Home»Math Guides»Double Integrals using Polar Coordinates (cylinder and ellipsoid region of integration)

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

We want to use polar coordinates to find the volume shared between two shapes. Most of the work involves correctly setting up the problem.

For example, a question you might find would sound like:

Example Problem: Find the volume contained in 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!

In general, when you see you have variables being squared, use polar coordinates!

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.

There does exist a technique where you “transform to spherical coordinates”, but in this particular kind of problem there’s a trick to simplify it significantly without the use of spherical coordinates.

This means that this is a perfectly valid question to ask in an earlier calculus course (spherical coordinates is done near the end of calculus 3 or “vector calculus”, but you could get asked this on a midterm before even hearing of spherical coordinates)

You can remove the “z” variable from the ellipsoid equation 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.

\[4{x^2} + 4{y^2} + {z^2} = 64\]

\[{z^2} = 64 – 4{x^2} – 4{y^2}\]

Remember that \({x^2} + {y^2} = {r^2}\)

\[{z^2} = 64 – 4{r^2}\]

\[z = \pm \sqrt {64 – 4{r^2}} \]

\[z = \pm \sqrt {4(16 – {r^2})} \]

\[z = \pm 2\sqrt {(16 – {r^2})} \]

The only problem is that we have a plus or minus symbol which gets messy and complicated quickly.

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 part, find the volume from that, and multiply it by a factor of two to get the true volume.

Think about it logically, we’re getting the volume only 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!

Don’t forget to change the boundaries when doing substitution.

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

The final answer above is a little messy, but it’s as simplified as can be.

Don’t think of doing something stupid like combining the (16-12) under the same root, you can’t do that! Don’t try to be a hero!

Try an example with different geometries, where you use double integrals to solve volumes contained in both a sphere and cylinder!

Click here to return to the Math Guides hubpage