Q 51 - Calculus

Question

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

$\int {\int }_{R}y\cos \left(xy\right)dA,$

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

Step-by-Step Solution

Answer

The value of double integral is :-

$\int {\int }_{R}y\cos \left(xy\right)dA=\frac{2}{\mathrm{\pi }}$

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

Step 1. Given Information

We have given the following double integral :-

$\int {\int }_{R}y\cos \left(xy\right)dA,$

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

We have to evaluate this double integral.

Step 2. Use iterated integrals

The given double integral is :-

$\int {\int }_{R}y\cos \left(xy\right)dA,$

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

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

$\int {\int }_{R}y\cos \left(xy\right)dA={\int }_0^1{\int }_0^{\frac{\mathrm{\pi }}{2}}y\cos \left(xy\right)dxdy$

Then by using iterated integrals, we have :-

${\int }_0^1{\int }_0^{\frac{\mathrm{\pi }}{2}}y\cos \left(xy\right)dxdy={\int }_0^1\left({\int }_0^{\frac{\mathrm{\pi }}{2}}y\cos \left(xy\right)dx\right)dy$

Now we can solve this integral as following :-

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