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=−√4x2+4y2 inside the cylinder x2+y2=3 with x≤0.
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Start SolutionOkay, 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=−√4x2+4y2 is the bottom “cap” on the cylinder. We only have half of the cylinder because of the x≤0 portion of the problem statement.
Show Step 2The volume of this solid is given by,
V=∭ Show Step 3So, 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 4For 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 5Plugging 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 6Okay, 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 7Next 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 8Finally, 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}}