Q. 57 - 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.

${\int }_0^{3\sqrt{2}/2}{\int }_x^{\sqrt{9-x^2}}dydx$

Step-by-Step Solution

Answer

The value of the integral is 

${\int }_0^{3\sqrt{2}/2}{\int }_x^{\sqrt{9-x^2}}dydx=\frac{9\pi }{8}$

Step 1: Given information

The integral is 

$I={\int }_0^{3\sqrt{2}/2}{\int }_x^{\sqrt{9-x^2}}dydx$

Here, $x=0$ and $x=\frac{3\sqrt{2}}{2},y=x$and $y=\sqrt{9-x^2}$

Step 2: Calculation

The region of integration $R$ is shown in  the figure

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

$$\begin{gathered} r\sin \theta =r\cos \theta \\ \tan \theta =1\Rightarrow \theta =\frac{\pi }{4} \end{gathered}$$

Substitute $x=r\cos \theta$ and $y=r\sin \theta$ in the upper limit of$y$.

$$\begin{gathered} r\sin \theta =\sqrt{9-r^2{\cos }^2\theta } \\ {r}^2{\sin }^2\theta =9-r^2{\cos }^2\theta \\ {r}^2{\sin }^2\theta +r^2{\cos }^2\theta =9 \\ {r}^2=9\Rightarrow r=\pm 3 \end{gathered}$$

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 $r=3$ and that of $\theta$ are$\frac{\pi }{4}$and $\frac{\pi }{2}$.

$dxdy=rdrd\theta$

Therefore,

$$\begin{gathered} I={\int }_0^{3\sqrt{2}/2}{\int }_x^{\sqrt{9-x^2}}dydx={\int }_{\pi /4}^{\pi /2}{\int }_0^{3}rdrd\theta \\ I={\int }_{\pi /4}^{\pi /2}\left[{\int }_0^{3}rdr\right]\theta \end{gathered}$$

Integrate with respect to $r$ first

$$\begin{gathered} I={\int }_{s/4}^{\pi /2}{\left[\frac{r^2}{2}\right]}_0^{3}d\theta \\ I={\int }_{\pi /4}^{\pi /2}\left[\frac{9}{2}\right]d\theta \\ I=\frac{9}{2}[\theta {]}_{n/4}^{\pi /2} \\ I=\frac{9}{2}\left[\frac{\pi }{2}-\frac{\pi }{4}\right] \\ I=\frac{9\pi }{8} \end{gathered}$$

Thus, the value of the integral is ${\int }_0^{3\sqrt{2}/2}{\int }_x^{\sqrt{9-x^2}}dydx=\frac{9\pi }{8}$