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July 31

Triple Double Trouble

Greetings once again, and thank you all very much for those of You who helped with my previous question on triple integration. I now have a second such question which also has me stumped, and I am perplexed, since I thought the technique we were given was meant to simplify things. We have been asked to express and evaluate the following triple integral in terms of Spherical Polar Coordinates, and I have had a number of goes at it, but cannot seem to figure it out.

$\underset{-1}{\overset{1}{\int }}\underset{0}{\overset{\sqrt{1-x^2}}{\int }}\underset{0}{\overset{\sqrt{1-x^2-y^2}}{\int }}dzdydx.$

of the function e^ -(x^2 + y^2 + z^2 ) ^ (3/2)

Now there seem to be a number of formulae about and I am not sure about the order of integration, but one I had been given was as follows, since  :

as x = ρcosθsinφ y = ρsinθsinφ and z = ρcosφ while dzdydx = ρ2sinφdρdφdθ then we would have

$\underset{-1}{\overset{1}{\int }}\underset{0}{\overset{\sqrt{1-{\rho }^2{\cos }^2\theta {\sin }^2\varphi }}{\int }}\underset{0}{\overset{\sqrt{1-{\rho }^2{\cos }^2\theta {\sin }^2\varphi -1-{\rho }^2{\sin }^2\theta {\sin }^2\varphi }}{\int }}{\rho }^2\sin \varphi d\rho d\varphi d\theta$

of the polar version of the exponential, which is e^ -(rho squared cos squared theta sin squared phi + rho squared sin squared phi sin squared theta + rho squared sin squared phi ) to the three over two.

But then if this is right, how do I proceed ? I have worked at it, and been able to simplify some of it, but then when I integrate I end up with what seem to be horrible integrals and I am not sure I have done it right. Thank You, Chris the Russian Christopher Lilly 08:20, 31 July 2011 (UTC)

The most obvious problem is you didn't change the limits of integration with the substitution. For example the outermost integral is still using the range for x, not the range for theta.--RDBury (talk) 19:33, 31 July 2011 (UTC)

---

The expression

$x^2+y^2+z^2$

should simply become

${\rho }^2.$

The bounds of integration need to be those for the appropriate corresponding variables. Thus $\rho$ goes from 0 to 1, and $\theta$ goes from 0 only to $\pi /2$ since you have only the upper hemisphere. And $\phi$, the "longitude", goes all the way around from 0 to $2\pi$. Hence you need

$\underset{0}{\overset{\pi /2}{\int }}\underset{0}{\overset{2\pi }{\int }}\underset{0}{\overset{1}{\int }}{e}^{-{\rho }^2}{\rho }^2\sin \varphi d\rho d\varphi d\theta .$

Since there's no $\theta$ in the function being integrated, this becomes

$\frac{\pi }{2}\underset{0}{\overset{2\pi }{\int }}\underset{0}{\overset{1}{\int }}{e}^{-{\rho }^2}{\rho }^2\sin \varphi d\rho d\varphi$

Then in the inner integral, $\sin \phi$ is a constant that pulls out:

$\frac{\pi }{2}\underset{0}{\overset{2\pi }{\int }}(\sin \phi \underset{0}{\overset{1}{\int }}{e}^{-{\rho }^2}{\rho }^{2}d\rho )d\varphi$

Finally, the now-inner integral is itself a constant factor that pulls out:

$\frac{\pi }{2}\left(\underset{0}{\overset{1}{\int }}{e}^{-{\rho }^2}{\rho }^{2}d\rho \right)(\underset{0}{\overset{2\pi }{\int }}\sin \phi d\varphi ).$

You largely missed the point of the transformation to spherical coordinates.

(Clearly something is wrong in the details above; I suspect I'm confusing the longitude and latitude. The answer should be a positive number.) Michael Hardy (talk) 19:40, 31 July 2011 (UTC)

Rather, the integral is over the interior of a quarter of a sphere of unit radius. In the spherical coordinate system, the limits of integration would be $\rho \in (0,1),\theta \in (0,\pi ),\phi \in (0,\pi /2)$. Note also the typo above: $x^2+y^2+z^2={\rho }^2$, so the function being integrated is $\exp \{-(x^2+y^2+z^2{)}^{3/2}\}=\exp \{-({\rho }^2{)}^{3/2}\}=\exp \{-{\rho }^3\}$. Nm420 (talk) 19:55, 31 July 2011 (UTC)

This is really excellent, and has helped enormously. Thank You all very much. Chris the Russian Christopher Lilly 01:05, 1 August 2011 (UTC)