Paul's Online Notes
Home / Calculus III / Multiple Integrals / Triple Integrals in Spherical 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 4-7 : Triple Integrals in Spherical Coordinates

5. Evaluate the following integral by first converting to an integral in spherical coordinates.

$\int_{{ - 1}}^{0}{{\int_{{ - \sqrt {1 - {x^{\,2}}} }}^{{\sqrt {1 - {x^{\,2}}} }}{{\int_{{\sqrt {6{x^{\,2}} + 6{y^{\,2}}} }}^{{\sqrt {7 - {x^{\,2}} - {y^{\,2}}} }}{{\,\,\,18y\,\,\,dz}}\,dy}}\,dx}}$

Show All Steps Hide All Steps

Start Solution

First let’s just get the Cartesian limits from the integral.

$\begin{array}{c} - 1 \le x \le 0\\ - \sqrt {1 - {x^2}} \le y \le \sqrt {1 - {x^2}} \\ \sqrt {6{x^2} + 6{y^2}} \le z \le \sqrt {7 - {x^2} - {y^2}} \end{array}$ Show Step 2

Now we need to convert the integral into spherical coordinates. Let’s first take care of the limits first.

From the upper $$z$$ limit we see that we are under $$z = \sqrt {7 - {x^2} - {y^2}}$$ (which is just the equation for the upper portion of a sphere of radius $$\sqrt 7$$).

From the lower $$z$$ limit we see that we are above $$z = \sqrt {6{x^2} + 6{y^2}}$$ (which is just the equation of a cone).

So, we appear to be inside an “ice cream cone” shaped region as we usually are when dealing with spherical coordinates.

This leads us to the following $$\rho$$ limits.

$0 \le \rho \le \sqrt 7$ Show Step 3

The limits for $$\varphi$$ we can get them from the equation of the cone that is the lower $$z$$ limit referenced in the previous step. First, we know that, in terms of cylindrical coordinates, $$\sqrt {{x^2} + {y^2}} = r$$ and we know that, in terms of spherical coordinates, $$r = \rho \sin \varphi$$. Therefore, if we convert the equation of the cone into spherical coordinates we get,

$\rho \cos \varphi = \sqrt 6 \rho \sin \varphi \hspace{0.25in} \to \hspace{0.25in} \tan \varphi = \frac{1}{{\sqrt 6 }} \hspace{0.25in} \to \hspace{0.25in}\varphi = {\tan ^{ - 1}}\left( {\frac{1}{{\sqrt 6 }}} \right) = 0.3876$

Because the region we are working on is above the cone we know that $$\varphi$$ must therefore range from 0 to 0.3876.

Show Step 4

Finally, let’s get the $$\theta$$ limits. For reference purposes here are the $$x$$ and $$y$$ limits we found in Step 1.

$\begin{array}{c} - 1 \le x \le 0\\ - \sqrt {1 - {x^2}} \le y \le \sqrt {1 - {x^2}} \end{array}$

All the $$y$$ limits tell us is that the region $$D$$ from the original Cartesian coordinates integral is a portion of the circle of radius 1. Note that this should make sense as this is also the intersection of the sphere and cone we get from the $$z$$ limits (we’ll leave it to you to verify this statement).

Now, from the $$x$$ limits we see that we must have the left side of the circle of radius 1 and so the limits for $$\theta$$ are then,

$\frac{\pi }{2} \le \theta \le \frac{{3\pi }}{2}$

The full set of spherical coordinate limits for the integral are then,

$\begin{array}{c} \displaystyle 0 \le \varphi \le {\tan ^{ - 1}}\left( {\frac{1}{{\sqrt 6 }}} \right) = 0.3876\\ \displaystyle \frac{\pi }{2} \le \theta \le \frac{{3\pi }}{2}\\0 \le \rho \le \sqrt 7 \end{array}$ Show Step 5

Okay, let’s convert the integral in to spherical coordinates.

