Question

Use spherical coordinates to show that $$\int_{-\infty}^{\infty} \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} \sqrt{x^{2}+y^{2}+z^{2}} e^{-\left(x^{2}+y^{2}+z^{2}\right)} d x d y d z=2 \pi$$

Short answer

After converting the exercise's integral from Cartesian into spherical coordinates, and performing the integral calculation, it's distinguished that it equates to \(2 \pi\).

Step-by-step solution

Transform to Spherical Coordinates

In spherical coordinates, \(x = \rho\sin\phi\cos\theta\), \(y = \rho\sin\phi\sin\theta\), and \(z = \rho\cos\phi\). The Jacobian determinant \(J = \rho^{2}\sin\phi\). Here, \(\sqrt{x^{2}+y^{2}+z^{2}}\) is \(\rho\), and \(e^{-\left(x^{2}+y^{2}+z^{2}\right)}\) is \(e^{-\rho^{2}}\).

Set Up the Integral

The integral becomes \(\int \sqrt{\rho^{2}} e^{-\rho^{2}} \rho^{2} \sin\phi \, d\rho\, d\phi\, d\theta\). This simplifies to \(\int_{0}^{2\pi} \int_{0}^{\pi} \int_{0}^{\infty} \rho^{3} e^{-\rho^{2}} \sin\phi \, d\rho\, d\phi\, d\theta\).

Calculation of the Integral

The result of the integral \(\int_{0}^{2\pi} d\theta\) is \(2\pi\). You can take this out of the integral, leaving \(\int_{0}^{\pi} \int_{0}^{\infty} \rho^{3} e^{-\rho^{2}} \sin\phi \, d\rho\, d\phi\). The result of the integral \(\int_{0}^{\pi} \sin\phi \, d\phi\) is \(2\), this can also be taken out of the integral, now leaving \(\int_{0}^{\infty} \rho^{3} e^{-\rho^{2}} \, d\rho\). Now perform the integral \(\int_{0}^{\infty} \rho^{3} e^{-\rho^{2}} \, d\rho\) using the substitution \(\rho^{2} = u\), which gives \(2\rho \, d\rho = du\), the integral now becomes \(\int_{0}^{\infty} \frac{1}{2}u e^{-u} \, du\). The result is \(\frac{1}{2}\). Multiply this by \(2\pi\) and \(2\), leading to \(2\pi\).

Summary

This proves that the triple integral of \(\sqrt{x^{2}+y^{2}+z^{2}} e^{-\left(x^{2}+y^{2}+z^{2}\right)}\) over all space in Cartesian coordinates does indeed yield \(2 \pi\), when converted to spherical coordinates.