Welcome to our community

Be a part of something great, join today!

Problem of the Week #14 - July 2nd, 2012

Status
Not open for further replies.
  • Thread starter
  • Moderator
  • #1

Chris L T521

Well-known member
Staff member
Jan 26, 2012
995
Thanks to those who participated in last week's POTW!! Here's this week's problem.

-----

Problem: Show that

\[\int_{-\infty}^{\infty} \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} \sqrt{x^2+y^2+z^2}e^{-(x^2+y^2+z^2)}\,dx\,dy\,dz = 2\pi\]

(Note that the improper triple integral is defined as the limit of a triple integral over a solid sphere as the radius of the sphere increases indefinitely.)

-----

Here's a hint for this week's problem.

Use spherical coordinates.

Remember to read the POTW submission guidelines to find out how to submit your answers!
 
Last edited:
  • Thread starter
  • Moderator
  • #2

Chris L T521

Well-known member
Staff member
Jan 26, 2012
995
This week's problem was correctly answered by Sudharaka and BAdhi. You can find BAdhi's solution below.

\(\displaystyle \int_{- \infty }^{ \infty }\int_{- \infty }^{ \infty }\int_{- \infty }^{ \infty } \sqrt{x^2+y^2+z^2} e^{-(x^2+y^2+z^2)} dxdydz\)

using polar coordinates,


\(\displaystyle x=r\sin \theta \cos \phi \)
\(\displaystyle y=r\sin \theta \sin \phi\)
\(\displaystyle z=r\cos\theta \)


since the integration occurs for all the values in 3d space,


\(\displaystyle 0<r<\infty\)
\(\displaystyle 0<\theta<\pi\)
\(\displaystyle 0<\phi<2\pi\)


and due to variable change,


\(\displaystyle dxdydz=r^2\sin \theta drd\theta d\phi\)


we know that,


\(\displaystyle x^2+y^2+z^2=r^2\)


because of that,


\(\displaystyle \int_{0}^{ \infty }\int_{0}^{\pi}\int_{0}^{2\pi } r e^{-r^2} r^2 \sin \theta drd\theta d\phi\)


\(\displaystyle =\int_{0}^{ \infty }\int_{0}^{\pi} r^3e^{-r^2}\sin \theta \left[\phi\right]^{2\pi}_{0} drd\theta\)


\(\displaystyle =\int_{0}^{ \infty }\int_{0}^{\pi} r^3 e^{-r^2} \sin \theta (2\pi) drd\theta\)


\(\displaystyle =(2\pi)\int_{0}^{ \infty } r^3 e^{-r^2} \left[-\cos \theta\right]^{\pi}_{0} dr\)


\(\displaystyle =(2\pi)\int_{0}^{ \infty } r^3 e^{-r^2} [2] dr\)


\(\displaystyle =(4\pi)\int_{0}^{ \infty } r^3 e^{-r^2} dr\)


now use the substitution,\(\displaystyle t=r^2\)




\(\displaystyle dt=2r dr\)
\(\displaystyle 0<t<\infty\)


then,


\(\displaystyle (2\pi)\int_{0}^{ \infty }t e^{-t}dt\)


\(\displaystyle =2\pi\left( \left[-te^{-t}\right]_{0}^{\infty}+\int_{0}^{\infty}e^{-t}dt\right)\)


by l'hopitals rule to \(\displaystyle e^{-t}\) we can get that, \(\displaystyle te^{-t}=0\) when t goes to infinity


then,


\(\displaystyle =2\pi\left( \left[0-0\right]+\int_{0}^{\infty}e^{-t}dt\right)\)


\(\displaystyle =2\pi\left( \int_{0}^{\infty}e^{-t}dt\right)\)


\(\displaystyle =2\pi\left(-e^{-t}\right)_{0}^{\infty}\)


\(\displaystyle =2\pi\left(0-(-1)\right)\)


\(\displaystyle =2\pi\)


because of this,


\(\displaystyle \int_{- \infty }^{ \infty }\int_{- \infty }^{ \infty }\int_{- \infty }^{ \infty } \sqrt{x^2+y^2+z^2} e^{-(x^2+y^2+z^2)} dxdydz=2\pi\)
 
Status
Not open for further replies.