91Ó°ÊÓ

$r=\sqrt{3}-2\sin \mathrm{θ}$

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

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

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

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

$\sqrt{3}-2\sin \mathrm{θ}=0$$2\sin \mathrm{θ}=\sqrt{3}$Then $\mathrm{θ}=\frac{\mathrm{π}}{3},\frac{2\mathrm{π}}{3}$

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

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

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

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

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

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

$$\begin{gathered} A={∫}_{\frac{\mathrm{π}}{2}}^{\frac{\mathrm{π}}{3}}(\sqrt{3}-2\sin \mathrm{θ}{)}^{2}d\mathrm{θ}-{∫}_{\frac{\mathrm{π}}{3}}^{\frac{\mathrm{π}}{2}}(\sqrt{3}-2\sin \mathrm{θ}{)}^{2}d\mathrm{θ}[\text{since}r=\sqrt{3}-2\sin \mathrm{θ}] \\ A={∫}_{\frac{\mathrm{π}}{2}}^{\frac{\mathrm{π}}{3}}\left(3+4{\sin }^2\mathrm{θ}-2\sqrt{3}\sin \mathrm{θ}\right)d\mathrm{θ}-{∫}_{\frac{\mathrm{π}}{3}}^{\frac{\mathrm{π}}{2}}\left(3+4{\sin }^2\mathrm{θ}-2\sqrt{3}\sin \mathrm{θ}\right)d\mathrm{θ} \\ A={∫}_{\frac{\mathrm{π}}{2}}^{\frac{\mathrm{π}}{3}}\left(3+4\frac{(1-\cos 2\mathrm{θ})}{2}-2\sqrt{3}\sin \mathrm{θ}\right)d\mathrm{θ}-{∫}_{\frac{\mathrm{π}}{3}}^{\frac{\mathrm{π}}{2}}\left(3+4\frac{(1-\cos 2\mathrm{θ})}{2}-2\sqrt{3}\sin \mathrm{θ}\right)d\mathrm{θ} \\ \left[\text{Since}\cos 2\mathrm{θ}=1-2{\sin }^2\mathrm{θ}⇒{\sin }^2\mathrm{θ}=\frac{1-\cos 2\mathrm{θ}}{2}\right] \end{gathered}$$

Then,

$$\begin{gathered} A={∫}_{-\frac{\mathrm{π}}{2}}^{\frac{\mathrm{π}}{3}}(3+2-2\cos 2\mathrm{θ}-2\sqrt{3}\sin \mathrm{θ})d\mathrm{θ}-{∫}_{\frac{\mathrm{π}}{3}}^{\frac{\mathrm{π}}{2}}(3+2-2\cos 2\mathrm{θ}-2\sqrt{3}\sin \mathrm{θ})d\mathrm{θ} \\ A={∫}_{-\frac{\mathrm{π}}{2}}^{\frac{\mathrm{π}}{3}}(5-2\cos 2\mathrm{θ}-2\sqrt{3}\sin \mathrm{θ})d\mathrm{θ}-{∫}_{\frac{\mathrm{π}}{3}}^{\frac{\mathrm{π}}{2}}(5-2\cos 2\mathrm{θ}-2\sqrt{3}\sin \mathrm{θ})d\mathrm{θ} \end{gathered}$$

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

$$\begin{gathered} {∫}_{\frac{\mathrm{π}}{2}}^{\frac{\mathrm{π}}{3}}(5-2\cos 2\mathrm{θ}-2\sqrt{3}\sin \mathrm{θ})d\mathrm{θ}={\left[5\mathrm{θ}-\frac{2\sin 2\mathrm{θ}}{2}+2\sqrt{3}\cos \mathrm{θ}\right]}_{-\frac{\mathrm{π}}{2}}^{\frac{\mathrm{π}}{3}} \\ =\left[5\frac{\mathrm{π}}{3}-\sin 2·\frac{\mathrm{π}}{3}+2\sqrt{3}\cos \frac{\mathrm{π}}{3}-\left(-\frac{5\mathrm{π}}{2}+\sin 2·\frac{\mathrm{π}}{2}+2\sqrt{3}·\cos \left(-\frac{\mathrm{π}}{2}\right)\right)\right] \end{gathered}$$

Thus,

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

Now take the integral,

${∫}_{\frac{\mathrm{π}}{3}}^{\frac{\mathrm{π}}{2}}(5-2\cos 2\mathrm{θ}-2\sqrt{3}\sin \mathrm{θ})d\mathrm{θ}={\left[5\mathrm{θ}-2\frac{\sin 2\mathrm{θ}}{2}+2\sqrt{3}\cos \mathrm{θ}\right]}_{\frac{\mathrm{π}}{3}}^{\frac{\mathrm{π}}{2}}$

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

Thus,

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

Now go back to equation one and change the values,

Thus

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

Therefore the area is $A=\frac{10\mathrm{Ï€}}{3}+\sqrt{3}$