91Ó°ÊÓ

We have given the following double integral :-

$∫{∫}_{R}xy\sin \left(x^2\right)dA,$

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

We have to evaluate this double integral.

The given double integral is :-

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

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

role="math" localid="1650412181225" $∫{∫}_{R}xy\sin \left(x^2\right)dA={∫}_0^1{∫}_0^{\sqrt{\mathrm{π}}}xy\sin \left(x^2\right)dxdy$

Then by using iterated integrals, we have :-

role="math" localid="1650412215486" ${∫}_0^1{∫}_0^{\sqrt{\mathrm{π}}}xy\sin \left(x^2\right)dxdy={∫}_0^1\left({∫}_0^{\sqrt{\mathrm{π}}}xy\sin \left(x^2\right)dx\right)dy$

Now we can solve this integral as following :-

${∫}_0^1\left({∫}_0^{\sqrt{\mathrm{π}}}xy\sin \left(x^2\right)dx\right)dy$

To solve put $x^2=t$,

Differentiate both sides, then we have :-

$$\begin{gathered} 2xdx=dt \\ ⇒xdx=\frac{dt}{2} \end{gathered}$$

Also when $x=0$, $t=0^2=0$and when $x=\sqrt{\mathrm{Ï€}}$, $t={\left(\sqrt{\mathrm{Ï€}}\right)}^2=\mathrm{Ï€}$

Put these vales in the our integral, then we have :-

$$\begin{gathered} {∫}_0^1\left({∫}_0^{\sqrt{\mathrm{Ï€}}}xy\sin \left(x^2\right)dx\right)dy \\ ={∫}_0^1\left({∫}_0^{\mathrm{Ï€}}\frac{1}{2}y\sin \left(t\right)dt\right)dy \\ ={∫}_0^1\frac{1}{2}y{{\left[-\cos t\right]}^{\mathrm{Ï€}}}_{0}dy \\ ={∫}_0^1\frac{1}{2}y\left[-\mathrm{³¦´Ç²õÏ€}-\left(-\cos 0\right)\right]dy \\ ={∫}_0^1\frac{1}{2}y\left[-(-1)+1\right]dy \\ ={∫}_0^1\frac{1}{2}y×2dy \\ ={∫}_0^{1}ydy \\ ={{\left[\frac{y^2}{2}\right]}^1}_0 \\ =\left[\frac{1}{2}-0\right] \\ =\frac{1}{2} \end{gathered}$$