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
Suppose \(S_1\) and \(S_2\) are two open surfaces which have the same boundary curve, \(\partial S_1=\partial S_2\). We say that \(S_1\) can be continuously deformed into \(S_2\) if there is a solid region \(V\) whose outward oriented boundary is \(\partial V=S_1-S_2\). This means that the boundary of \(V\) with outward normal consists of \(S_1\) with its normal unchanged and \(S_2\) with its normal reversed.
In the figure, notice that the normal \(\vec N_1\) points out of the volume \(V\) while \(\vec N_2\) points in. So \(\vec N_2\), needs to be reversed to complete the boundary of \(V\) with the outward normal.
![def_surfindep_open](graphics/def_surfindep_open.jpg)
Similarly, suppose \(S_1\) and \(S_2\) are two closed surfaces. We say that \(S_1\) can be continuously deformed into \(S_2\) if there is a solid region \(V\) whose outward oriented boundary is \(\partial V=S_1-S_2\). This means that the boundary of \(V\) with outward normal (but inward in holes) consists of \(S_1\) with its normal unchanged and \(S_2\) with its normal reversed.
In the figure, \(S_1\) is an ellipsoid with radii \(6\), \(5\) and \(4\) and \(S_2\) is a sphere with radius \(2\) inside it. Further, \(V\) is the soild between the ellipse and the sphere. The normals \(\vec N_1\) and \(\vec N_2\) both point outward. So \(\vec N_2\) needs to be reversed to complete the boundary of \(V\) with the outward normal (but inward in holes).
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 \((*)\).
Compute the integral
\(\displaystyle \iint_P \vec F\cdot d\vec S\)
for
\[
\vec F=\langle 2xy^2,2yx^2,x^2z+y^2z\rangle
\]
over the paraboloid, \(P\), given by \(z=9-x^2-y^2\),
for \(z \ge 0\), oriented upward, in two different ways:
(1) directly and
(2) using the relation \((*)\) where \(S_1\) is the
paraboloid \(P\) and \(S_2\) is the disk, \(D\), given by
\(x^2+y^2 \le 9\) and \(z=0\) also oriented upward.
Comparison: Notice the answers are the same. Clearly the integrals over the volume and disk were easier than the surface integral over the paraboloid.
Compute the integral \(\displaystyle \iint_H \vec F\cdot d\vec S\) for \(\displaystyle \vec F=\langle xz^2,yz^2,x^2+y^2\rangle\) over the hemisphere \(z=\sqrt{4-x^2-y^2 }\), oriented upward.
![ex_surfindep_hemisph_prob](graphics/ex_surfindep_hemisph_prob.jpg)
Use the relation \((*)\) where \(S_1\) is the hemisphere \(H\) and \(S_2\) is the disk \(D\) given by \(x^2+y^2 \le 4\) and \(z=0\) also oriented upward.
\(\displaystyle \iint_H \vec F\cdot d\vec S=\dfrac{248}{15}\pi\)
Gauss' Theorem in the form \((*)\) says: \[ \iint_{H} \vec F\cdot d\vec S =\iint_{D} \vec F\cdot d\vec S+\iiint_V \vec\nabla\cdot\vec F\,dV \] where \(H\) is the hemisphere surface oriented up, \(D\) is the disk at the bottom also oriented up and \(V\) is the solid hemisphere volume between them.
