91Ó°ÊÓ

The expression is

$2{∫}_{-\mathrm{π}/4}^{\mathrm{π}/2}{∫}_0^{(\sqrt{2}/2)+\sin \mathrm{θ}}rdrd\mathrm{θ}-2{∫}_{-\mathrm{π}/2}^{-\mathrm{π}/4}{∫}_0^{(\sqrt{2}/2)+\sin \mathrm{θ}}rdrd\mathrm{θ}$

Here, $r=0,r=(\sqrt{2}/2)+\sin \mathrm{θ}$and $\mathrm{θ}=-\mathrm{π}/4,\mathrm{θ}=-\mathrm{π}/2,\mathrm{θ}=\mathrm{π}/2$

$\begin{array}{|ll|}\hline \mathrm{θ}& r=(1/2)+\cos \mathrm{θ}\\ 0& 1.5\\ \mathrm{π}/4& 1.2071\\ \mathrm{π}/2& 0.5\\ 3\mathrm{π}/4& -0.2071\\ \mathrm{π}& -0.5\\ 5\mathrm{π}/4& -0.2071\\ 4\mathrm{π}/3& 0\\ \hline\end{array}$

To sketch the region use the above table

Plot of $r=(\sqrt{2}/2)+\sin \mathrm{θ}$

$I=2{∫}_{-\mathrm{π}/4}^{\mathrm{π}/2}{∫}_0^{(\sqrt{2}/2)+\sin \mathrm{θ}}rdrd\mathrm{θ}-2{∫}_{-\mathrm{π}/2}^{-\mathrm{π}/4}{∫}_0^{(\sqrt{2}/2)+\sin \mathrm{θ}}rdrd\mathrm{θ}$

$I=I_1-I_2$

Where,

$I_1=2{∫}_{-\mathrm{π}/4}^{\mathrm{π}/2}{∫}_0^{(\sqrt{2}/2)+sis\mathrm{θ}}rdrd\mathrm{θ}$

$I_1=2{∫}_{-\mathrm{π}/4}^{\mathrm{π}/2}{\left[\frac{r^2}{2}\right]}_0^{(\sqrt{2}/2)·\sin \mathrm{θ}}d\mathrm{θ}$

$I_1=2{∫}_{-\mathrm{π}/4}^{\mathrm{π}/2}\left[\frac{\left\{\right(\sqrt{2}/2)+\sin \mathrm{θ}{\}}^2-0}{2}\right]\mathrm{θ}$

$I_1=2{∫}_{-\mathrm{π}/4}^{\mathrm{π}/2}\left[\frac{\left\{\frac{1}{2}+\sqrt{2}\sin \mathrm{θ}+{\sin }^2\mathrm{θ}\right\}}{2}d\mathrm{θ}\right.$

$I_1=2{∫}_{-\mathrm{π}/4}^{\mathrm{π}/2}\left[\frac{\left\{\frac{1}{2}+\sqrt{2}\sin \mathrm{θ}+\frac{1}{2}(1-\cos 2\mathrm{θ})\right\}}{2}d\mathrm{θ}\right.$

Integrate with respect to $\mathrm{θ}\text{.}$

localid="1651553993945" $$\begin{gathered} {\left.\left.I_1=2{\left[\frac{{\left.\left\{\frac{\mathrm{θ}}{2}-\sqrt{2}\cos \mathrm{θ}+\frac{1}{2}\left(\mathrm{θ}-\frac{1}{2}\sin 2\mathrm{θ}\right)\right\}\right]}_{-\mathrm{π}/4}^{\mathrm{π}/2}}{2}\right]}_1\right]-\frac{{\left.\left\{\mathrm{θ}-\sqrt{2}\cos \mathrm{θ}-\frac{1}{4}\sin 2\mathrm{θ}\right\}\right]}_{-\mathrm{π}/4}^{\mathrm{π}/2}}{2}\right]}_1=2\left[\frac{\left\{\frac{\mathrm{π}}{2}-\sqrt{2}\cos \left(\frac{\mathrm{π}}{2}\right)-\frac{1}{4}\sin \left(\mathrm{π}\right)\right\}-\left\{-\frac{\mathrm{π}}{4}-\sqrt{2}\cos \left(-\frac{\mathrm{π}}{4}\right)-\frac{1}{4}\sin \left(-\frac{\mathrm{π}}{2}\right)\right\}}{2}\right] \\ {I}_1=2\left[\frac{\left.\left\{\frac{\mathrm{π}}{2}\right\}-\left\{-\frac{\mathrm{π}}{4}-1+\frac{1}{4}\right\}\right]}{2}\right] \\ {I}_1=2\left[\frac{[\mathrm{π}+1)}{I_1}=\frac{3}{4}\left[\frac{[\mathrm{π}}{2}\right]\right. \end{gathered}$$

now,

$$\begin{gathered} I_2=2{∫}_{-\mathrm{π}/2}^{-\mathrm{π}/4}{∫}_0^{(\sqrt{2}/2)+\sin \mathrm{θ}}rdrd\mathrm{θ} \\ {I}_2=2{∫}_{-\mathrm{π}/2}^{-\mathrm{π}/4}{\left[\frac{r^2}{2}\right]}_0^{(\sqrt{2}/2)+\sin \mathrm{θ}}d\mathrm{θ} \\ {I}_2=2{∫}_{-\mathrm{π}/2}^{-\mathrm{π}/4}\left[\frac{\left\{\frac{1}{2}+\sqrt{2}\sin \mathrm{θ}+\frac{1}{2}(1-\cos 2\mathrm{θ})\right\}}{2}d\mathrm{θ}\right. \\ {I}_2=2{\left[\frac{\left\{\mathrm{θ}-\sqrt{2}\cos \mathrm{θ}-\frac{1}{4}\sin 2\mathrm{θ}\right\}}{2}\right]}_{-\mathrm{π}/2}^{-\mathrm{π}/4} \\ {I}_2=2\left[\frac{\left\{-\frac{\mathrm{π}}{4}-\sqrt{2}\cos \left(-\frac{\mathrm{π}}{4}\right)-\frac{1}{4}\sin \left(-\frac{\mathrm{π}}{2}\right)\right\}-\left\{-\frac{\mathrm{π}}{2}-\sqrt{2}\cos \left(-\frac{\mathrm{π}}{2}\right)-\frac{1}{4}\sin (-\mathrm{π})\right\}}{2}\right] \\ {I}_2=2\left[\frac{\left\{-\frac{\mathrm{π}}{4}-1+\frac{1}{4}\right\}-\left\{-\frac{\mathrm{π}}{2}\right\}}{2}\right] \\ {I}_2=\frac{3}{4}(\mathrm{π}-1) \end{gathered}$$

Therefore,

$$\begin{gathered} I=\frac{3}{4}(\mathrm{Ï€}+1)-\frac{3}{4}(\mathrm{Ï€}-1) \\ I=\frac{3}{2} \end{gathered}$$

Thus, the value of integral is

$2{∫}_{-\mathrm{π}/4}^{\mathrm{π}/2}{∫}_0^{(\sqrt{2}/2)·\sin \mathrm{θ}}rdrd\mathrm{θ}-2{∫}_{-\mathrm{π}/2}^{-\mathrm{π}/4}{∫}_0^{(\sqrt{2}/2)·\sin \mathrm{θ}}rdrd\mathrm{θ}=\frac{3}{2}$