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Section 6-6 : Divergence Theorem

In this section we are going to relate surface integrals to triple integrals. We will do this with the Divergence Theorem.

Divergence Theorem

Let \(E\) be a simple solid region and \(S\) is the boundary surface of \(E\) with positive orientation. Let \(\vec F\) be a vector field whose components have continuous first order partial derivatives. Then,

\[\iint\limits_{S}{{\vec F\centerdot d\vec S}} = \iiint\limits_{E}{{{\mathop{\rm div}\nolimits} \vec F\,dV}}\]

Let’s see an example of how to use this theorem.

Example 1 Use the divergence theorem to evaluate \(\displaystyle \iint\limits_{S}{{\vec F\centerdot d\vec S}}\) where \(\vec F = xy\,\vec i - \frac{1}{2}{y^2}\,\vec j + z\,\vec k\) and the surface consists of the three surfaces, \(z = 4 - 3{x^2} - 3{y^2}\), \(1 \le z \le 4\) on the top, \({x^2} + {y^2} = 1\), \(0 \le z \le 1\) on the sides and \(z = 0\) on the bottom.
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Let’s start this off with a sketch of the surface.

This is a graph with the standard 3D coordinate system.  The positive z-axis is straight up, the positive x-axis moves off to the left and slightly downward and positive y-axis moves off the right and slightly downward.  The walls of the solid in this graph are the cylinder given in the problem statement and whose “cap” is the elliptic paraboloid that starts at z=4 and opens along the z-axis in the negative z direction until it hits the cylinder at z=1.

The region \(E\) for the triple integral is then the region enclosed by these surfaces. Note that cylindrical coordinates would be a perfect coordinate system for this region. If we do that here are the limits for the ranges.

\[\begin{array}{c}0 \le z \le 4 - 3{r^2}\\ 0 \le r \le 1\\ 0 \le \theta \le 2\pi \end{array}\]

We’ll also need the divergence of the vector field so let’s get that.

\[{\mathop{\rm div}\nolimits} \vec F = y - y + 1 = 1\]

The integral is then,

\[\begin{align*}\iint\limits_{S}{{\vec F\centerdot d\vec S}} & = \iiint\limits_{E}{{{\mathop{\rm div}\nolimits} \vec F\,dV}}\\ & = \int_{{\,0}}^{{\,2\pi }}{{\int_{{\,0}}^{{\,1}}{{\int_{{\,0}}^{{4 - 3{r^2}}}{{r\,dz}}\,dr}}\,d\theta }}\\ & = \int_{{\,0}}^{{\,2\pi }}{{\int_{{\,0}}^{{\,1}}{{4r - 3{r^3}\,dr}}\,d\theta }}\\ & = \int_{{\,0}}^{{\,2\pi }}{{\left. {\left( {2{r^2} - \frac{3}{4}{r^4}} \right)} \right|_0^1\,d\theta }}\\ & = \int_{{\,0}}^{{\,2\pi }}{{\frac{5}{4}\,d\theta }}\\ & = \frac{5}{2}\pi \end{align*}\]