Quantum Chemistry/Integration by parts

Example Problem:

Evaluate the triple integral shown below in spherical polar coordinates using the given boundary conditions: $0\leq {r}\leq 1,$ $0\leq {\theta }\leq \pi ,$ $0\leq \phi \leq 2\pi$

$\int ^{}\int ^{\ }\int ^{\ }r^{2}\sin \theta ~{\textrm {d}}r~{\textrm {d}}\theta ~{\textrm {d}}\phi$

Solution:

The triple integral can be rewritten as an iterated integral under the boundary conditions as follows:

$\int ^{}\int ^{\ }\int ^{\ }r^{2}\sin \theta ~{\textrm {d}}r~{\textrm {d}}\theta ~{\textrm {d}}\phi$ = $\int _{0}^{\pi }\int _{0}^{1}\int _{0}^{2\pi }r^{2}\sin \theta ~{\textrm {d}}r~{\textrm {d}}\theta ~{\textrm {d}}\phi$

The integral can then be evaluated using Fubini's theorem to separate the multivariable integrals into single variable integrals that can be solved accordingly as follows:

$\int _{0}^{\pi }\int _{0}^{1}\int _{0}^{2\pi }r^{2}\sin \theta ~{\textrm {d}}r~{\textrm {d}}\theta ~{\textrm {d}}\phi$ = $\int _{0}^{\pi }\sin \theta ~{\textrm {d}}\theta ~$ $\int _{0}^{1}\ r^{2}{\textrm {d}}r$ $\int _{0}^{2\pi }{\textrm {d}}\phi$

= $(-cos\theta \mid _{0}^{\pi })$ $\left({\frac {r^{3}}{3}}\mid _{0}^{1}\right)$ $(\phi \mid _{0}^{2\pi })$ = $(2)$ $\left({\frac {1}{3}}\right)$ $(2\pi )$

Multiplying these results together then gives the final solution:

(2) $\left({\frac {1}{3}}\right)$ $(2\pi )$ = $\left({\frac {4\pi }{3}}\right)$

Therefore,

$\int _{0}^{\pi }\int _{0}^{1}\int _{0}^{2\pi }r^{2}\sin \theta ~{\textrm {d}}r~{\textrm {d}}\theta ~{\textrm {d}}\phi$ = $\left({\frac {4\pi }{3}}\right)$