Quantum Chemistry/Integration by parts

Example Problem:

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

${\int }^{}{\int }^{}{\int }^{}{r}^2\sin \theta \text{d}r\text{d}\theta \text{d}\varphi$

Solution:

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

${\int }^{}{\int }^{}{\int }^{}{r}^2\sin \theta \text{d}r\text{d}\theta \text{d}\varphi$ = $\underset{0}{\overset{\pi }{\int }}\underset{0}{\overset{1}{\int }}\underset{0}{\overset{2\pi }{\int }}{r}^2\sin \theta \text{d}r\text{d}\theta \text{d}\varphi$

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:

$\underset{0}{\overset{\pi }{\int }}\underset{0}{\overset{1}{\int }}\underset{0}{\overset{2\pi }{\int }}{r}^2\sin \theta \text{d}r\text{d}\theta \text{d}\varphi$ = $\underset{0}{\overset{\pi }{\int }}\sin \theta \text{d}\theta$ $\underset{0}{\overset{1}{\int }}{r}^2\text{d}r$ $\underset{0}{\overset{2\pi }{\int }}\text{d}\varphi$

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

Multiplying these results together then gives the final solution:

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

Therefore,

$\underset{0}{\overset{\pi }{\int }}\underset{0}{\overset{1}{\int }}\underset{0}{\overset{2\pi }{\int }}{r}^2\sin \theta \text{d}r\text{d}\theta \text{d}\varphi$ = $\left(\frac{4\pi }{3}\right)$

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