Q. 32 - Calculus

Question

In Exercises 29–38, find an iterated integral in polar coordinates that represents the area of the given region in the polar plane and then evaluate the integral. 

The region between the two loops of the limac¸on  $\mathrm{r}=\sqrt{3}-2\sin \mathrm{\theta }$.

Step-by-Step Solution

Answer

The integral value is $\mathrm{A}=\mathrm{\pi }-1$

Step 1 : Given Information

The loops of the limacon is $\mathrm{r}=\sqrt{3}-2\sin \mathrm{\theta }$

Step 2: Calculations

The goal of this issue is to find and evaluate an iterated integral in polar coordinates that reflects the area of a given region in the polar plane.

Make a map of the limacon is$\mathrm{r}=\sqrt{3}-2\sin \mathrm{\theta }$

$\mathrm{r}=\sqrt{3}-2\sin \mathrm{\theta }$ plotted.

The limacon has the value $r=\sqrt{3}-2\sin \theta$. When $\mathrm{r}=0,\sqrt{3}-2\sin \mathrm{\theta }=0\Rightarrow \sin \mathrm{\theta }=\frac{\sqrt{3}}{2}$

At the pole, the tangents are $\mathrm{\theta }=\frac{\mathrm{\pi }}{4}\text{ and }\mathrm{\theta }=\frac{3\mathrm{\pi }}{4}$

For the area of the little loop $\frac{5\mathrm{\pi }}{4}\le \mathrm{\theta }\le \frac{7\mathrm{\pi }}{4}$

For the area of the huge loop $\frac{\pi }{4}\le \theta \le \frac{3\pi }{4}$

The area of the limacon's region between the two loops can be stated as 

A= Area of big loop - Area of small loop

$\mathrm{A}={\int }_{\mathrm{n}/4}^{3\mathrm{\pi }/4} {\int }_0^{\sqrt{3}-2\sin \mathrm{\theta }} \mathrm{rdrd\theta }-{\int }_{5\mathrm{\pi }/4}^{7\mathrm{\pi }/4} {\int }_0^{\sqrt{3}-2\sin \mathrm{\theta }} \mathrm{rdrd\theta }$

Integrate first with regard to r

$$\begin{gathered} \begin{array}{r}\mathrm{A}={\int }_{\mathrm{\pi }/4}^{3\mathrm{\pi }/4} {\left[\frac{{\mathrm{r}}^2}{2}\right]}_0^{\sqrt{3}-2\sin \mathrm{\theta }}\mathrm{d\theta }-{\int }_{5\mathrm{\pi }/4}^{7\mathrm{\pi }/4} {\left[\frac{{\mathrm{r}}^2}{2}\right]}_0^{\sqrt{3}-2\sin \mathrm{\theta }}\mathrm{d\theta }\\ \mathrm{A}={\int }_{\mathrm{\pi }/4}^{3\mathrm{\pi }/4} \left[\frac{(\sqrt{3}-2\sin \mathrm{\theta }{)}^2}{2}\right]\mathrm{d\theta }-{\int }_{5\mathrm{\pi }/4}^{7\mathrm{\pi }/4} \left[\frac{\left.(\sqrt{3}-2\sin \mathrm{\theta }{)}^2\right]}{2}\mathrm{d\theta }\right.\\ \mathrm{A}={\int }_{\mathrm{\pi }/4}^{3\mathrm{\pi }/4} \left[\frac{\left.\left(3-4\sqrt{3}\sin \mathrm{\theta }+4{\sin }^2\mathrm{\theta }\right)\right]}{2}\right]\mathrm{\theta }-{\int }_{5\mathrm{\pi }/4}^{7\mathrm{\pi }/4} \left[\frac{\left(3-4\sqrt{3}\sin \mathrm{\theta }+4{\sin }^2\mathrm{\theta }\right)}{2}\right]\mathrm{\theta }\end{array} \\ \begin{array}{r}\mathrm{A}={\int }_{\mathrm{\pi }/4}^{3\mathrm{\pi }/4} \left[\frac{(3-4\sqrt{3}\sin \mathrm{\theta }+2(1-\cos 2\mathrm{\theta }\left)\right)}{2}\right]\mathrm{d\theta }-{\int }_{5\mathrm{\pi }/4}^{7\mathrm{\pi }/4} \left[\frac{(3-4\sqrt{3}\sin \mathrm{\theta }+2(1-\cos 2\mathrm{\theta }\left)\right)}{2}\right]\mathrm{d\theta }\\ \mathrm{A}={\mathrm{A}}_1-{\mathrm{A}}_2\end{array} \end{gathered}$$

Since,

${\mathrm{A}}_1={\int }_{\mathrm{\pi }/4}^{3\mathrm{\pi }/4} \left[\frac{(3-4\sqrt{3}\sin \mathrm{\theta }+2(1-\cos 2\mathrm{\theta }\left)\right)}{2}\right]\mathrm{d\theta }$

