Can You Solve This Week's 3D Improper Integral Challenge?

[Chris L T521]

Thanks to those who participated in last week's POTW! Here's this week's problem.

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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.)

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

Spoiler

Use spherical coordinates.

 

[Chris L T521]

This week's problem was correctly answered by Sudharaka and BAdhi. You can find BAdhi's solution below.

Spoiler

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

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 infinitythen,\(\displaystyle =2\pi\left( \left[0-0\right]+\int_{0}^{\infty}e^{-t}dt\right)\)\(\displaystyle =2\pi\left( \int_{0}^{\infty}e^{-t}dt\right)\)\(\displaystyle =2\pi\left(-e^{-t}\right)_{0}^{\infty}\)\(\displaystyle =2\pi\left(0-(-1)\right)\)\(\displaystyle =2\pi\)because of this,\(\displaystyle \int_{- \infty }^{ \infty }\int_{- \infty }^{ \infty }\int_{- \infty }^{ \infty } \sqrt{x^2+y^2+z^2} e^{-(x^2+y^2+z^2)} dxdydz=2\pi\)