Q 20. - Calculus

Question

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

Step-by-Step Solution

Answer

$A=\frac{10\pi }{3}+\sqrt{3}$

Step 1: Given information

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

Step 2: Calculation

Consider the polar function, $r=\sqrt{3}-2\sin \theta$

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

Calculate the shaded region's area using the function $r=\sqrt{3}-2\sin \theta$

To find the limits equate $r=\sqrt{3}-2\sin \theta$ to zero then,

$\sqrt{3}-2\sin \theta =0$ $2\sin \theta =\sqrt{3}$ Then $\theta =\frac{\pi }{3},\frac{2\pi }{3}$

The corresponding limits for the inner loop are $\frac{\pi }{3}$ to $\frac{\pi }{2}$

Thus the limits for the inner loop are $\frac{\pi }{3}$ to $\frac{\pi }{2}$ And the outer loop region limits are from $-\frac{\pi }{2}$ to $\frac{\pi }{3}$

Thus the interval is $\left[-\frac{\pi }{2},\frac{\pi }{3}\right]\left[\frac{\pi }{3},\frac{\pi }{2}\right]$

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

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

The area of the function's outer loop is determined as follows: 

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

Then,

$$\begin{gathered} A={\int }_{-\frac{\pi }{2}}^{\frac{\pi }{3}}(3+2-2\cos 2\theta -2\sqrt{3}\sin \theta )d\theta -{\int }_{\frac{\pi }{3}}^{\frac{\pi }{2}}(3+2-2\cos 2\theta -2\sqrt{3}\sin \theta )d\theta \\ A={\int }_{-\frac{\pi }{2}}^{\frac{\pi }{3}}(5-2\cos 2\theta -2\sqrt{3}\sin \theta )d\theta -{\int }_{\frac{\pi }{3}}^{\frac{\pi }{2}}(5-2\cos 2\theta -2\sqrt{3}\sin \theta )d\theta \end{gathered}$$

Here, we'll calculate the integrals individually before subtracting the values.

$$\begin{gathered} {\int }_{\frac{\pi }{2}}^{\frac{\pi }{3}}(5-2\cos 2\theta -2\sqrt{3}\sin \theta )d\theta ={\left[5\theta -\frac{2\sin 2\theta }{2}+2\sqrt{3}\cos \theta \right]}_{-\frac{\pi }{2}}^{\frac{\pi }{3}} \\ =\left[5\frac{\pi }{3}-\sin 2·\frac{\pi }{3}+2\sqrt{3}\cos \frac{\pi }{3}-\left(-\frac{5\pi }{2}+\sin 2·\frac{\pi }{2}+2\sqrt{3}·\cos \left(-\frac{\pi }{2}\right)\right)\right] \end{gathered}$$

Step 3: Calculation

Thus,

$$\begin{gathered} {\int }_{-\frac{\pi }{2}}^{\frac{\pi }{3}}(5-2\cos 2\theta -2\sqrt{3}\sin \theta )d\theta =\left[5\frac{\pi }{3}-\frac{\sqrt{3}}{2}+2\sqrt{3}·\frac{1}{2}+\frac{5\pi }{2}-\sin \pi -2\sqrt{3}·\cos \left(-\frac{\pi }{2}\right)\right] \\ =\left[5\frac{\pi }{3}-\frac{\sqrt{3}}{2}+\sqrt{3}+\frac{5\pi }{2}\right] \\ \frac{\frac{\pi }{3}}{{\int }_{\frac{\pi }{2}}^2}(5-2\cos 2\theta -2\sqrt{3}\sin \theta )d\theta =\left[\frac{25\pi }{6}+\frac{\sqrt{3}}{2}\right] \end{gathered}$$

Now take the integral,

${\int }_{\frac{\pi }{3}}^{\frac{\pi }{2}}(5-2\cos 2\theta -2\sqrt{3}\sin \theta )d\theta ={\left[5\theta -2\frac{\sin 2\theta }{2}+2\sqrt{3}\cos \theta \right]}_{\frac{\pi }{3}}^{\frac{\pi }{2}}$

$$\begin{gathered} =\left[5\frac{\pi }{2}-2\frac{\sin 2\frac{\pi }{2}}{2}+2\sqrt{3}\cos \frac{\pi }{2}-\left(5\frac{\pi }{3}-2\frac{\sin 2\frac{\pi }{3}}{2}+2\sqrt{3}\cos \frac{\pi }{3}\right)\right] \\ =\left[\frac{5\pi }{2}-0-0-\frac{5\pi }{3}+\frac{\sqrt{3}}{2}-2\sqrt{3}·\frac{1}{2}\right] \end{gathered}$$

Thus,

$$\begin{gathered} {\int }_{\frac{\pi }{3}}^{\frac{\pi }{2}}(5-2\cos 2\theta -2\sqrt{3}\sin \theta )d\theta =\left[\frac{5\pi }{2}-\frac{5\pi }{3}+\frac{\sqrt{3}}{2}\right. \\ {\int }_{\frac{\pi }{3}}^{\frac{\pi }{2}}(5-2\cos 2\theta -2\sqrt{3}\sin \theta )d\theta =\left[\frac{5\pi }{6}-\frac{\sqrt{3}}{2}\right] \end{gathered}$$

Now go back to equation one and change the values,

$$\begin{gathered} A={\int }_{\frac{\pi }{2}}^{\frac{\pi }{3}}(\sqrt{3}-2\sin \theta {)}^{2}d\theta -{\int }_{\frac{\pi }{3}}^{\frac{\pi }{2}}(\sqrt{3}-2\sin \theta {)}^{2}d\theta \\ =\left[\frac{25\pi }{6}+\frac{\sqrt{3}}{2}\right]-\left[\frac{5\pi }{6}-\frac{\sqrt{3}}{2}\right] \end{gathered}$$

Thus

$A=\left[\frac{25\pi }{6}-\frac{5\pi }{6}+\frac{\sqrt{3}}{2}+\frac{\sqrt{3}}{2}\right]$

The value of the integral is $A=\frac{20\pi }{6}+\sqrt{3}$

$A=\frac{10\pi }{3}+\sqrt{3}$

Therefore the area is $A=\frac{10\pi }{3}+\sqrt{3}$