Calculus III - Triple Integrals

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Section 15.5 : Triple Integrals

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9. Use a triple integral to determine the volume of the region that is below $z = 8 - {x^2} - {y^2}$ above $z = - \sqrt {4{x^2} + 4{y^2}}$ and inside ${x^2} + {y^2} = 4$ .

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Okay, let’s start off with a quick sketch of the region we want the volume of so we can get a feel for what we’re dealing with. We’ll call this region $E$ .

We’ve given the sketches with a set of “traditional” axes as well as a set of “box” axes to help visualize the surface and region.

The top of the region (the orange colored surface) is the portion of the graph of the elliptic paraboloid $z = 8 - {x^2} - {y^2}$ that is inside the cylinder ${x^2} + {y^2} = 4$ . The bottom of the region is the portion of the graph of the cone $z = - \sqrt {4{x^2} + 4{y^2}}$ that is inside the cylinder ${x^2} + {y^2} = 4$ . The walls of the region (which are translucent to show the bottom portion) is the cylinder ${x^2} + {y^2} = 4$ .

Show Step 2

The volume of this solid is given by,

$V = \iiint\limits_{E}{{dV}}$ Show Step 3

So, we now need to get the limits set up for the integral. From the sketch above we can see that we’ll need to integrate with respect to z first so here are those limits.

$- \sqrt {4{x^2} + 4{y^2}} \le z \le 8 - {x^2} - {y^2}$

We’ll also need limits for $D$ . In this case $D$ is just the disk given by ${x^2} + {y^2} \le 4$ (i.e. the portion of the $xy$ -plane that is inside the cylinder. This is the region in the $xy$ -plane that we need to graph the paraboloid and cone and so is $D$ .

Because $D$ is a disk it makes sense to use polar coordinates for integrating over $D$ . Here are the limits for $D$ .

$\begin{array}{c}0 \le \theta \le 2\pi \\ 0 \le r \le 2\end{array}$

Now, let’s set up the volume integral as follows.

$V = \iiint\limits_{E}{{dV}} = \iint\limits_{D}{{\left[ {\int_{{ - \sqrt {4{x^2} + 4{y^2}} }}^{{8 - {x^2} - {y^2}}}{{\,dz}}} \right]\,dA}}$

Because we know that we’ll need to do the outer double integral in polar coordinates we’ll hold off putting those limits in until we have the $z$ integration done.

Show Step 4

Okay, let’s do the $z$ integration.

$\begin{align*}V & = \iint\limits_{D}{{\left. z \right|_{ - \sqrt {4{x^2} + 4{y^2}} }^{8 - {x^2} - {y^2}}\,dA}}\\ & = \iint\limits_{D}{{8 - {x^2} - {y^2} + \sqrt {4{x^2} + 4{y^2}} \,dA}}\end{align*}$ Show Step 5

Now let’s convert the integral over to polar coordinates. Don’t forget that ${x^2} + {y^2} = {r^2}$ and that $dA = r\,dr\,d\theta$ .

The volume integral in terms of polar coordinates is then,

$\begin{align*}V & = \int_{0}^{{2\pi }}{{\int_{0}^{2}{{\left[ {8 - {r^2} + 2r} \right]r\,dr}}\,d\theta }}\\ & = \int_{0}^{{2\pi }}{{\int_{0}^{2}{{8r - {r^3} + 2{r^2}\,dr}}\,d\theta }}\end{align*}$ Show Step 6

The $r$ integration is then,

$V = \int_{0}^{{2\pi }}{{\left. {\left( {4{r^2} - \frac{1}{4}{r^4} + \frac{2}{3}{r^3}} \right)} \right|_0^2\,d\theta }} = \int_{0}^{{2\pi }}{{\frac{{52}}{3}\,d\theta }}$ Show Step 7

Finally, we can compute the very simple $\theta$ integral to get the volume of the region.

$V = \int_{0}^{{2\pi }}{{\frac{{52}}{3}\,d\theta }} = \require{bbox} \bbox[2pt,border:1px solid black]{{\frac{{104}}{3}\pi }}$