Q. 29 - Calculus

Question

Evaluate the iterated integral :

${\int }_0^{\pi /2} {\int }_0^{\cos y} {\int }_0^{\sin y} (2x+y)dzdxdy$

Step-by-Step Solution

Answer

$\frac{\pi }{8}+\frac{1}{3}$

Step 1. Given information.

Given integral is :

${\int }_0^{\pi /2} {\int }_0^{\cos y} {\int }_0^{\sin y} (2x+y)dzdxdy$

We have to evaluate it.

Step 2. Evaluate.

Given 

${\int }_0^{\pi /2} {\int }_0^{\cos y} {\int }_0^{\sin y} (2x+y)dzdxdy$

$\begin{array}{r}={\int }_0^{\frac{\pi }{2}} {\int }_0^{\cos y} \left[{\int }_0^{\sin y} (2x+y)dz\right]dxdy\\ ={\int }_0^{\frac{\pi }{2}} {\int }_0^{\cos y} \left[2x{\int }_0^{\sin y} dz+y{\int }_0^{\sin y} dz\right]dxdy \\ ={\int }_0^{\frac{\pi }{2}} {\int }_0^{\cos y} \left[2x(z{)}_0^{\sin y}+y(z{)}_0^{\sin y}\right]dxdy \\ ={\int }_0^{\frac{\pi }{2}} {\int }_0^{\cos y} [2x\sin y+y\sin y]dxdy \end{array}$

Step 3. Integrate with respect to x.

Integrate with respect to x

$\begin{array}{r}={\int }_0^{\frac{\pi }{2}} \left[{\int }_0^{\cos y} 2x\sin ydx+{\int }_0^{\cos y} y\sin ydx\right]dy\\ ={\int }_0^{\frac{\pi }{2}} \left[2\sin y{\int }_0^{\cos y} xdx+y\sin y{\int }_0^{\cos y} dx\right]dy\\ ={\int }_0^{\frac{\pi }{2}} \left[2\sin y{\left(\frac{x^2}{2}\right)}_0^{\cos y}+y\sin y(x{)}_0^{\cos y}\right]dy \\ ={\int }_0^{\frac{\pi }{2}} \left[2\sin y\left(\frac{{\cos }^{2}y}{2}\right)+y\sin y(\cos y)\right]dy \end{array}$

Integrate with respect to y

$\begin{array}{r}={\int }_0^{\frac{\pi }{2}} \left(\sin y{\cos }^{2}y+y\sin y\cos y\right)dy \\ ={\int }_0^{\frac{\pi }{2}} \sin y{\cos }^{2}y+\frac{1}{2}{\int }_0^{\frac{\pi }{2}} y\sin 2ydy\\ ={\left(\frac{-{\cos }^{3}y}{3}\right)}_0^{\frac{\pi }{2}}+\frac{1}{2}\left[{\left(\frac{-y\cos 2y}{2}+\frac{\sin 2y}{4}\right)}_0^{\frac{\pi }{2}}\right]\\ =\frac{-1}{3}(0-1)+\frac{1}{2}\left[\frac{-1}{2}\left(\frac{\pi }{2}\cos \pi -0\right)+\frac{1}{4}(\sin \pi -\cos 0)\right] \\ =\frac{1}{3}+\frac{1}{2}\left[\frac{\pi }{4}\right]\\ =\frac{\pi }{8}+\frac{1}{3} \end{array}$