Divergence theorem

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We introduce the divergence theorem.

The divergence theorem

The divergence theorem states that certain volume integrals are equal to certain surface integrals. Let’s see the statement.

Divergence Theorem Suppose that the components of $\vec {F}:\R ^3\to \R ^3$ have continuous partial derivatives. If $R$ is a solid bounded by a surface $\partial R$ oriented with the normal vectors pointing outside, then: $\iiint _R \divergence \vec {F} \d V = \oiint _{\partial R} \vec {F}\dotp \uvec {n}\d S$

Integrals of the type above arise any time we wish to understand “fluid flow” through a surface. The “fluid” in question could be a real fluid like air or water, or it could be an electromagnetic field, or something else entirely. Unfortunately, many of the “real” applications of the divergence theorem require a deeper understanding of the specific context where the integral arises. For our part, we will focus on using the divergence theorem as a tool for transforming one integral into another (hopefully easier!) integral.

Let $R = \{(x,y,z):-2\le x\le 1, 1\le y\le 3, -5\le z\le 1\}$ and $\vec {F}(x,y,z) = \vector {\sin (z),y^2,e^x},$ compute: $\oiint _{\partial R} \vec {F}\dotp \uvec {n} \d S$

To compute this integral, we’ll use the divergence theorem. Since our region is a box, the limits of the triple integral will be easy to work with. Moreover, computing the divergence of $\vec {F}$ we see $\divergence \vec {F}(x,y,z) = \answer [given]{2y}.$ So by the divergence theorem, we have $\begin{align*} \oiint_{\partial R} \vec{F}\dotp \uvec{n} \d S &= \iiint_R \divergence \vec{F} \d V \\ &= \int_{-2}^{\answer[given]{1}} \int_{-5}^{\answer[given]{1}} \int_{1}^{\answer[given]{3}} \answer[given]{2y} \d \answer[given]{y}\d \answer[given]{z}\d \answer[given]{x}\\ &= \answer[given]{144}. \end{align*}$

Now let’s see another example:

Let $\begin{align*} R = \{(x,y,z):&0\le x\le1, \\ &0\le y\le 2-2x, \\ &0\le z\le 3-3x-3y/2\} \end{align*}$ and $\vec {F}(x,y,z) = \vector {x,y,z},$ compute: $\oiint _{\partial R} \vec {F}\dotp \uvec {n} \d S$

To compute this integral, we’ll use the divergence theorem. This time our region is a tetrahedron, we’ll work in cartesian coordinates. Moreover, computing the divergence of $\vec {F}$ we see $\divergence \vec {F}(x,y,z) = \answer [given]{3}.$ So by the divergence theorem, we have $\begin{align*} \oiint_{\partial R} \vec{F}\dotp \uvec{n} \d S &= \iiint_R \divergence \vec{F} \d V \\ &= \int_{\answer[given]{0}}^{\answer[given]{1}} \int_{\answer[given]{0}}^{\answer[given]{2-2x}} \int_{\answer[given]{0}}^{\answer[given]{3-3x-3y/2}} \answer[given]{3} \d z \d y \d x\\ &= \answer[given]{3}. \end{align*}$

With our next two examples, we cannot help but flex our mathematical muscles a bit:

Let $\begin{align*} R = \{(x,y,z):&-1\le x\le1, \\ &-1\le y\le 1,\\ &0\le z\le \sqrt{1-x^2-y^2}\} \end{align*}$ and $\vec {F}(x,y,z) = \vector {x,2y,3z},$ compute: $\oiint _{\partial R} \vec {F}\dotp \uvec {n} \d S$

To compute this integral, we’ll use the divergence theorem. Computing the divergence of $\vec {F}$ we see $\divergence \vec {F}(x,y,z) = \answer [given]{6}.$ So by the divergence theorem, we have $\begin{align*} \oiint_{\partial R} \vec{F}\dotp \uvec{n} \d S &= \iiint_R \divergence \vec{F} \d V \\ &= \iiint_R \answer[given]{6} \d V \\ &= \answer[given]{6} \iiint_R \d V \end{align*}$ But $\iiint _R \d V$ is just the volume of the hemisphere of radius $1$ , which we know is $\answer [given]{2\pi /3}$ . Hence: $\oiint _{\partial R} \vec {F}\dotp \uvec {n} \d S = \answer [given]{4\pi }$

Above, we used the fact that we know that the volume of a sphere is $4\pi r^3/3$ . When you know the volume that’s great! If not you have to compute the integral.

Let $R$ be the cone whose base is a disk of radius $2$ in the plane $z=1$ and whose vertex is a the origin. Compute the flux of $\vec {F}(x,y,z) = \vector {x-\sin (y), 2y +e^x, \cos (xy)-4z}$ across the boundary of $R$ .

Again, we will use the divergence theorem. Computing the divergence of $\vec {F}$ we see: $\divergence \vec {F}(x,y,z) = \answer [given]{-1}$ Now write with me $\begin{align*} \oiint_{\partial R} \vec{F}\dotp \uvec{n} \d S &= \iiint_R \divergence \vec{F} \d V \\ &= \iiint_R \answer[given]{-1} \d V \\ &= \answer[given]{-1} \iiint_R \d V \end{align*}$ If you know the volume of the cone described above, you can be done! If not, do not despair, we’ll simply use cylindrical coordinates to compute it. Write with me: $\begin{align*} \iiint_R \d V &= \int_{\answer[given]{0}}^{\answer[given]{2\pi}} \int_{\answer[given]{0}}^{\answer[given]{2}} \int_{\answer[given]{r/2}}^{\answer[given]{1}} r \d z \d r \d\theta\\ &=\int_{\answer[given]{0}}^{\answer[given]{2\pi}} \int_{\answer[given]{0}}^{\answer[given]{2}} \answer[given]{r - r^2/2}\d r \d\theta\\ &=\int_{\answer[given]{0}}^{\answer[given]{2\pi}} \answer[given]{2/3} \d\theta\\ &=\answer[given]{4\pi/3} \end{align*}$ Hence: $\oiint _{\partial R} \vec {F}\dotp \uvec {n} \d S = \answer [given]{-4\pi /3}$

A new fundamental theorem of calculus

How is the divergence theorem a fundamental theorem of calculus? Well consider this:

Are there more fundamental theorems of calculus? Absolutely, and we’re ready for the last one of this course. Read on young mathematician!