![ex_surfindep_hemisph_sol](graphics/ex_surfindep_hemisph_sol.jpg)
Surface integral over the Disk \(D\): The disk \(D\) is parameterized as: \(\vec R=(r\cos\theta,r\sin\theta,0)\). The normal vector is: \(\vec N=(0,0,r)\) which is correctly oriented upward. The vector field on the disk is: \[ \vec F =\langle xz^2,yz^2,x^2+y^2\rangle =\langle 0,0,r^2\rangle \] So the surface integral over \(D\) is \[\begin{aligned} \iint_D &\vec F\cdot d\vec S =\int_0^{2\pi}\int_0^2 \vec F\cdot\vec N\,dr\,d\theta \\ &=\int_0^{2\pi}\int_0^2 r^3\,dr\,d\theta=0 =2\pi\left[\dfrac{r^4}{4}\right]_0^2 =8\pi \end{aligned}\]
Triple integral over the volume \(V\) between the Disk and the Hemisphere: The divergence in rectangular and then spherical coordinates is: \[ \vec\nabla\cdot\vec F =z^2+z^2=2z^2=2\rho^2\cos^2\phi \] The spherical volume element is \(dV=\rho^2\sin\phi\,d\rho\,d\phi\,d\theta\). So the integral over the volume \(V\) is: \[\begin{aligned} \iiint_V &\vec\nabla\cdot\vec F\,dV =\int_0^{2\pi}\int_0^{\pi/2}\int_0^2 2\rho^2\cos^2\phi\,\rho^2\sin\phi\,d\rho\,d\phi\,d\theta\\ &=2\pi\left[-\,\dfrac{\cos^3\phi}{3}\right]_0^{\pi/2} \left[\dfrac{2\rho^5}{5}\right]_0^2 =2\pi\left(\dfrac{1}{3}\right)\dfrac{2^6}{5} =\dfrac{128}{15}\pi \end{aligned}\]
Total: Finally we add the two integrals to get our answer: \[\begin{aligned} \iint_{H} \vec F\cdot d\vec S &=\iint_{D} \vec F\cdot d\vec S+\iiint_V \vec\nabla\cdot\vec F\,dV \\ &=8\pi+\dfrac{128}{15}\pi=\dfrac{248}{15}\pi \end{aligned}\]
We check by doing the surface integral over the hemisphere directly, which is actually harder. The hemisphere may be parametrized in spherical coordinates as: \[\begin{aligned} \vec R(\phi,&\theta)=(2\sin\phi\cos\theta,2\sin\phi\sin\theta,2\cos\phi) \\ &\text{for} \quad 0 \le \phi \le \dfrac{\pi}{2} \quad \text{and} \quad 0 \le \theta \le 2\pi \end{aligned}\] The normal vector is: \[\begin{aligned} \vec N &=\vec e_\phi\times\vec e_\theta \\ &=\langle 4\sin^2\phi\cos\theta,4\sin^2\phi\sin\theta,4\sin\phi\cos\phi\rangle \end{aligned}\] The normal points upward as required. On the surface, the vector field is: \[\begin{aligned} \vec F&=\langle xz^2,yz^2,x^2+y^2\rangle \\ &=\langle 8\sin\phi\cos^2\phi\cos\theta, 8\sin\phi\cos^2\phi\sin\theta, 4\sin^2\phi\rangle \\ \end{aligned}\] And its dot product with the normal is: \[\begin{aligned} \vec F\cdot\vec N &=32\sin^3\phi\cos^2\phi\cos^2\theta +32\sin^3\phi\cos^2\phi\sin^2\theta \\ &\quad+16\sin^3\phi\cos\phi \\ &=32\sin^3\phi\cos^2\phi+16\sin^3\phi\cos\phi \end{aligned}\] So the integral is: \[\begin{aligned} \iint_H &\vec F\cdot d\vec S =\int_0^{2\pi}\int_0^{\pi/2} \vec F\cdot\vec N\,d\phi\,d\theta \\ &=\int_0^{2\pi}\int_0^{\pi/2} (32\sin^3\phi\cos^2\phi+16\sin^3\phi\cos\phi)\,d\phi\,d\theta \\ &=2\pi\int_0^{\pi/2} (32(1-\cos^2\phi)\cos^2\phi\sin\phi+16\sin^3\phi\cos\phi)\,d\phi \\ &=2\pi\left[ 32\left(-\,\dfrac{\cos^3\phi}{3}+\,\dfrac{\cos^5\phi}{5}\right)+16\dfrac{\sin^4\phi}{4} \right]_0^{\pi/2} \\ &=2\pi\left[16\cdot\dfrac{1}{4}\right] -2\pi\left[32\left(-\,\dfrac{1}{3}+\,\dfrac{1}{5}\right)\right] =\dfrac{248}{15}\pi \end{aligned}\]
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.
-
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: \[\begin{aligned} \vec\nabla\cdot\vec F &=\partial_x(xz^2)+\partial_y(-yz^2)+\partial_z(x^2+y^2) \\ &=z^2-z^2+0 =0 \end{aligned}\]
-
Compute the integral
\(\displaystyle \iint_S \vec F\cdot d\vec S\)
for each of the folowing surfaces:
- the paraboloid \(z=4-x^2-y^2 \ge 0\), oriented upward. (shown in cyan)
- the hemisphere \(z=\sqrt{4-x^2-y^2}\), oriented upward. (shown in orange)
- 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: \[\begin{aligned} \iint_S &\vec F\cdot d\vec S =\iint_S \vec F\cdot\vec N\,dr\,d\theta \\ &=\int_0^{2\pi}\int_0^2 r^3\,dr\,d\theta =2\pi\left[\dfrac{r^4}{4}\right]_0^2 =8\pi \end{aligned}\]
-
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 \(\vec F\).
-
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.
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