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### 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 $$x \le 0$$.

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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.

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