MHB 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

[math]\int_{- \infty }^{ \infty }\int_{- \infty }^{ \infty }\int_{- \infty }^{ \infty } \sqrt{x^2+y^2+z^2} e^{-(x^2+y^2+z^2)} dxdydz[/math]

using polar coordinates,[math] x=r\sin \theta \cos \phi [/math]
[math] y=r\sin \theta \sin \phi[/math]
[math] z=r\cos\theta [/math]since the integration occurs for all the values in 3d space, [math] 0<r<\infty[/math]
[math]0<\theta<\pi[/math]
[math]0<\phi<2\pi[/math]and due to variable change,[math]dxdydz=r^2\sin \theta drd\theta d\phi[/math]we know that,[math]x^2+y^2+z^2=r^2[/math]because of that,[math]\int_{0}^{ \infty }\int_{0}^{\pi}\int_{0}^{2\pi } r e^{-r^2} r^2 \sin \theta drd\theta d\phi[/math][math]=\int_{0}^{ \infty }\int_{0}^{\pi} r^3e^{-r^2}\sin \theta \left[\phi\right]^{2\pi}_{0} drd\theta[/math][math]=\int_{0}^{ \infty }\int_{0}^{\pi} r^3 e^{-r^2} \sin \theta (2\pi) drd\theta[/math][math]=(2\pi)\int_{0}^{ \infty } r^3 e^{-r^2} \left[-\cos \theta\right]^{\pi}_{0} dr[/math][math]=(2\pi)\int_{0}^{ \infty } r^3 e^{-r^2} [2] dr[/math][math]=(4\pi)\int_{0}^{ \infty } r^3 e^{-r^2} dr[/math]now use the substitution,[math]t=r^2[/math]

[math]dt=2r dr[/math]
[math]0<t<\infty[/math]then,[math](2\pi)\int_{0}^{ \infty }t e^{-t}dt[/math][math]=2\pi\left( \left[-te^{-t}\right]_{0}^{\infty}+\int_{0}^{\infty}e^{-t}dt\right)[/math]by l'hopitals rule to [math]e^{-t}[/math] we can get that, [math]te^{-t}=0[/math] when t goes to infinitythen,[math]=2\pi\left( \left[0-0\right]+\int_{0}^{\infty}e^{-t}dt\right)[/math][math]=2\pi\left( \int_{0}^{\infty}e^{-t}dt\right)[/math][math]=2\pi\left(-e^{-t}\right)_{0}^{\infty}[/math][math]=2\pi\left(0-(-1)\right)[/math][math]=2\pi[/math]because of this,[math]\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[/math]