Q 48 - Calculus

Question

Evaluate each of the double integrals in Exercises $37-54$ as iterated integrals.

$\int {\int }_R\sin \left(2x+y\right)dA,$

where $R=\left\{\left(x,y\right)|0\le x\le \frac{\mathrm{\pi }}{2} and 0\le y\le \frac{\mathrm{\pi }}{2}\right\}$

Step-by-Step Solution

Answer

The value of double integral is :-

$\int {\int }_R\sin \left(2x+y\right)dA=1$

 where,$R=\left\{\left(x,y\right)|0\le x\le \frac{\mathrm{\pi }}{2} and 0\le y\le \frac{\mathrm{\pi }}{2}\right\}$

Step 1. Given Information

We have given the following double integral :-

$\int {\int }_R\sin \left(2x+y\right)dA,$

 where$R=\left\{\left(x,y\right)|0\le x\le \frac{\mathrm{\pi }}{2} and 0\le y\le \frac{\mathrm{\pi }}{2}\right\}$.

We have to evaluate this  double integral.

Step 2. Use iterated integrals

The given double integral is :-

$\int {\int }_R\sin \left(2x+y\right)dA,$

where $R=\left\{\left(x,y\right)|0\le x\le \frac{\mathrm{\pi }}{2} and 0\le y\le \frac{\mathrm{\pi }}{2}\right\}$

We know that :-

$\sin \left(A+B\right)=\sin A\cos B+\cos A\sin B$.

Then we have :-

$\int {\int }_R\sin \left(2x+y\right)dA=\int {\int }_R\left(\sin 2x\cos y+\cos 2x\sin y\right)dA$

Then we can separate this into two double integrals as following :-

$\int {\int }_R\left(\sin 2x\cos y+\cos 2x\sin y\right)dA=\int {\int }_R\sin 2x\cos ydA+\int {\int }_R\cos 2x\sin ydA$

where $R=\left\{\left(x,y\right)|0\le x\le \frac{\mathrm{\pi }}{2} and 0\le y\le \frac{\mathrm{\pi }}{2}\right\}$

Then by using Fubini's Theorem, we can writ this double integral as following :-

$\int {\int }_R\sin 2x\cos ydA+\int {\int }_R\cos 2x\sin ydA={\int }_0^{\mathrm{\pi }/2}{\int }_0^{\mathrm{\pi }/2}\sin 2x\cos ydxdy+{\int }_0^{\mathrm{\pi }/2}{{\int }_0^{\mathrm{\pi }/2}}_R\cos 2x\sin ydxdy$

Then by using iterated integrals, we have :-

${\int }_0^{\mathrm{\pi }/2}{\int }_0^{\mathrm{\pi }/2}\sin 2x\cos ydxdy+{\int }_0^{\mathrm{\pi }/2}{{\int }_0^{\mathrm{\pi }/2}}_R\cos 2x\sin ydxdy={\int }_0^{\mathrm{\pi }/2}\left({\int }_0^{\mathrm{\pi }/2}\sin 2x\cos ydx\right)dy+{\int }_0^{\mathrm{\pi }/2}\left({{\int }_0^{\mathrm{\pi }/2}}_R\cos 2x\sin ydx\right)dy$

Now we can solve this integral as following :-

$$\begin{gathered} {\int }_0^{\mathrm{\pi }/2}\left({\int }_0^{\mathrm{\pi }/2}\sin 2x\cos ydx\right)dy+{\int }_0^{\mathrm{\pi }/2}\left({{\int }_0^{\mathrm{\pi }/2}}_R\cos 2x\sin ydx\right)dy \\ ={\int }_0^{\mathrm{\pi }/2}{{\left[\frac{-\cos y\cos 2x}{2}\right]}^{\mathrm{\pi }/2}}_{0}dy+{\int }_0^{\mathrm{\pi }/2}{{\left[\frac{\sin y\sin 2x}{2}\right]}^{\mathrm{\pi }/2}}_{0}dy \\ ={\int }_0^{\mathrm{\pi }/2}\left[\frac{-\cos y\cos 2{\displaystyle \frac{\mathrm{\pi }}{2}}}{2}-\frac{-\cos y\cos 0}{2}\right]dy+{\int }_0^{\mathrm{\pi }/2}\left[\frac{\sin y\sin 2{\displaystyle \frac{\mathrm{\pi }}{2}}}{2}-\frac{\sin y\sin 0}{2}\right]dy \\ ={\int }_0^{\mathrm{\pi }/2}\left[\frac{-\cos y(-1)}{2}+\frac{\cos y\left(1\right)}{2}\right]dy+{\int }_0^{\mathrm{\pi }/2}\left[0-0\right]dy \\ ={\int }_0^{\mathrm{\pi }/2}\left[\frac{\cos y}{2}+\frac{\cos y}{2}\right]dy \\ ={\int }_0^{\mathrm{\pi }/2}\cos ydy \\ ={{\left[\sin y\right]}^{\mathrm{\pi }/2}}_0 \\ =\left[\sin \frac{\mathrm{\pi }}{2}-\sin 0\right] \\ =1-0 \\ =1 \end{gathered}$$