91Ó°ÊÓ

The integral is

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

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

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

Substitute $x=r\cos \mathrm{θ}$and $y=r\sin \mathrm{θ}$in the lower limit of $y$.

$$\begin{gathered} r\sin \mathrm{θ}=r\cos \mathrm{θ} \\ \tan \mathrm{θ}=1⇒\mathrm{θ}=\frac{\mathrm{π}}{4} \end{gathered}$$

Substitute $x=r\cos \mathrm{θ}$and $y=r\sin \mathrm{θ}$in the upper limit of$y$.

$$\begin{gathered} r\sin \mathrm{θ}=\sqrt{9-r^2{\cos }^2\mathrm{θ}} \\ {r}^2{\sin }^2\mathrm{θ}=9-r^2{\cos }^2\mathrm{θ} \\ {r}^2{\sin }^2\mathrm{θ}+r^2{\cos }^2\mathrm{θ}=9 \\ {r}^2=9⇒r=Â\pm 3 \end{gathered}$$

Substitute $x=r\cos \mathrm{θ}$in the lower limit of $x$.

$r\cos \mathrm{θ}=0⇒r=0,\mathrm{θ}=\frac{\mathrm{π}}{2}$

Thus, the limits of $r$are$r=0$and $r=3$and that of $\mathrm{θ}$are$\frac{\mathrm{π}}{4}$and $\frac{\mathrm{π}}{2}$.

$dxdy=rdrd\mathrm{θ}$

Therefore,

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

Integrate with respect to $r$first

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

Thus, the value of the integral is localid="1651390111195" ${∫}_0^{3\sqrt{2}/2}{∫}_x^{\sqrt{9-x^2}}dydx=\frac{9\mathrm{π}}{8}$