Divergence Theorem

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Objectives:

1.: Know the statement of the Divergence Theorem.
2.: Be able to apply the Divergence Theorem to solve flux integrals.
3.: Know how to close the surface and use divergence theorem.
4.: Understand where the Divergence Theorem fits into your toolbox for flux integrals.

Recap Video

Here is a video highlights the main points of the section.

To summarize:

Let $S$ be a closed surface, with $S$ being the boundary of a 3D solid $R$ . Give $S$ the outward orientation. Let $\textbf {F}$ be a vector field whose components have continuous partials on $R$ . Then $\iint _S \textbf {F} \cdot d\textbf {S} = \iiint _R \text {div}(\textbf {F})dV.$

Example Video

Here is an example of using the Divergence Theorem.

Let $S$ be the cylinder $x^2+y^2=4$ for $0 \leq z \leq 3$ coupled with the disc $x^2+y^2 \leq 4$ in the plane $z=0$ , all oriented outward (i.e. cylinder outward and disc downward). If $\textbf {F} = \left \langle 2e^{y^2}+z^2+x,z^{10}-y^2,xy+z \right \rangle$ , evaluate $\iint _S \textbf {F} \cdot d\textbf {S}.$

Problems

Let $S$ be the outer faces of the box $B: [0,2] \times [0,3] \times [0,1]$ , oriented with outward normals, and let $\textbf {F} = \left \langle x^2,z^4,e^z \right \rangle$ . Evaluate $\displaystyle \iint _S \textbf {F} \cdot d\textbf {S}$ .

To use the Divergence Theorem, we need to verify the continuity of the (partials of the) components of $\textbf {F}$ and check whether $S$ is closed. In this case the components of $\textbf {F}$ areare not defined and continuous everywhere, and $S$ isis not closed. Therefore, we can apply the Divergence Theorem. Notice $S$ isis not correctly oriented. So

$\begin{eqnarray*} \iint_S \textbf{F} \cdot d\textbf{S} &=& \iiint_B \text{div}(\textbf{F})dV\\ &=& \int_{\answer{0}}^{\answer{2}}\int_{\answer{0}}^{\answer{3}}\int_{\answer{0}}^{\answer{1}} \answer{2x+e^z}dz\,dy\,dx \\ &=& \answer{6e+6}. \end{eqnarray*}$

Let $\textbf {F} = \left \langle x+xy^2+z^2,\sin (x+z)+\cos (x+z),e^{-y}-zy^2 \right \rangle$ , and let $S$ be the sphere $x^2+y^2+z^2 = 1$ , oriented inward. Evaluate $\displaystyle \iint _S \textbf {F} \cdot d\textbf {S}$ .

Notice that $\text {div}(\textbf {F}) = \answer {1}$ . To use the Divergence Theorem, we need to verify that $\textbf {F}$ has continuous partials and that $S$ is closed. In this case the components of $\textbf {F}$ areare not defined and continuous everywhere, and $S$ isis not closed. Let $R$ be the sphere $x^2+y^2+z^2\leq 1$ , so $S$ encloses $R$ . Therefore, we can apply the Divergence Theorem. In this problem, the surface $S$ isis not correctly oriented. We can get around this by adding a negative sign:

$\begin{eqnarray*} \iint_S \textbf{F} \cdot d\textbf{S} &=& -\iiint_R \text{div}(\textbf{F})dV \\ &=& -\iiint_R \answer{1}dV \\ &=&- \text{vol}(R) \\ &=& \answer{-4\pi/3}. \end{eqnarray*}$

Let $S$ be the portion of $z = 4-x^2-y^2$ above $z = 1$ , oriented outward, and let $\textbf {F} = \left \langle x+y^2+z,-2y+z+\sin (x^2),3z \right \rangle$ . Evaluate $\displaystyle \iint _S \textbf {F} \cdot d\textbf {S}$ .

Notice that in this problem, $S$ isis not closed. Also notice that $\text {div}(\textbf {F}) = \answer {2} \neq 0$ , so Stokes’ theorem doesn’t apply. Our options are to either do the flux integral directly, or to close the surface and use the Divergence Theorem. We will close it and use the Divergence Theorem, since the flux integral seems hard to do directly. A convenient choice to close with is the disc $x^2+y^2 \leq \answer {3}$ in the plane $z = 1$ , call this $S'$ . Since we are adding this surface ourselves, we should pick an orientation, so let us orient $S'$ upward (arbitrary choice).

Notice that $S$ and $S'$ together form a closed surface. For the Divergence Theorem, the surface $S$ is correctlyincorrectly oriented and $S'$ is correctlyincorrectly oriented. Therefore, $\iint _S \textbf {F} \cdot d\textbf {S} - \iint _{S'} \textbf {F} \cdot d\textbf {S} = \iiint _R \text {div}(\textbf {F})dV,$ where $R$ is the region inside our (now) closed surface. Now we have two pieces to compute.

