Q 50 - Calculus

Question

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

$\int {\int }_{R}xy\sin \left(x^2\right)dA,$

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

Step-by-Step Solution

Answer

The value of double integral is :-

$\int {\int }_{R}xy\sin \left(x^2\right)dA=\frac{1}{2}$

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

Step 1. Given Information

We have given the following double integral :-

$\int {\int }_{R}xy\sin \left(x^2\right)dA,$

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

We have to evaluate this double integral.

Step 2. Use iterated integrals

The given double integral is :-

$\int {\int }_{R}xy\sin \left(x^2\right)dA,$

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

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

$\int {\int }_{R}xy\sin \left(x^2\right)dA={\int }_0^1{\int }_0^{\sqrt{\mathrm{\pi }}}xy\sin \left(x^2\right)dxdy$

Then by using iterated integrals, we have :-

${\int }_0^1{\int }_0^{\sqrt{\mathrm{\pi }}}xy\sin \left(x^2\right)dxdy={\int }_0^1\left({\int }_0^{\sqrt{\mathrm{\pi }}}xy\sin \left(x^2\right)dx\right)dy$

Now we can solve this integral as following :-

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

To solve put $x^2=t$,

Differentiate both sides, then we have :-

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

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

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

$$\begin{gathered} {\int }_0^1\left({\int }_0^{\sqrt{\mathrm{\pi }}}xy\sin \left(x^2\right)dx\right)dy \\ ={\int }_0^1\left({\int }_0^{\mathrm{\pi }}\frac{1}{2}y\sin \left(t\right)dt\right)dy \\ ={\int }_0^1\frac{1}{2}y{{\left[-\cos t\right]}^{\mathrm{\pi }}}_{0}dy \\ ={\int }_0^1\frac{1}{2}y\left[-\mathrm{cos\pi }-\left(-\cos 0\right)\right]dy \\ ={\int }_0^1\frac{1}{2}y\left[-(-1)+1\right]dy \\ ={\int }_0^1\frac{1}{2}y\times 2dy \\ ={\int }_0^{1}ydy \\ ={{\left[\frac{y^2}{2}\right]}^1}_0 \\ =\left[\frac{1}{2}-0\right] \\ =\frac{1}{2} \end{gathered}$$