Q. 31 - Calculus

Question

Find the area of the region between the loops of the limacon $r=1+\sqrt{2}\cos \theta$

Step-by-Step Solution

Answer

The region between the limacon's two loops has a surface area is $A=\mathrm{\pi }-1$

Step 1: Given information

Consider the limacon is $r=1+\sqrt{2}\cos \theta$

Step 2: Evaluating the integral

The goal of this issue is to discover an iterated integral in polar coordinates that represents the area of a given location in the polar plane, then evaluate it.

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

Step 3: Pointing the Graph of integral

Graph of $r=1+\sqrt{2}\cos \theta$

The integral of limicon is $r=1+\sqrt{2}\cos \theta$

For,  $\mathrm{r}=0,1+\sqrt{2}\cos \mathrm{\theta }=0\Rightarrow \cos \mathrm{\theta }=-\frac{1}{\sqrt{2}}$

Pole tangents are $\theta =\frac{\pi }{4}\text{ and }\theta =\frac{3\pi }{4}$

Area of small loop is $-\frac{\pi }{4}\le \theta \le \frac{\pi }{4}$

Area of large loop is $-\frac{3\pi }{4}\le \theta \le \frac{3\pi }{4}$

The area of the limacon's region between the two loops is equal to A= Area of Big loop - Area of small loop

$\mathrm{A}={\int }_{-3\mathrm{\pi }/4}^{3\mathrm{\pi }/4} {\left[\frac{{\mathrm{r}}^2}{2}\right]}_0^{3+\sqrt{2}\cos \mathrm{\theta }}\mathrm{d\theta }-{\int }_{-\mathrm{\pi }/4}^{\mathrm{\pi }/4} {\left[\frac{{\mathrm{r}}^2}{2}\right]}_{\mathrm{\theta }}^{3+\sqrt{2}\cos \mathrm{\theta }}\mathrm{d\theta }$

Step 4 : Calculations

Integrate for the first time with respect to r

$$\begin{gathered} \mathrm{A}={\int }_{-3\mathrm{\pi }/4}^{3\mathrm{\theta }/4} {\left[\frac{{\mathrm{r}}^2}{2}\right]}_0^{1+\sqrt{2}\cos \mathrm{\theta }}\mathrm{d\theta }-{\int }_{-\mathrm{\pi }/4}^{\mathrm{\pi }/4} {\left[\frac{{\mathrm{r}}^2}{2}\right]}_0^{1+\sqrt{2}\cos \mathrm{\theta }}\mathrm{d\theta } \\ \mathrm{A}={\int }_{-3\mathrm{\pi }/4}^{3\mathrm{\pi }/4} \left[\frac{(1+\sqrt{2}\cos \mathrm{\theta }{)}^2}{2}\mathrm{d\theta }-{\int }_{-\mathrm{\pi }/4}^{\mathrm{\pi }/4} \left[\frac{(1+\sqrt{2}\cos \mathrm{\theta }{)}^2}{2}\right]\mathrm{\theta }\right. \\ \mathrm{A}={\int }_{-3\mathrm{\pi }/4}^{3\mathrm{\pi }/4} \left[\frac{\left(1+2\sqrt{2}\cos \mathrm{\theta }+2{\cos }^2\mathrm{\theta }\right)}{2}\right]\mathrm{\theta }-{\int }_{-\mathrm{\pi }/4}^{\mathrm{\pi }/4} \left[\frac{\left(1+2\sqrt{2}\cos \mathrm{\theta }+2{\cos }^2\mathrm{\theta }\right)}{2}\right]\mathrm{d\theta } \\ \mathrm{A}={\int }_{-3\mathrm{\pi }/4}^{3\mathrm{\pi }/4} \left[\frac{(1+2\sqrt{2}\cos \mathrm{\theta }+1+\cos 2\mathrm{\theta })}{2}\right]\mathrm{d\theta }-{\int }_{-\mathrm{\pi }/4}^{\mathrm{\pi }/4} \left[\frac{(1+2\sqrt{2}\cos \mathrm{\theta }+1+\cos 2\mathrm{\theta })}{2}\right]\mathrm{d\theta } \end{gathered}$$