To compute the triple integral, we can use cylindrical coordinates. In cylindrical, we get $\iiint _R \text {div}(\textbf {F})dV = \int _{\answer {0}}^{\answer {2\pi }}\int _{\answer {0}}^{\answer {\sqrt {3}}}\int _{\answer {1}}^{\answer {4-r^2}} \answer {2r}\,dz\,dr\,d\theta .$ This integral evaluates to $\answer {9\pi }$ .
To do the flux integral over $S'$ : we have to parametrize $S'$ , which we will do as $\textbf {r}(x,y) = \left \langle x,y,1\right \rangle$ for $x^2+y^2 \leq \answer {3}$ (call this domain $D$ ). Then $\textbf {r}_x \times \textbf {r}_y = \left \langle \answer {0},\answer {0},\answer {1} \right \rangle ,$ which is the normal direction. Therefore, the flux is $\iint _{S'} \textbf {F} \cdot d\textbf {S} = \iint _D \answer {3}dA = \answer {9\pi }.$

Putting all this together, we get $\iint _S \textbf {F} \cdot d\textbf {S} = \answer {18\pi }.$

Let $\textbf {F} = \left \langle x^2+yz,-2xy+z^2,1+x+y \right \rangle$ , and let $S$ be the ellipsoid $x^2+4y^2+9z^2=36$ , oriented outward. Then $\displaystyle \iint _S \textbf {F} \cdot d\textbf {S} = \answer {0}$ .

Notice $\text {div}(\textbf {F}) = 0$ and $S$ is closed.

Let $\textbf {F} = \left \langle x^2+y,y^2+z,x+y \right \rangle$ , and let $S$ be the sphere $x^2+y^2+z^2 = 1$ for $z \geq 0$ , oriented outward, coupled with the disc $x^2+y^2 \leq 1$ in the plane $z=0$ , oriented downward. Then $\displaystyle \iint _S \textbf {F} \cdot d\textbf {S} = \answer {0}$ .

Notice $\text {div}(\textbf {F}) = \answer {2x+2y}$ . The surface $S$ is closed, and $\textbf {F}$ has components which have continuous partials everywhere, so we can use the Divergence Theorem. Let $R$ be the region $x^2+y^2+z^2 \leq 1$ for $z \geq 0$ , so $S$ encloses $R$ . Then $\iint _S \textbf {F} \cdot d\textbf {S} = \iiint _R (2x+2y)dV.$ This is best in spherical: $\iiint _R (2x+2y) = \int _0^{\answer {2\pi }} \int _0^{\answer {\pi /2}} \int _0^{\answer {1}} (2\rho \cos (\theta )\sin (\phi )+2\rho \sin (\theta )\sin (\phi )\rho ^2\sin (\phi )d\rho \,d\phi \,d\theta .$

Let $\textbf {F} = \left \langle y^3,x^3,z \right \rangle$ , and let $S$ be the portion of $z = \sqrt {x^2+y^2}$ with $0 \leq z \leq 1$ , oriented with outward normals. Then $\displaystyle \iint _S \textbf {F} \cdot d\textbf {S} = \answer {-2\pi /3}$ .

Let $S'$ be the disc $x^2+y^2 \leq 1$ in the plane $z=1$ , oriented upward. Then $\iint _S \textbf {F} \cdot d\textbf {S}+\iint _{S'} \textbf {F} \cdot d\textbf {S}= \iiint _R \text {div}(\textbf {F})dV.$ Notice $\text {div}(\textbf {F}) =\answer {1}$ . Thus, the triple integral can be written as $\int _0^{\answer {2\pi }}\int _0^{\answer {1}} \int _{\answer {r}}^1 r\,dz\,dr\,d\theta = \frac {\pi }{3}.$ The flux integral over $S'$ : parametrize the surface as $\textbf {r}(x,y) = \left \langle x,y,1 \right \rangle$ over the domain $D$ given by $x^2+y^2 \leq 1$ . Then $\textbf {r}_x \times \textbf {r}_y = \left \langle 0,0,\answer {1} \right \rangle$ . The flux becomes $\iint _D 1dA = \answer {\pi }.$ Thus, $\iint _S \textbf {F} \cdot d\textbf {S} = \answer {-\frac {2\pi }{3}}.$

If $\textbf {F} = \text {curl}(\textbf {G})$ , explain how surface independence of $\displaystyle \iint _S \textbf {F} \cdot d\textbf {S}$ follows from both Stokes’ theorem and Divergence theorem.

True/False

Suppose $\textbf {F}$ is a three-dimensional vector field whose components have continuous partials everywhere, and $S$ is a closed surface. Then it must be true that $\displaystyle \iint _S \textbf {F} \cdot d\textbf {S} = 0$ .

True False

Suppose $\textbf {F}$ is a three-dimensional vector field whose components have continuous partials everywhere, and $\textbf {F} = \text {curl}(\textbf {G})$ for some vector field $\textbf {G}$ . Then it must be true that $\displaystyle \iint _S \textbf {F} \cdot d\textbf {S} = 0$ .

True False

Suppose $\textbf {F}$ is a three-dimensional vector field whose components have continuous partials everywhere, and $\text {div}(\textbf {F}) \neq 0$ . Then it must be true that $\displaystyle \iint _S \textbf {F} \cdot d\textbf {S} \neq 0$ .

True False