### Examples of Double Integrals in Polar Coordinates in Cylinders and Ellipsoids (Example 4)

We want to use polar coordinates to find the volume inside both the cylinder \({x^2} + {y^2} = 4\) and the ellipsoid \(4{x^2} + 4{y^2} + {z^2} = 64\).

We can just use polar coordinates to do this, even though it's a 3-dimensional problem! We just need to find a way to convert these x,y, and z coordinates into radius and theta coordinates. This is not always possible; however, in this problem it is possible to do so.

You can remove the "z" variable by isolating the equation in terms of "z", then sticking it inside the integral. Then you convert x and y normally using polar coordinates.

So first we need to isolate the "z" variable in the equation for the ellipsoid. Be careful when factoring inside of a square root. When you remove a constant from a square root, you need to square root it as shown in the subsequent steps.

\[\begin{array}{l}4{x^2} + 4{y^2} + {z^2} = 64\\{z^2} = 64 - 4{x^2} - 4{y^2}\\{z^2} = 64 - 4{r^2}\\z = \pm \sqrt {64 - 4{r^2}} \\z = \pm \sqrt {4(16 - {r^2})} \\z = \pm 2\sqrt {(16 - {r^2})} \end{array}\]

The only problem is that we have a plus or minus symbol. The key here is that the ellipsoid is symmetric with respect to the "z" axis. Think about it, the plus part of the equation is above the z axis, and the minus part of the equation is below the z axis, but aside from the plus or minus symbols it's the same equation so it's symmetric. So we can just take the positive and multiply it by a factor of two. Because we'll be getting the volume above the z axis, then just multiply it by two to acquire the total volume.

We then need to figure out the boundaries of radius and theta. In the question we're told we need to find the volume that lies both within the cylinder and the ellipsoid. So the radius is from 0 to 2 (2 is the root of 4 in the cylinder). And it's just a regular 3D shape with no restrictions on the quadrants, it's in all the quadrants, so theta is from 0 to \(2\pi \). And we multiply by an extra "r" term when converting to polar coordinates. Don't forget we multiply by a factor of two as well! Let V represent the volume we're asked to find.

\[V = \int_0^{2\pi } {\int\limits_0^2 {\left( {2 \cdot 2\sqrt {\left( {16 - {r^2}} \right)} } \right)rdrd\theta } } \]

And we use regular substitution to solve the problem!

\[\begin{array}{l}u = 16 - {r^2}\\du = - 2rdr\\u(0) = 16\\u(2) = 16 - 4 = 12\\V = 2\int_0^{2\pi } {\int\limits_{16}^{12} {\left( { - \sqrt u } \right)dud\theta } } \\V = 2\int_0^{2\pi } {\frac{2}{3}} {\left( {16} \right)^{\frac{3}{2}}} - \frac{2}{3}{\left( {12} \right)^{\frac{3}{2}}}d\theta \\V = 2 \cdot \frac{2}{3}{\int_0^{2\pi } {\left( {16} \right)} ^{\frac{3}{2}}} - {\left( {12} \right)^{\frac{3}{2}}}d\theta \\V = \frac{{8\pi }}{3}\left[ {{{(16)}^{\frac{3}{2}}} - {{\left( {12} \right)}^{\frac{3}{2}}}} \right]\end{array}\]

The final answer above is a little messy, but that should be reasonable enough. Don't think of doing something stupid like combining (16-12) under the same root, you can't do that!

Please see other examples of solving double integrals in polar coordinates below:

Double Integrals in Polar Coordinates (Ex1)

Double Integrals in Polar Coordinates (Ex2)

Double Integrals in Polar Coordinates (Ex3)

Double Integrals in Polar Coordinates (Ex5 Spheres and Cylinders)

Alternatively, return to the solving double integrals using polar coordinates hub page:

Double Integrals in Polar Coordinates