Question

Evaluate the iterated integral :

${\int }_0^1 {\int }_0^3 {\int }_0^{{\sin }^{-1}x} (x+y)dzdydx$

Short answer

$\frac{21\pi }{8}-\frac{9}{2}$

Step-by-step solution

Step 1. Given information.

Given integral is :

${\int }_0^1 {\int }_0^3 {\int }_0^{{\sin }^{-1}x} (x+y)dzdydx$

We have to evaluate it.

Step 2. Evaluate.

${\int }_0^1 {\int }_0^3 {\int }_0^{{\sin }^{-1}x} (x+y)dzdydx$

$\begin{array}{r}={\int }_0^1 {\int }_0^3 \left[{\int }_0^{{\sin }^{-1}x} (x+y)dz\right]dydx\\ ={\int }_0^1 {\int }_0^3 \left[(x+y){\int }_0^{{\sin }^{-1}x} dz\right]dydx\\ ={\int }_0^1 {\int }_0^3 \left[(x+y)[z{]}_0^{{\sin }^{-1}x}\right]dydx\\ ={\int }_0^1 {\int }_0^3 \left[(x+y)\left[{\sin }^{-1}x\right]\right]dydx\end{array}$

Step 3. Integrate with respect to y.

Integrate with respect to y

$\begin{array}{r}={\int }_0^1 {\int }_0^3 \left[(x+y)\left[{\sin }^{-1}x\right]\right]dydx\\ ={\int }_0^1 \left[x{\sin }^{-1}x{\int }_0^3 dy+{\sin }^{-1}x{\int }_0^3 ydy\right]dx\\ ={\int }_0^1 \left[x{\sin }^{-1}x[y{]}_0^3+{\sin }^{-1}x{\left[\frac{y^2}{2}\right]}_0^3\right]dx\\ ={\int }_0^1 \left[x{\sin }^{-1}x\left[3\right]+{\sin }^{-1}x\left[\frac{3^2}{2}\right]\right]dx\\ =3{\int }_0^1 x{\sin }^{-1}xdx+\frac{9}{2}{\int }_0^1 {\sin }^{-1}xdx\end{array}$

Use integration by parts :

$\begin{array}{r}=3\left[\frac{\pi }{8}\right]+\frac{9}{2}\left[\frac{\pi }{2}-1\right]\\ =\left[\frac{3\pi }{8}+\frac{9\pi }{4}-\frac{9}{2}\right]\\ =\frac{21\pi }{8}-\frac{9}{2}\end{array}$