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

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#### Chris L T521

##### Well-known member
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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.

Use spherical coordinates.

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#### Chris L T521

##### Well-known member
Staff member

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

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,

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

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