Q. 36 - Calculus

Question

Areas of regions bounded by polar functions: Find the areas of the following regions. The area between the inner and outer loops of $r=1-\sqrt{2} \cos \theta$

Step-by-Step Solution

Answer

The required area is $\frac{4\mathrm{\pi }}{3}-\frac{\sqrt{3}}{4}+\sqrt{6} \mathrm{units}$

Step 1: Given information.

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

Step 2: Area

The graph of the function $r=1-\sqrt{2} \cos \theta$ is 

Now, find the limits of integration.

$$\begin{gathered} 0=1-\sqrt{2} \cos \theta \\ \Rightarrow \cos \theta =\frac{1}{\sqrt{2} } \\ \Rightarrow \theta ={\cos }^{-1}\left(\frac{1}{\sqrt{2} }\right) \\ \Rightarrow \theta =\frac{\mathrm{\pi }}{3},\frac{5\mathrm{\pi }}{3} \end{gathered}$$

Now, the area between the inner and outer loops is,

$$\begin{gathered} A=\frac{1}{2}{\int }_{\alpha }^{\beta }{r}^2 d\theta \\ A=\frac{1}{2}{\int }_{\frac{\mathrm{\pi }}{3}}^{\frac{5\mathrm{\pi }}{3}}{\left(1-\sqrt{2}\cos \theta \right)}^2 d\theta \\ A=\frac{1}{2}{\int }_{\frac{\mathrm{\pi }}{3}}^{\frac{5\mathrm{\pi }}{3}}\left(1+2{\cos }^2\theta -2\sqrt{2}\cos \theta \right)d\theta \\ A=\frac{1}{2}{\int }_{\frac{\mathrm{\pi }}{3}}^{\frac{5\mathrm{\pi }}{3}}\left(1+1+\cos 2\theta -2\sqrt{2}\cos \theta \right)d\theta \\ A=\frac{1}{2}{\int }_{\frac{\mathrm{\pi }}{3}}^{\frac{5\mathrm{\pi }}{3}}\left(2+\cos 2\theta -2\sqrt{2}\cos \theta \right)d\theta \\ A=\frac{1}{2}{\left(2\theta +\frac{\sin 2\theta }{2}-2\sqrt{2}\sin \theta \right)}_{\frac{\mathrm{\pi }}{3}}^{\frac{5\mathrm{\pi }}{3}} \\ A=\frac{1}{2}\left\{2\left(\frac{5\mathrm{\pi }}{3}\right)+\frac{\sin 2\left(\frac{5\mathrm{\pi }}{3}\right)}{2}-2\sqrt{2}\sin \left(\frac{5\mathrm{\pi }}{3}\right)-\left[2\left(\frac{\mathrm{\pi }}{3}\right)+\frac{\sin 2\left(\frac{\mathrm{\pi }}{3}\right)}{2}-2\sqrt{2}\sin \left(\frac{\mathrm{\pi }}{3}\right)\right]\right\} \\ A=\frac{4\mathrm{\pi }}{3}-\frac{\sqrt{3}}{8}+\frac{\sqrt{6}}{2}-\frac{\sqrt{3}}{8}+\frac{\sqrt{6}}{2} \\ A=\frac{4\mathrm{\pi }}{3}-\frac{\sqrt{3}}{4}+\sqrt{6} \mathrm{units} \end{gathered}$$

The required area is $A=\frac{4\mathrm{\pi }}{3}-\frac{\sqrt{3}}{4}+\sqrt{6} \mathrm{units}$