We introduce the divergence theorem.
1 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\le 1, \\ &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\le 1, \\ &-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:
\[ \oiint _{\partial R} \vec {F}\dotp \uvec {n} \d S = \answer [given]{-4\pi /3} \]
2 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!
![[Picture]](digInDivergenceTheorem.online-7502759621de32106811354b58ef0ca8.svg)