In this section we are going to relate surface integrals to triple integrals. We will do this with the 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,
\[\iint\limits_{S}{{\vec F\centerdot d\vec S}} = \iiint\limits_{E}{{{\mathop{\rm div}\nolimits} \vec F,円dV}}\]Let’s see an example of how to use this theorem.
Let’s start this off with a sketch of the surface.
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,
\[\begin{align*}\iint\limits_{S}{{\vec F\centerdot d\vec S}} & = \iiint\limits_{E}{{{\mathop{\rm div}\nolimits} \vec F,円dV}}\\ & = \int_{{,0円}}^{{,2円\pi }}{{\int_{{,0円}}^{{,1円}}{{\int_{{,0円}}^{{4 - 3{r^2}}}{{r,円dz}},円dr}},円d\theta }}\\ & = \int_{{,0円}}^{{,2円\pi }}{{\int_{{,0円}}^{{,1円}}{{4r - 3{r^3},円dr}},円d\theta }}\\ & = \int_{{,0円}}^{{,2円\pi }}{{\left. {\left( {2{r^2} - \frac{3}{4}{r^4}} \right)} \right|_0^1,円d\theta }}\\ & = \int_{{,0円}}^{{,2円\pi }}{{\frac{5}{4},円d\theta }}\\ & = \frac{5}{2}\pi \end{align*}\]