27. Gauss' Theorem

a. The Theorem

1. Statement

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 \[ \iiint\limits_V \vec\nabla\cdot\vec F\,dV =\iint\limits_{\partial V} \vec F\cdot d\vec S \] If the solid has no holes, then its boundary must be oriented outward.

Notice that the orientation of the boundary is crucial. If you reverse the orientation of the boundary, then the integral on the right will switch signs and the theorem will no longer be true.

The vector field \(\vec F\) is nice if it has continuous first partial derivatives. Similarly, the solid \(V\) and its boundary \(\partial V\) are nice if the boundary is a parametric surface with continuous first partial derivatives.

In words, Gauss' theorem says the volume integral of the divergence of a vector field \(\vec F\) over a solid \(V\) is equal to the flux (or expansion) of \(\vec F\) out through the boundary surface \(\partial V\). This makes sense in that the divergence is supposed to tell us how much the stuff is expanding at each point while the flux tells us how much stuff is flowing out the surrounding surface.

Notice that on the left hand side (LHS) of the theorem, we are integrating the divergence of the vector field, \(\vec\nabla\cdot\vec F\) (and not just \(\vec F\)). Since \(\vec\nabla\cdot\vec F\) is a scalar field (i.e. a function), the integral on the left is an ordinary triple integral. On the other hand, on the right hand side (RHS) of the theorem, we are integrating \(\vec F\) (and not its divergence). Since \(\vec F\) is a vector field and we are integrating it over the boundary surface, we need to dot it into the vector differential of surface area, \(d\vec S=\vec N\,du\,dv\).

It is recommended that you first compute \(\vec\nabla\cdot\vec F\) in rectangular coordinates, and then convert it to the coordinates needed for the volume integral. Computing \(\vec\nabla\cdot\vec F\) in other coordinate systems is beyond the scope of this course.

Recall that Green's theorem related an integral over a region in the plane to an integral over its boundary curve. In fact, Green's theorem can be regarded as a 2D analogue of Gauss' theorem, by writing the LHS as a 2D divergence and the right side as a 2D normal integral. See the page on the 2D Gauss' Theorem.

Gauss' Theorem is usually applied to regions without holes. So that is where we start:

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 cylinder \(x^2+y^2=9\) for \(0 \le z \le 4\) with outward normal. Compute the integral without and with Gauss' Theorem.

When we say the complete surface of a solid, we mean all the pieces which completely surround the solid. Then we need to integrate over all the pieces and add them up. For the solid cylinder, this means the cylindrical sides and the two disks at the ends.

Surface Integral:

We compute the surface integrals over the cylindrical sides and the two disks at the ends and add the results.

