91Ó°ÊÓ

We have given the following double integral :-

$∫{∫}_{R}y\sin xdA,$

where $R=\left\{\left(x,y\right)|0≤x≤\frac{\mathrm{π}}{2}and0≤y≤1\right\}$

We have to evaluate this double integral

The given double integral is :-

$∫{∫}_{R}y\sin xdA,$

where $R=\left\{\left(x,y\right)|0≤x≤\frac{\mathrm{π}}{2}and0≤y≤1\right\}$.

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

$∫{∫}_{R}y\sin xdA={∫}_0^1{∫}_0^{\frac{\mathrm{π}}{2}}y\sin xdxdy$.

Then by using iterated integrals we have :-

${∫}_0^1{∫}_0^{\frac{\mathrm{π}}{2}}y\sin xdxdy={∫}_0^1\left({∫}_0^{\frac{\mathrm{π}}{2}}y\sin xdx\right)dy$

Now we can solve this integral as following :-

$$\begin{gathered} {∫}_0^1\left({∫}_0^{\frac{\mathrm{π}}{2}}y\sin xdx\right)dy={∫}_0^1\left(y{∫}_0^{\frac{\mathrm{π}}{2}}\sin xdx\right)dy \\ ⇒{∫}_0^1\left({∫}_0^{\frac{\mathrm{π}}{2}}y\sin xdx\right)dy={∫}_0^1\left(y{{\left[-\cos x\right]}^{\raisebox{1ex}{$\mathrm{π}$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}_0\right)dy \\ ⇒{∫}_0^1\left({∫}_0^{\frac{\mathrm{π}}{2}}y\sin xdx\right)dy={∫}_0^{1}y\left[-\cos \frac{\mathrm{π}}{2}-\left(-\cos 0\right)\right]dy \\ ⇒{∫}_0^1\left({∫}_0^{\frac{\mathrm{π}}{2}}y\sin xdx\right)dy={∫}_0^{1}y\left(0+1\right)dy \\ ⇒{∫}_0^1\left({∫}_0^{\frac{\mathrm{π}}{2}}y\sin xdx\right)dy={∫}_0^{1}ydy \\ ⇒{∫}_0^1\left({∫}_0^{\frac{\mathrm{π}}{2}}y\sin xdx\right)dy={{\left[\frac{y^2}{2}\right]}^1}_0 \\ ⇒{∫}_0^1\left({∫}_0^{\frac{\mathrm{π}}{2}}y\sin xdx\right)dy=\left[\frac{1}{2}-0\right] \\ ⇒{∫}_0^1\left({∫}_0^{\frac{\mathrm{π}}{2}}y\sin xdx\right)dy=\frac{1}{2} \end{gathered}$$