Question

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

role="math" localid="1650327788023" $\int {\int }_{R}y\sin xdA,$

whererole="math" localid="1650327080219" $R=\left\{\left(x,y\right)|0\le x\le \frac{\mathrm{\pi }}{2}and0\le y\le 1\right\}$

Short answer

The value of double integral is :-

$\int {\int }_{R}y\sin xdA=\frac{1}{2}$,

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

Step-by-step solution

Step 1. Given Information

We have given the following double integral :-

$\int {\int }_{R}y\sin xdA,$

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

We have to evaluate this double integral

Step 2. Use iterated integrals

The given double integral is :-

$\int {\int }_{R}y\sin xdA,$

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

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

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

Then by using iterated integrals we have :-

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

Now we can solve this integral as following :-

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