Question

Show that

\(\int_{ - \infty }^\infty {\int_{ - \infty }^\infty {\int_{ - \infty }^\infty {\sqrt {{x^2} + {y^2} + {z^2}} } } } {e^{ - \left( {{x^2} + {y^2} + {z^2}} \right)}}dxdydz = 2\pi \)(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.)

Short answer

The integral is found to be \(2\pi \)

Step-by-step solution

Definethe spherical coordinate

Three coordinates in three-dimensional space that define the location of a point The length of its radius vector r, the angle between the vertical plane containing this vector and the x-axis, and the angle between this vector and the horizontal x–y plane are the three parameters.

Find the Integral

Our region of integration is all of three-dimensional space, so if we integrate over this region in spherical coordinates, we will let \(0 \le \theta \le 2\pi ,\)

\(0 \le \phi \le \pi ,\)and \(0 \le \rho \le \infty .\)

We know that\(\frac{{\partial (x,y,z)}}{{\partial (\rho ,\phi ,\theta )}} = {\rho ^2}\sin (\phi )\)

Also, the integrand above becomes \(\rho {e^{ - {\rho ^2}}}\)in spherical.

Therefore, the integral is

\(\begin{array}{l}\int_0^{2\pi } {\int_0^\pi {\int_0^\infty {{\rho ^3}} } } \sin (\phi ){e^{ - {\rho ^2}}}\;{\rm{d}}\rho {\rm{d}}\phi {\rm{d}}\theta \\ = \left. {\int_0^{2\pi } {\int_0^\pi {\sin } } (\phi )\frac{{ - \left( {{\rho ^2} + 1} \right){e^{ - {\rho ^2}}}}}{2}} \right|_0^\infty {\rm{d}}\phi {\rm{d}}\theta \end{array}\)

\(\begin{array}{l} = \int_0^{2\pi } {\int_0^\pi {\frac{{\sin (\phi )}}{2}} } \;{\rm{d}}\phi {\rm{d}}\theta \\ = \int_0^{2\pi } 1 \;{\rm{d}}\theta \\ = 2\pi \end{array}\)

Thus, the integral is \(2\pi \)