Wikipedia:Reference desk/Archives/Mathematics/2011 July 31

= 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.


 * $$\int_{-1}^1\int_{0}^{\sqrt{1-x^2}}\int_{0}^{\sqrt{1-x^2-y^2}}\,dz\,dy\,dx.$$

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


 * $$\int_{-1}^1\int_{0}^{\sqrt{1-\rho^2 \cos^2 \theta \sin^2 \phi}}\int_{0}^{\sqrt{1-\rho^2 \cos^2 \theta \sin^2 \phi-1-\rho^2 \sin^2 \theta \sin^2 \phi}}\,\rho^2 \sin \phi\, d\rho\,d\phi\,d\theta$$

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 $$\varphi$$, the "longitude", goes all the way around from 0 to $$2\pi$$.  Hence you need
 * $$ \int_0^{\pi/2} \int_0^{2\pi} \int_0^1 e^{-\rho^2} \rho^2 \sin \phi\, d\rho\,d\phi\,d\theta. $$
 * Since there's no $$\theta$$ in the function being integrated, this becomes
 * $$ \frac\pi2 \int_0^{2\pi} \int_0^1 e^{-\rho^2} \rho^2 \sin \phi\, d\rho\,d\phi $$
 * Then in the inner integral, $$ \sin\varphi$$ is a constant that pulls out:
 * $$ \frac\pi2 \int_0^{2\pi} \left(\sin\varphi \int_0^1 e^{-\rho^2} \rho^2 \, d\rho\right) \, d\phi $$
 * Finally, the now-inner integral is itself a constant factor that pulls out:
 * $$ \frac\pi2 \left(\int_0^1 e^{-\rho^2} \rho^2 \, d\rho\right)\left( \int_0^{2\pi} \sin\varphi \, d\phi\right). $$
 * 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 $$\rho\in(0,1), \theta\in(0,\pi), \varphi\in(0,\pi/2)$$. 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)