Question

The region between the two loops of the limac¸ on $r=1+\sqrt{2}\cos \theta$

Short answer

The area is$2\left(\frac{1}{2}{\int }_0^{\frac{3\pi }{4}}(1+\sqrt{2}\cos \theta {)}^{2}d\theta -\frac{1}{2}{\int }_{\frac{3\pi }{4}}^{\pi }(1+\sqrt{2}\cos \theta {)}^{2}d\theta =\pi +3\right)$

Step-by-step solution

Given information

$r=1+\sqrt{2}\cos \theta$

Calculation

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

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

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

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

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

Formula to find the area is $A={\int }_a^{\beta }\frac{1}{2}\left(f\right(\theta ){)}^{2}d\theta$or $A={\int }_{\alpha }^{\beta }\frac{1}{2}{r}^{2}d\theta$

Calculation

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,

$$\begin{gathered} A=\frac{1}{2}{\int }_0^{\frac{3\pi }{4}}(1+\sqrt{2}\cos \theta {)}^{2}d\theta [\text{since}r=1+\sqrt{2}\cos \theta ] \\ A=\frac{1}{2}{\int }_0^{\frac{3\pi }{4}}\left(1+2{\cos }^2\theta +2\sqrt{2}\cos \theta \right)d\theta \\ A=\frac{1}{2}{\int }_0^{\frac{3\pi }{4}}\left(1+2\frac{(1+\cos 2\theta )}{2}+2\sqrt{2}\cos \theta \right)d\theta \\ \left[\text{Since}\cos 2\theta =2{\cos }^2\theta -1\Rightarrow {\cos }^2\theta =\frac{1+\cos 2\theta }{2}\right] \end{gathered}$$

Thus,

$$\begin{gathered} A=\frac{1}{2}{\int }_0^{\frac{3\pi }{4}}\left(1+2^{\text{'}}\frac{(1+\cos 2\theta )}{2}+2\sqrt{2}\cos \theta \right)d\theta \\ A=\frac{1}{2}{\int }_0^{\frac{3\pi }{4}}(2+\cos 2\theta +2\sqrt{2}\cos \theta )d\theta \\ A=\frac{1}{2}{\left(2\theta +\frac{\sin 2\theta }{2}+2\sqrt{2}\sin \theta \right)}_0^{\frac{3\pi }{4}} \end{gathered}$$

By applying the limits,

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

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

Calculation

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

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

Thus,

$$\begin{gathered} A=\frac{1}{2}{\int }_{\frac{3\pi }{4}}^{\pi }(2+\cos 2\theta +2\sqrt{2}\cos \theta )d\theta \\ A=\frac{1}{2}{\left(2\theta +\frac{\sin 2\theta }{2}+2\sqrt{2}\sin \theta \right)}_{\frac{3\pi }{4}}^{\pi } \end{gathered}$$

By applying the limits,

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

Then,

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

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

Thus,

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

Then the simplified value is,

$$\begin{gathered} A=2·\frac{2\pi }{4}+2·2·\frac{3}{4} \\ A=\frac{4\pi }{4}+\frac{12}{4} \\ A=\pi +3 \end{gathered}$$

Therefore, the area is $2\left(\frac{1}{2}{\int }_0^{\frac{3\pi }{4}}(1+\sqrt{2}\cos \theta {)}^{2}d\theta -\frac{1}{2}{\int }_{\frac{3\pi }{4}}^{\pi }(1+\sqrt{2}\cos \theta {)}^{2}d\theta =\pi +3\right)$