Cylindrical Sides:  We parameterize the cylinder of radius \(3\) as: \[\begin{aligned} &\vec R(\theta,z)=(3\cos\theta,3\sin\theta,z) \\ \text{for} &\quad 0 \le \theta \le 2\pi \quad \text{and} \quad 0 \le z \le 4 \end{aligned}\] We compute the normal vector \[\begin{aligned} \vec N &=\vec e_\theta\times\vec e_z= \begin{vmatrix} \hat{\imath} & \hat{\jmath} & \hat{k} \\ -3\sin\theta & 3\cos\theta & 0 \\ 0 & 0 & 1 \end{vmatrix} \\ &=\hat{\imath}(3\cos\theta) -\hat{\jmath}(-3\sin\theta) +\hat{k}(0) \\ &=\langle 3\cos\theta,3\sin\theta,0\rangle \end{aligned}\] The normal is horizontal and points outward as needed. On the cylinder the vector field, \(\vec F=\langle xz^2,yz^2,z(x^2+y^2)\rangle\), is \[ \vec F =\langle 3\cos\theta z^2,3\sin\theta z^2,9z\rangle \] So the dot product with the normal is \[ \vec F\cdot\vec N =9\cos^2\theta z^2+9\sin^2\theta z^2 =9z^2 \] We are ready to do the integral: \[\begin{aligned} \iint\limits_\text{cyl} \vec F\cdot d\vec S =\int_0^4\int_0^{2\pi} 9z^2\,d\theta\,dz =2\pi\left[\rule{0pt}{10pt}3z^3\right]_0^4 =384\pi \end{aligned}\] Upper Disk:  We parameterize the upper disk as: \[\begin{aligned} &\vec R(r,\theta)=(r\cos\theta,r\sin\theta,4) \\ \text{for} &\quad 0 \le r \le 3 \quad \text{and} \quad 0 \le \theta \le 2\pi \end{aligned}\] 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 \[ \vec F =\langle 16r\cos\theta,16r\sin\theta,4r^2\rangle \] So the dot product with the normal is just \[ \vec F\cdot\vec N=4r^3 \] We do the integral: \[ \iint\limits_\text{upper} \vec F\cdot d\vec S =\int_0^{2\pi}\int_0^3 4r^3\,dr\,d\theta =2\pi\left[\rule{0pt}{10pt}r^4\right]_0^3 =162\pi \] Lower Disk:  We parameterize the lower disk as: \[\begin{aligned} &\vec R(r,\theta)=(r\cos\theta,r\sin\theta,0) \\ \text{for} &\quad 0 \le r \le 3 \quad \text{and} \quad 0 \le \theta \le 2\pi \end{aligned}\] On the lower disk the vector field, \(\vec F=\langle xz^2,yz^2,z(x^2+y^2)\rangle\), is \[ \vec F =\langle 0,0,0\rangle \] So the integral is trivially: \[ \iint\limits_\text{lower} \vec F\cdot d\vec S=0 \] Total:  We complete the computation of the surface integral by adding together the surface integrals from the three pieces of the boundary: \[\begin{aligned} \iint\limits_{\partial C} \vec F\cdot d\vec S &=\iint\limits_\text{cyl} \vec F\cdot d\vec S +\iint\limits_\text{upper} \vec F\cdot d\vec S +\iint\limits_\text{lower} \vec F\cdot d\vec S \\ &=384\pi+162\pi+0=546\pi \end{aligned}\]

Volume Integral:

Gauss' Theorem says that \(\displaystyle \iint_{\partial C} \vec F\cdot d\vec S\) is equal to \(\displaystyle \iiint_C \vec\nabla\cdot\vec F\,dV\). So we now compute this integral over the solid cylinder. We compute the divergence of \(\vec F=\langle xz^2,yz^2,z(x^2+y^2)\rangle\) in rectangular coordinates and convert to cylindrical coordinates: \[ \vec\nabla \cdot\vec F=z^2+z^2+x^2+y^2=r^2+2z^2 \] We also recall the cylindrical volume element: \[ dV=r\,dr\,d\theta\,dz \] Now we integrate: \[\begin{aligned} \iiint_C &\vec\nabla\cdot\vec F\,dV =\int_0^4\int_0^{2\pi}\int_0^3 (r^2+2z^2)r\,dr\,d\theta\,dz \\ &=2\pi\left(\left[\dfrac{r^4}{4}\right]_0^3\left[\rule{0pt}{10pt}z\right]_0^4 +2\left[\dfrac{r^2}{2}\right]_0^3\left[\dfrac{z^3}{3}\right]_0^4\right) \\ &=2\pi(3^4+3\cdot4^3) =546\pi \end{aligned}\]

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 surface of the sphere \(x^2+y^2+z^2=16\) with outward normal.

