Paul's Online Notes
Paul's Online Notes
Home / Calculus III / Multiple Integrals / Triple Integrals in Cylindrical Coordinates
Show Mobile Notice Show All Notes Hide All Notes
Mobile Notice
You appear to be on a device with a "narrow" screen width (i.e. you are probably on a mobile phone). Due to the nature of the mathematics on this site it is best views in landscape mode. If your device is not in landscape mode many of the equations will run off the side of your device (should be able to scroll to see them) and some of the menu items will be cut off due to the narrow screen width.

Section 15.6 : Triple Integrals in Cylindrical Coordinates

4. Use a triple integral to determine the volume of the region below \(z = 6 - x\), above \(z = - \sqrt {4{x^2} + 4{y^2}} \) inside the cylinder \({x^2} + {y^2} = 3\) with \(x \le 0\).

Show All Steps Hide All Steps

Start Solution

Okay, let’s start off with a quick sketch of the region \(E\) so we can get a feel for what we’re dealing with.

Here then is the sketch of \(E\).

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 plane \(z = 6 - x\) is the top “cap” on the cylinder and the cone \(z = - \sqrt {4{x^2} + 4{y^2}} \) is the bottom “cap” on the cylinder. We only have half of the cylinder because of the \(x \le 0\) portion of the problem statement.

Show Step 2

The volume of this solid is given by,

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

So, from the sketch it is hopefully clear that the region \(D\) will be in the \(xy\)-plane and so we’ll need to integrate with respect to \(z\) first. That means that the \(z\) limits are,

\[ - \sqrt {4{x^2} + 4{y^2}} \le z \le 6 - x\] Show Step 4

For this problem \(D\) is simply the portion of the disk \({x^2} + {y^2} \le \sqrt 3 \) with \(x \le 0\). Here is a quick sketch of \(D\) to maybe help with the limits.

Since \(D\) is clearly a portion of a disk it makes sense that we’ll be using cylindrical coordinates. So, here are the cylindrical coordinates for this problem.

\[\begin{array}{c}\displaystyle \frac{\pi }{2} \le \theta \le \frac{{3\pi }}{2}\\ 0 \le r \le \sqrt 3 \\ - 2r \le z \le 6 - r\cos \theta \end{array}\]

Don’t forget to convert the \(z\) limits into cylindrical coordinates.

Show Step 5

Plugging these limits into the integral and converting to cylindrical coordinates gives,

\[V = \iiint\limits_{E}{{\,dV}} = \int_{{\frac{\pi }{2}}}^{{\frac{{3\pi }}{2}}}{{\int_{0}^{{\sqrt 3 }}{{\int_{{ - 2r}}^{{6 - r\cos \theta }}{{r\,dz}}\,dr}}\,d\theta }}\]

Don’t forget that \(dV = r\,dx\,dr\,d\theta \) and so we pick up an \(r\) when converting the \(dV\) to cylindrical coordinates.

Show Step 6

Okay, now all we need to do is evaluate the integral. Here is the \(z\) integration.

\[\begin{align*}V & = \int_{{\frac{\pi }{2}}}^{{\frac{{3\pi }}{2}}}{{\int_{0}^{{\sqrt 3 }}{{\left. {\left( {rz} \right)} \right|_{ - 2r}^{6 - r\cos \theta }\,dr}}\,d\theta }}\\ & = \int_{{\frac{\pi }{2}}}^{{\frac{{3\pi }}{2}}}{{\int_{0}^{{\sqrt 3 }}{{6r - {r^2}\cos \theta + 2{r^2}\,dr}}\,d\theta }}\end{align*}\] Show Step 7

Next let’s do the \(r\) integration.

\[\begin{align*}V & = \int_{{\frac{\pi }{2}}}^{{\frac{{3\pi }}{2}}}{{\left. {\left( {3{r^2} - \frac{1}{3}{r^3}\cos \theta + \frac{2}{3}{r^3}} \right)} \right|_0^{\sqrt 3 }\,d\theta }}\\ & = \int_{{\frac{\pi }{2}}}^{{\frac{{3\pi }}{2}}}{{9 - \sqrt 3 \cos \theta + 2\sqrt 3 \,d\theta }}\end{align*}\] Show Step 8

Finally, we’ll do the \(\theta \) integration.

\[V = \left. {\left( {9\theta + 2\sqrt 3 \theta - \sqrt 3 \sin \theta } \right)} \right|_{\frac{\pi }{2}}^{\frac{{3\pi }}{2}} = \require{bbox} \bbox[2pt,border:1px solid black]{{2\sqrt 3 + \left( {9 + 2\sqrt 3 } \right)\pi = 42.6212}}\]