Q 49 - Calculus

Question

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

$\int {\int }_R{y}^2\sin xdA,$

where $R=\left\{\left(x,y\right)|0\le x\le \mathrm{\pi } \mathrm{and} 0\le \mathrm{y}\le 3\right\}$

Step-by-Step Solution

Answer

The value of double integral is :-

$\int {\int }_R{y}^2\sin xdA=18$

 where $R=\left\{\left(x,y\right)|0\le x\le \mathrm{\pi } \mathrm{and} 0\le \mathrm{y}\le 3\right\}$

Step 1. Given Information

We have given the following double integral :-

$\int {\int }_R{y}^2\sin xdA,$

 where $R=\left\{\left(x,y\right)|0\le x\le \mathrm{\pi } \mathrm{and} 0\le \mathrm{y}\le 3\right\}$

We have to evaluate this double integral. 

Step 2. Use iterated integrals

The given double integral is :-

$\int {\int }_R{y}^2\sin xdA,$

where $R=\left\{\left(x,y\right)|0\le x\le \mathrm{\pi } \mathrm{and} 0\le \mathrm{y}\le 3\right\}$

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

$\int {\int }_R{y}^2\sin xdA={\int }_0^3{\int }_0^{\mathrm{\pi }}{y}^2\sin xdxdy$

Then by using iterated integrals, we have :- 

${\int }_0^3{\int }_0^{\mathrm{\pi }}{y}^2\sin xdxdy={\int }_0^3\left({\int }_0^{\mathrm{\pi }}{y}^2\sin xdx\right)dy$

Now we can solve this integral as following :-

$$\begin{gathered} {\int }_0^3\left({\int }_0^{\mathrm{\pi }}{y}^2\sin xdx\right)dy \\ ={\int }_0^3{{\left[-y^2\cos x\right]}^{\mathrm{\pi }}}_{0}dy \\ ={\int }_0^3\left[-y^2\mathrm{cos\pi }-\left(-y^2\cos 0\right)\right]dy \\ ={\int }_0^3\left[-y^2\left(-1\right)+y^2\left(1\right)\right]dy \\ ={\int }_0^3\left[y^2+y^2\right]dy \\ ={\int }_0^{3}2y^{2}dy \\ ={{\left[\frac{2y^3}{3}\right]}^3}_0 \\ =\left[\frac{2\left(3^3\right)}{3}-0\right] \\ =2\times 9 \\ =18 \end{gathered}$$