Integrate in relation to $\theta$,

$$\begin{gathered} {\mathrm{A}}_1={\left[\frac{\left\{3\mathrm{\theta }+4\sqrt{3}\cos \mathrm{\theta }+2\left(\mathrm{\theta }-\frac{1}{2}\sin 2\mathrm{\theta }\right)\right\}}{2}\right]}_{\mathrm{\pi }/4}^{3\mathrm{\pi }/4} \\ \begin{array}{l}{\mathrm{A}}_1=\left[\begin{array}{l}\frac{\left\{9\mathrm{\pi }/4+4\sqrt{3}\cos (3\mathrm{\pi }/4)+2\left(3\mathrm{\pi }/4-\frac{1}{2}\sin (3\mathrm{\pi }/2)\right)\right\}}{2}-\\ \frac{\left\{3\mathrm{\pi }/4+4\sqrt{3}\cos (\mathrm{\pi }/4)+2\left(\mathrm{\pi }/4-\frac{1}{2}\sin (\mathrm{\pi }/2)\right)\right\}}{2}\end{array}\right]\\ {\mathrm{A}}_1=\left[\frac{\left\{\frac{9\mathrm{\pi }}{4}-\frac{4\sqrt{3}}{\sqrt{2}}+2\left(\frac{3\mathrm{\pi }}{4}+\frac{1}{2}\right)\right\}}{2}-\frac{\left\{\frac{3\mathrm{\pi }}{4}+\frac{4\sqrt{3}}{\sqrt{2}}+2\left(\frac{\mathrm{\pi }}{4}-\frac{1}{2}\right)\right\}}{2}\right]\\ {\mathrm{A}}_1=\frac{5\mathrm{\pi }}{4}-4\sqrt{\frac{3}{2}}+1\end{array} \end{gathered}$$

$$\begin{gathered} {\mathrm{A}}_2={\int }_{5\mathrm{\pi }/4}^{7\mathrm{\pi }/4} \left[\frac{(3-4\sqrt{3}\sin \mathrm{\theta }+2(1-\cos 2\mathrm{\theta }\left)\right)}{2}\right]\mathrm{d\theta } \\ {\mathrm{A}}_2={\left[\frac{\left\{3\mathrm{\theta }+4\sqrt{3}\cos \mathrm{\theta }+2\left(\mathrm{\theta }-\frac{1}{2}\sin 2\mathrm{\theta }\right)\right\}}{2}\right]}_{5\mathrm{\pi }/4}^{7\mathrm{\pi }/4} \\ {\mathrm{A}}_2=\left[\frac{{\displaystyle \left\{21\mathrm{\pi }/4+4\sqrt{3}\cos (7\mathrm{\pi }/4)+2\left(7\mathrm{\pi }/4-\frac{{\displaystyle 1}}{{\displaystyle 2}}\sin (7\mathrm{\pi }/2)\right)\right\}}}{{\displaystyle 2}}-\frac{{\displaystyle \left\{15\mathrm{\pi }/4+4\sqrt{3}\cos (5\mathrm{\pi }/4)+2\left(5\mathrm{\pi }/4-\frac{{\displaystyle 1}}{{\displaystyle 2}}\sin (5\mathrm{\pi }/2)\right)\right\}}}{2}\right] \\ {\mathrm{A}}_2=\left[\frac{\left\{\frac{21\mathrm{\pi }}{4}+\frac{4\sqrt{3}}{\sqrt{2}}+2\left(\frac{7\mathrm{\pi }}{4}+\frac{1}{2}\right)\right\}}{2}-\frac{\left\{\frac{15\mathrm{\pi }}{4}-\frac{4\sqrt{3}}{\sqrt{2}}+2\left(\frac{5\mathrm{\pi }}{4}-\frac{1}{2}\right)\right\}}{2}\right] \\ {\mathrm{A}}_2=\frac{5\mathrm{\pi }}{4}+4\sqrt{\frac{3}{2}}+1 \end{gathered}$$

As a result, the region between the loops is

$\begin{array}{r}\mathrm{A}={\mathrm{A}}_1-{\mathrm{A}}_2\\ \mathrm{A}=\left(\frac{5\mathrm{\pi }}{4}-4\sqrt{\frac{3}{2}}+1\right)-\left(\frac{5\mathrm{\pi }}{4}+4\sqrt{\frac{3}{2}}+1\right)\\ \mathrm{A}=-8\sqrt{\frac{3}{2}}\\ \mathrm{A}=-4\sqrt{6}\end{array}$

As a result, the area of the limacon's zone between the two loops is

$\begin{array}{r}\left|\mathrm{A}\right|=4\sqrt{6}\\ \mathrm{A}={\left[\mathrm{\theta }+\sqrt{2}\sin \mathrm{\theta }+\frac{1}{4}\sin 2\mathrm{\theta }\right]}_{-3\mathrm{\pi }/4}^{3\mathrm{\pi }/4}-{\left[\mathrm{\theta }+\sqrt{2}\sin \mathrm{\theta }+\frac{1}{4}\sin 2\mathrm{\theta }\right]}_{-\mathrm{\pi }/4}^{\mathrm{\pi }/4}\end{array}$

Plotting the limits,

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