Integrate in relation to $\mathrm{\theta }$,

$$\begin{gathered} \mathrm{A}={\int }_{-3\mathrm{\pi }/4}^{3/4} \left[\frac{\left(\mathrm{\theta }+2\sqrt{2}\sin \mathrm{\theta }+\mathrm{\theta }+\frac{1}{2}\sin 2\mathrm{\theta }\right)}{2}\right]\mathrm{\theta }-{\int }_{-\mathrm{\pi }/4}^{\mathrm{\pi }/4} \left[\frac{\left(\mathrm{\theta }+2\sqrt{2}\sin \mathrm{\theta }+\mathrm{\theta }+\frac{1}{2}\sin 2\mathrm{\theta }\right)}{2}\right]\mathrm{\theta } \\ \mathrm{A}={\left[\frac{\left(2\mathrm{\theta }+2\sqrt{2}\sin \mathrm{\theta }+\frac{1}{2}\sin 2\mathrm{\theta }\right)}{2}\right]}_{-3\mathrm{\pi }/4}^{3\mathrm{\pi }/4}-{\left[\frac{\left(2\mathrm{\theta }+2\sqrt{2}\sin \mathrm{\theta }+\frac{1}{2}\sin 2\mathrm{\theta }\right)}{2}\right]}_{-\mathrm{\pi }/4}^{\mathrm{\pi }/4} \\ \mathrm{A}={\left[\mathrm{\theta }+\sqrt{2}\sin \mathrm{\theta }+\frac{1}{4}\sin 2\mathrm{\theta }\right]}_{-3\mathrm{\pi }/4}^{3\mathrm{\pi }/4}-{\left[\mathrm{\theta }+\sqrt{2}\sin \mathrm{\theta }+\frac{1}{4}\sin 2\mathrm{\theta }\right]}_{-\mathrm{\pi }/4}^{\mathrm{\pi }/4} \end{gathered}$$

Pointing the limits,

$\begin{array}{r}\mathrm{A}=\left[\left\{\frac{3\mathrm{\pi }}{4}+\sqrt{2}\sin \left(\frac{3\mathrm{\pi }}{4}\right)+\frac{1}{4}\sin \left(\frac{3\mathrm{\pi }}{2}\right)\right\}-\left\{-\frac{3\mathrm{\pi }}{4}+\sqrt{2}\sin \left(-\frac{3\mathrm{\pi }}{4}\right)+\frac{1}{4}\sin \left(-\frac{3\mathrm{\pi }}{2}\right)\right\}\right]\\ -\left[\left\{\frac{\mathrm{\pi }}{4}+\sqrt{2}\sin \left(\frac{\mathrm{\pi }}{4}\right)+\frac{1}{4}\sin \left(\frac{\mathrm{\pi }}{2}\right)\right\}-\left\{-\frac{\mathrm{\pi }}{4}+\sqrt{2}\sin \left(-\frac{\mathrm{\pi }}{4}\right)+\frac{1}{4}\sin \left(-\frac{\mathrm{\pi }}{2}\right)\right\}\right]\\ \mathrm{A}=\left[\left\{\frac{3\mathrm{\pi }}{4}+1-\frac{1}{4}\right\}-\left\{-\frac{3\mathrm{\pi }}{4}-1+\frac{1}{4}\right\}\right]-\left[\left\{\frac{\mathrm{\pi }}{4}+1+\frac{1}{4}\right\}-\left\{-\frac{\mathrm{\pi }}{4}-1-\frac{1}{4}\right\}\right]\\ \mathrm{A}=\mathrm{\pi }-1\end{array}$

As a result, the area of the reaction between the two limacon loops is $A=\pi -1$