91Ó°ÊÓ

We have given the following double integral :-

$∫{∫}_R\sin \left(2x+y\right)dA,$

where$R=\left\{\left(x,y\right)|0≤x≤\frac{\mathrm{π}}{2}and0≤y≤\frac{\mathrm{π}}{2}\right\}$.

We have to evaluate this double integral.

The given double integral is :-

$∫{∫}_R\sin \left(2x+y\right)dA,$

where$R=\left\{\left(x,y\right)|0≤x≤\frac{\mathrm{π}}{2}and0≤y≤\frac{\mathrm{π}}{2}\right\}$

We know that :-

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

Then we have :-

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

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

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

where $R=\left\{\left(x,y\right)|0≤x≤\frac{\mathrm{π}}{2}and0≤y≤\frac{\mathrm{π}}{2}\right\}$

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

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

Then by using iterated integrals, we have :-

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

Now we can solve this integral as following :-

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