Volume and the divergence theorem

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We compute volumes using the divergence theorem.

Work in groups of 3–4, writing your answers on a separate sheet of paper.

So far, we’ve been using the divergence theorem to simplify the computations of surface integrals. However, as we will see, we can also use the divergence theorem to compute volumes of solid regions. Specifically, we will now try to compute the volume of an ellipsoid: $\vec {E}(\theta ,\phi ) = \vector {a \cos (\theta )\sin (\phi ), b \sin (\theta )\sin (\phi ),c \cos (\phi )}$ for $0\le \theta < 2\pi$ and $0\le \phi \le \pi$ .

Give a careful sketch of the graph of $\vec {E}$ :

Describe in pictures, words, interpretative dance, how the ellipsoid is drawn by $\vec {E}$ as $\theta$ runs from $0$ to $2\pi$ , and $\phi$ runs from $0$ to $\pi$ .

As a gesture of friendship, I will tell you that the implicit formula for an ellipsoid is: $\frac {x^2}{a^2} + \frac {y^2}{b^2} + \frac {z^2}{c^2} =1$ Use the vector-valued formula $\vec {E}$ to confirm this formula.

If you were to set-up an iterated integral to compute the volume of the ellipsoid, which coordinates would be easiest to use? Why?

Set-up an iterated integral that will compute the volume of the ellipsoid.

Now we will try to use the divergence theorem to compute volume. As a gesture of friendship, we remind you of the statement of the divergence theorem: $\iiint _R \divergence \vec {F} \d V = \oiint _{\partial R} \vec {F}\dotp \uvec {n}\d S$ The first thing we need are vector fields such that $\divergence \vec {F} = 1$ .

Find $7$ (simple!) vector fields $\vec {F}: \R ^3\to \R ^3$ such that $\divergence \vec {F} =1$ .

Setting $\vec {F} = \vector {x/3,y/3,z/3}$ , use the divergence theorem to compute the volume of the ellipsoid.