\begin{align*}\int_{{ - 1}}^{0}{{\int_{{ - \sqrt {1 - {x^{\,2}}} }}^{{\sqrt {1 - {x^{\,2}}} }}{{\int_{{\sqrt {6{x^{\,2}} + 6{y^{\,2}}} }}^{{\sqrt {7 - {x^{\,2}} - {y^{\,2}}} }}{{\,\,\,18y\,\,\,dz}}\,dy}}\,dx}} & = \int_{0}^{{0.3876}}{{\int_{{\frac{\pi }{2}}}^{{\frac{{3\pi }}{2}}}{{\int_{0}^{{\sqrt 7 }}{{\left( {18\rho \sin \varphi \sin \theta } \right)\left( {{\rho ^2}\sin \varphi } \right)\,d\rho }}\,d\theta }}\,d\varphi }}\\ & = \int_{0}^{{0.3876}}{{\int_{{\frac{\pi }{2}}}^{{\frac{{3\pi }}{2}}}{{\int_{0}^{{\sqrt 7 }}{{18{\rho ^3}{{\sin }^2}\varphi \sin \theta \,d\rho }}\,d\theta }}\,d\varphi }}\end{align*}

Don’t forget to convert the $$y$$ into spherical coordinates. Also, don’t forget that the $$dz\,dy\,dx$$ come from the $$dV$$ in the original triple integral. We also know that, in terms of spherical coordinates,$$dV = {\rho ^2}\sin \varphi \,d\rho \,d\theta \,d\varphi$$ and so we in turn know that,

$dz\,dy\,dx = {\rho ^2}\sin \varphi \,d\rho \,d\theta \,d\varphi$ Show Step 6

Okay, now all we need to do is evaluate the integral. Here is the $$\rho$$ integration.

\begin{align*}\int_{{ - 1}}^{0}{{\int_{{ - \sqrt {1 - {x^{\,2}}} }}^{{\sqrt {1 - {x^{\,2}}} }}{{\int_{{\sqrt {6{x^{\,2}} + 6{y^{\,2}}} }}^{{\sqrt {7 - {x^{\,2}} - {y^{\,2}}} }}{{\,\,\,18y\,\,\,dz}}\,dy}}\,dx}} & = \int_{0}^{{0.3876}}{{\int_{{\frac{\pi }{2}}}^{{\frac{{3\pi }}{2}}}{{\left. {\left( {\frac{9}{2}{\rho ^4}{{\sin }^2}\varphi \sin \theta } \right)} \right|_0^{\sqrt 7 }\,d\theta }}\,d\varphi }}\\ & = \int_{0}^{{0.3876}}{{\int_{{\frac{\pi }{2}}}^{{\frac{{3\pi }}{2}}}{{\frac{{441}}{2}{{\sin }^2}\varphi \sin \theta \,d\theta }}\,d\varphi }}\end{align*} Show Step 7

Next let’s do the $$\theta$$ integration.

\begin{align*}\int_{{ - 1}}^{0}{{\int_{{ - \sqrt {1 - {x^{\,2}}} }}^{{\sqrt {1 - {x^{\,2}}} }}{{\int_{{\sqrt {6{x^{\,2}} + 6{y^{\,2}}} }}^{{\sqrt {7 - {x^{\,2}} - {y^{\,2}}} }}{{\,\,\,18y\,\,\,dz}}\,dy}}\,dx}} & = \int_{0}^{{0.3876}}{{\left. {\left( { - \frac{{441}}{2}{{\sin }^2}\varphi \cos \theta } \right)} \right|_{\frac{\pi }{2}}^{\frac{{3\pi }}{2}}\,d\varphi }}\\ & = \int_{0}^{{0.3876}}{{0\,d\varphi }}\\ & = \require{bbox} \bbox[2pt,border:1px solid black]{0}\end{align*}

So, as noted above once we got the integrand down to zero there was no reason to continue integrating as the answer will continue to be zero for the rest of the problem.

Don’t get excited about it when these kinds of things happen. They will on occasion and all it means is that we get to stop integrating a little sooner that we would have otherwise.