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.

d. Applications

3. Surface Independence

In either case, \(\partial V=S_1-S_2\) and so Gauss' Theorem states \[ \iiint_V \vec\nabla\cdot\vec F\,dV =\iint_{\partial V} \vec F\cdot d\vec S =\iint_{S_1} \vec F\cdot d\vec S -\iint_{S_2} \vec F\cdot d\vec S \] This may be rewritten as \[ \iint_{S_1} \vec F\cdot d\vec S =\iint_{S_2} \vec F\cdot d\vec S +\iiint_V \vec\nabla\cdot\vec F\,dV \qquad (*) \] In other words, to compute \(\displaystyle \iint_{S_1} \vec F\cdot d\vec S\), you may alternatively compute \(\displaystyle \iint_{S_2} \vec F\cdot d\vec S\) and \(\displaystyle \iiint_V \vec\nabla\cdot\vec F\,dV\), if that is easier.

Suppose that \(\vec F\) is a vector field satisfying \[ \iint_{S_1} \vec{F}\cdot d\vec S =\iint_{S_2} \vec{F}\cdot d\vec S \] whenever \(S_1\) can be continuously deformed into \(S_2\). Then we say \(\vec F\) has surface independent surface integrals.

In the special case that \(\vec\nabla\cdot\vec{F}=\vec 0\), then equation (*) says \(\vec F\) has surface independent surface integrals.

Let's compute a surface integral in two different ways: Once directly and once using Gauss' Theorem and the relation \((*)\).

Comparison: Notice the answers are the same. Clearly the integrals over the volume and disk were easier than the surface integral over the paraboloid.

Consider the surface integral \(\displaystyle \iint_S \vec F\cdot d\vec S\) where \(\displaystyle \vec F=\langle xz^2,-yz^2,x^2+y^2\rangle\) for various surfaces.

1. Verify the vector field \(\vec F\) has surface independent integrals. In other words, the integral \(\displaystyle \iint_S \vec F\cdot d\vec S\) is the same for any two surfaces \(S\) which have the same boundary \(\partial S\).

   Compute the divergence of \(\vec F\).

   \(\displaystyle \vec\nabla\cdot\vec F=0\)

   The vector field has surface independent integrals if it is divergencefree. Here: \[\begin{aligned} \vec\nabla\cdot\vec F &=\partial_x(xz^2)+\partial_y(-yz^2)+\partial_z(x^2+y^2) \\ &=z^2-z^2+0 =0 \end{aligned}\]

2. Compute the integral \(\displaystyle \iint_S \vec F\cdot d\vec S\) for each of the folowing surfaces:
   1. the paraboloid \(z=4-x^2-y^2 \ge 0\), oriented upward. (shown in cyan)
   2. the hemisphere \(z=\sqrt{4-x^2-y^2}\), oriented upward. (shown in orange)
   3. the disk \(x^2+y^2 \le 4\) and \(z=0\),oriented upward. (shown in red)

   

   Check the \(3\) surfaces have the same boundary curve. Then compute the surface integral you think will be easiest.

   \(\displaystyle \iint_S \vec F\cdot d\vec S=8\pi\) for all \(3\) surfaces.

   All \(3\) surfaces have the boundary curve \(x^2+y^2=4\) with \(z=0\). Since \(\vec F\) has surface independent integrals, all \(3\) integrals give the same value.

   The simplest is the disk which may be parametrized as \(\vec R(r,\theta)=(r\cos\theta,r\sin\theta,0)\). Its normal is \(\vec N=(0,0,r)\) which points up as required. On the disk, the vector field is \(\vec F=\langle 0,0,r^2\rangle\). So the integral is: \[\begin{aligned} \iint_S &\vec F\cdot d\vec S =\iint_S \vec F\cdot\vec N\,dr\,d\theta \\ &=\int_0^{2\pi}\int_0^2 r^3\,dr\,d\theta =2\pi\left[\dfrac{r^4}{4}\right]_0^2 =8\pi \end{aligned}\]

3. Verify the vector field \(\vec A=\langle -yx^2,xy^2,xyz^2\rangle\) is a vector potential for \(\vec F=\langle xz^2,-yz^2,x^2+y^2\rangle\).

   Compute the curl of \(\vec A\).

   \(\displaystyle \vec\nabla\times\vec A=\vec F\)

   Since: \[\begin{aligned} \vec\nabla\times\vec A &= \begin{vmatrix} \hat{\imath} & \hat{\jmath} & \hat{k} \\ \partial_x & \partial_y & \partial_z \\ -yx^2 & xy^2 & xyz^2 \end{vmatrix} \\ &=\hat{\imath}(xz^2) -\hat{\jmath}(-yz^2) +\hat{k}(y^2+x^2) =\vec F \end{aligned}\] we conclude \(\vec A\) is a vector potential for \(\vec F\).

4. Use Stokes' Theorem and the vector field \(\vec A=\langle -yx^2,xy^2,xyz^2\rangle\) to recompute the surface integrals in part (b).

   Stokes' Theorem states: \[ \iint_S \vec\nabla\times\vec A\cdot d\vec S =\oint_{\partial S} \vec A\cdot d\vec s \]

   \(\displaystyle \iint_S \vec F\cdot d\vec S =\oint_{\partial S} \vec A\cdot d\vec s=8\pi\)

   By Stokes' Theorem: \[ \iint_S \vec F\cdot d\vec S =\iint_S \vec\nabla\times\vec A\cdot d\vec S =\oint_{\partial S} \vec A\cdot d\vec s \] The boundary curve for all \(3\) surfaces is the circle \(x^2+y^2=4\) with \(z=0\) which may be parametrized by \(\vec r(\theta)=(2\cos\theta,2\sin\theta,0)\). On the circle: \[ \vec A=\langle -yx^2,xy^2,xyz^2\rangle =\langle -8\cos^2\theta\sin\theta,8\cos\theta\sin^2\theta,0\rangle \] The tangent vector is \[ \vec v=\langle -2\sin\theta,2\cos\theta,0\rangle \] So the line integral is \[\begin{aligned} \oint_{\partial S} &\vec A\cdot d\vec s =\int_0^{2\pi} \vec A\cdot\vec v\,d\theta =\int_0^{2\pi} 32\cos^2\theta\sin^2\theta\,d\theta \\ &=\int_0^{2\pi} 8\sin^2(2\theta)\,d\theta =\int_0^{2\pi} 4(1-\cos(4\theta))\,d\theta \\ &=\left[\rule{0pt}{10pt}4\theta-\sin(4\theta)\right]_0^{2\pi} =8\pi \end{aligned}\]

   The answers are the same. This method may be easier, provided you have or can find a vector potential.