Q. 59 - Calculus

Question

Sketch the region of integration for each of integrals in Exercises $57–60$, and then evaluate the integral by converting to polar coordinates. 

${\int }_0^4{\int }_0^{\sqrt{16-x^2}}{e}^{x^2}{e}^{y^2}dydx$

Step-by-Step Solution

Answer

The value of the integral is ${\int }_0^4{\int }_0^{\sqrt{16-x^2}}{e}^{x^2}{e}^{y^2}dydx=\frac{\pi }{4}\left(e^{16}-1\right)$

Step 1: Given information

The integral is $I={\int }_0^4{\int }_0^{\sqrt{16-x^2}}{e}^{x^2}{e}^{y^2}dydx$

Here, $x=0$ and $x=4, y=0$ and $y=\sqrt{16-x^2}$

Step 2: Calculation

The region  of integration R is shown in the figure

$$\begin{gathered} r^2{\sin }^2\theta +r^2{\cos }^2\theta =16 \\ {r}^2=16\Rightarrow r=4 \end{gathered}$$

Substitute $x=r\cos \theta$ in the lower limit of $x$.

$r\cos \theta =0\Rightarrow r=0,\theta =\frac{\pi }{2}$

Thus, the limits of $r$ are $r=0$ and $r=4$ and that of $\theta$ are $0$ and $\frac{\pi }{2}.$

$dxdy=rdrd\theta$

Therefore,

$$\begin{gathered} I={\int }_0^4{\int }_0^{\sqrt{16-x^2}}{e}^{x^2}{e}^{y^2}dydx={\int }_0^{x/2}{\int }_0^4{e}^{r^2}rdrd\theta \\ I={\int }_0^{\pi /2}\left[{\int }_0^{4}r{e}^{r^2}dr\right]d\theta \end{gathered}$$

Integrate with respect to$r$ first

Put $r^2=t$

$$\begin{gathered} 2rdr=dt\Rightarrow rdr=\frac{dt}{2} \\ I={\int }_0^{\pi /2}\left[{\int }_0^{16}{e}^t\frac{dt}{2}\right]d\theta \\ I=\frac{1}{2}{\int }_0^{\pi /2}{\left[e^t\right]}_0^{16}d\theta \\ I=\frac{1}{2}{\int }_0^{\pi /2}\left[e^{16}-e^0\right]d\theta \\ I=\frac{1}{2}{\int }_0^{\pi /2}\left[e^{16}-1\right]d\theta \\ I=\frac{1}{2}\left(e^{16}-1\right)[\theta {]}_0^{\pi /2} \\ I=\frac{1}{2}\left(e^{16}-1\right)\left[\frac{\pi }{2}\right] \\ I=\frac{\pi }{4}\left(e^{16}-1\right) \end{gathered}$$

Thus, the value of the integral is

${\int }_0^4{\int }_0^{\sqrt{16-x^2}}{e}^{x^2}{e}^{y^2}dydx=\frac{\pi }{4}\left(e^{16}-1\right)$