  1. Surface Integral:  First solve it directly as a surface integral.

\(\displaystyle \iint_{\partial S} \vec F\cdot d\vec S =\dfrac{4096\pi}{3}\)

We parameterize the sphere of radius \(4\) as: \[\begin{aligned} \vec R(\phi,&\theta)=(4\sin\phi\cos\theta,4\sin\phi\sin\theta,4\cos\phi) \\ &\text{for} \quad 0 \le \phi \le \pi \quad \text{and} \quad 0 \le \theta \le 2\pi \end{aligned}\] We compute the normal vector: \[\begin{aligned} \vec N &=\vec e_\phi\times\vec e_\theta= \begin{vmatrix} \hat{\imath} & \hat{\jmath} & \hat{k} \\ 4\cos\phi\cos\theta & 4\cos\phi\sin\theta & -4\sin\phi \\ -4\sin\phi\sin\theta & 4\sin\phi\cos\theta & 0 \end{vmatrix} \\ &=\hat{\imath}(16\sin^2\phi\cos\theta) -\hat{\jmath}(-16\sin^2\phi\sin\theta) \\ &\quad+\hat{k}(16\sin\phi\cos\phi\cos^2\theta+16\sin\phi\cos\phi\sin^2\theta) \\ &=\langle 16\sin^2\phi\cos\theta,16\sin^2\phi\sin\theta,16\sin\phi\cos\phi\rangle \end{aligned}\] The normal points outward and is, as expected, a multiple of the position vector \[ \vec N=4\sin\phi\vec R(\phi,\theta) \] On the surface, the vector field, \(\vec F=\langle x z^2, y z^2, z^3\rangle\) becomes: \[ \vec F=\langle 64\sin\phi\cos^2\phi\cos\theta, 64\sin\phi\cos^2\phi\sin\theta, 64\cos^3\phi\rangle \] And its dot product with the normal is: \[\begin{aligned} \vec F\cdot\vec N &=1024\left(\sin^3\phi\cos^2\phi\cos^2\theta +\sin^3\phi\cos^2\phi\sin^2\theta +\sin\phi\cos^4\phi\right) \\ &=1024\left(\sin^3\phi\cos^2\phi +\sin\phi\cos^4\phi\right) =\sin\phi\cos^2\phi \end{aligned}\] So the integral is: \[\begin{aligned} \iint_{\partial S} &\vec F\cdot d\vec S =\int_0^{2\pi}\int_0^{\pi} \vec F\cdot\vec N\,d\phi\,d\theta \\ &=\int_0^{2\pi}\int_0^{\pi} 1024\sin\phi\cos^2\phi\,d\phi\,d\theta \\ &=2048\pi\left[-\,\dfrac{\cos^3\phi}{3}\right]_0^{\pi} =\dfrac{4096\pi}{3} \end{aligned}\]

  1. Volume Integral:  Next use Gauss' Theorem to convert the surface integral into a volume integral inside the sphere.

\(\displaystyle \iiint_S \vec\nabla\cdot\vec F\,dV =\dfrac{4096\pi}{3}\)

Gauss' Theorem says that \(\displaystyle \iint_{\partial S} \vec F\cdot d\vec S\) is equal to \(\displaystyle \iiint_S \vec\nabla\cdot\vec F\,dV\). So we now compute this integral over the solid. We compute the divergence of \(\vec F=\langle xz^2,yz^2,z^3\rangle\) in rectangular coordinates and convert to spherical coordinates: \[ \vec\nabla \cdot\vec F=z^2+z^2+3z^2=5z^2=5\rho^2\cos^2\phi \] We also recall the spherical volume element: \[ dV=\rho^2\sin\phi\,d\rho\,d\phi\,d\theta \] Now we integrate: \[\begin{aligned} \iiint_S &\vec\nabla\cdot\vec F\,dV =\int_0^{2\pi}\int_0^\pi\int_0^4 5\rho^2\cos^2\phi\,\rho^2\sin\phi\,d\rho\,d\phi\,d\theta \\ &=2\pi\left[\rule{0pt}{10pt}\rho^5\right]_0^4\left[-\,\dfrac{\cos^3\phi}{3}\right]_0^\pi =2\pi4^5\cdot\dfrac{2}{3} =\dfrac{4096\pi}{3} \end{aligned}\]

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. This may not always be the case.

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