91Ó°ÊÓ

$r=1+\sqrt{2}\cos \mathrm{θ}$

Consider the polar function, $r=1+\sqrt{2}\cos \mathrm{θ}$

The goal is to find the region between the function's inner and outer loops.

Calculate the shaded region's area using the function $r=1+\sqrt{2}\cos \mathrm{θ}$

The corresponding limits of the shaded region are $0$to $\frac{3\mathrm{Ï€}}{4}$for inner loop and $\frac{3\mathrm{Ï€}}{4}$to $\mathrm{Ï€}$

Thus the interval is $\left[0,\frac{3\mathrm{Ï€}}{4}\right]\left[\frac{3\mathrm{Ï€}}{4},\mathrm{Ï€}\right]$

Formula to find the area is $A={∫}_a^{\mathrm{β}}\frac{1}{2}\left(f\right(\mathrm{θ}){)}^{2}d\mathrm{θ}$or $A={∫}_{\mathrm{Î\pm }}^{\mathrm{β}}\frac{1}{2}{r}^{2}d\mathrm{θ}$

The graphical representation is as follows,

The area between the inner and outer loops of the function $A=2($outer loop area-inner loop area )

Area

The area of the outer loop of the function is calculated as below,

Thus,

By applying the limits,

Thus, the area of the outer loop is $A=\frac{3\mathrm{Ï€}}{4}+\frac{3}{4}$

The inner loop area of the function is calculated as follows:

$$\begin{gathered} A=\frac{1}{2}{∫}_{\frac{3\mathrm{π}}{4}}^{\mathrm{π}}(1+\sqrt{2}\cos \mathrm{θ}{)}^{2}d\mathrm{θ}[\text{since}r=1+\sqrt{2}\cos \mathrm{θ}] \\ A=\frac{1}{2}{∫}_{\frac{3\mathrm{π}}{4}}^{\mathrm{π}}\left(1+2{\cos }^2\mathrm{θ}+2\sqrt{2}\cos \mathrm{θ}\right)d\mathrm{θ} \\ A=\frac{1}{2}{∫}_{\frac{3\mathrm{π}}{4}}^{\mathrm{π}}\left(1+2\frac{(1+\cos 2\mathrm{θ})}{2}+2\sqrt{2}\cos \mathrm{θ}\right)d\mathrm{θ} \\ A=\frac{1}{2}{∫}_{\frac{3\mathrm{π}}{4}}^{\mathrm{π}}\left(1+2\frac{(1+\cos 2\mathrm{θ})}{2}+2\sqrt{2}\cos \mathrm{θ}\right)d\mathrm{θ} \end{gathered}$$

Thus,

$$\begin{gathered} A=\frac{1}{2}{∫}_{\frac{3\mathrm{π}}{4}}^{\mathrm{π}}(2+\cos 2\mathrm{θ}+2\sqrt{2}\cos \mathrm{θ})d\mathrm{θ} \\ A=\frac{1}{2}{\left(2\mathrm{θ}+\frac{\sin 2\mathrm{θ}}{2}+2\sqrt{2}\sin \mathrm{θ}\right)}_{\frac{3\mathrm{π}}{4}}^{\mathrm{π}} \end{gathered}$$

By applying the limits,

$$\begin{gathered} A=\frac{1}{2}\left(2\mathrm{π}+\frac{1}{2}·\sin 2·\mathrm{π}+2\sqrt{2}\sin \mathrm{π}-\left(2\frac{3\mathrm{π}}{4}+\frac{1}{2}·\sin 2·\frac{3\mathrm{π}}{4}+2\sqrt{2}\sin \frac{3\mathrm{π}}{4}\right)\right) \\ A=\frac{1}{2}\left(2\mathrm{π}+0+0-\left(\frac{3\mathrm{π}}{2}+\frac{1}{2}·\sin \frac{3\mathrm{π}}{2}+2\sqrt{2}\sin \frac{3\mathrm{π}}{4}\right)\right) \\ A=\frac{1}{2}\left(2\mathrm{π}-\left(\frac{3\mathrm{π}}{2}+\frac{1}{2}·-1+2\sqrt{2}·\frac{1}{\sqrt{2}}\right)\right) \end{gathered}$$

Then,

$$\begin{gathered} A=\frac{1}{2}\left(2\mathrm{π}-\left(\frac{3\mathrm{π}}{2}+\frac{1}{2}·-1+2\right.\right. \\ A=\frac{1}{2}\left(2\mathrm{π}-\left(\frac{3\mathrm{π}}{2}+\frac{3}{2}\right)\right) \\ A=\frac{\mathrm{π}}{4}-\frac{3}{4} \end{gathered}$$

The inner loop area is $A=\frac{\mathrm{Ï€}}{4}-\frac{3}{4}$,outer loop area is $A=\frac{3\mathrm{Ï€}}{4}+\frac{3}{4}$. The required area between the inner and outer loop is, $A=2(outerlooparea-innerlooparea)$

Thus,

$$\begin{gathered} A=2\left(\left(\frac{3\mathrm{π}}{4}+\frac{3}{4}\right)-\left(\frac{\mathrm{π}}{4}-\frac{3}{4}\right)\right) \\ A=2\left(\frac{3\mathrm{π}}{4}+\frac{3}{4}-\frac{\mathrm{π}}{4}+\frac{3}{4}\right) \\ A=2\left(\frac{2\mathrm{π}}{4}+2·\frac{3}{4}\right) \end{gathered}$$

Then the simplified value is,

Therefore, the area is $2\left(\frac{1}{2}{∫}_0^{\frac{3\mathrm{π}}{4}}(1+\sqrt{2}\cos \mathrm{θ}{)}^{2}d\mathrm{θ}-\frac{1}{2}{∫}_{\frac{3\mathrm{π}}{4}}^{\mathrm{π}}(1+\sqrt{2}\cos \mathrm{θ}{)}^{2}d\mathrm{θ}=\mathrm{π}+3\right)$