =\dfrac{3968\pi}{3} \end{aligned}\]

Both methods gave the same answer. This checks both computations and also verifies Gauss' Theorem for this vector field and solid. Notice how much easier the volume integral was.

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