91影视

The loops of the limacon is$\mathrm{r}=\sqrt{3}鈭�2\sin 鈦�\mathrm{胃}$

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$\mathrm{r}=\sqrt{3}鈭�2\sin 鈦�\mathrm{胃}$

$\mathrm{r}=\sqrt{3}鈭�2\sin 鈦�\mathrm{胃}$plotted.

The limaconhas the value $r=\sqrt{3}鈭�2\sin 鈦�胃$. When $\mathrm{r}=0,\sqrt{3}鈭�2\sin 鈦�\mathrm{胃}=0鈬�\sin 鈦�\mathrm{胃}=\frac{\sqrt{3}}{2}$

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

For the area of the little loop $\frac{5\mathrm{蟺}}{4}鈮�\mathrm{胃}鈮�\frac{7\mathrm{蟺}}{4}$

For the area of the huge loop $\frac{蟺}{4}鈮�胃鈮�\frac{3蟺}{4}$

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

A= Area of big loop - Area of small loop

$\mathrm{A}={鈭�}_{\mathrm{n}/4}^{3\mathrm{蟺}/4}鈥�{鈭�}_0^{\sqrt{3}鈭�2\sin 鈦�\mathrm{胃}}鈥�\mathrm{谤诲谤诲胃}鈭�{鈭�}_{5\mathrm{蟺}/4}^{7\mathrm{蟺}/4}鈥�{鈭�}_0^{\sqrt{3}鈭�2\sin 鈦�\mathrm{胃}}鈥�\mathrm{谤诲谤诲胃}$

Integrate first with regard to r

$$\begin{gathered} \begin{array}{r}\mathrm{A}={鈭�}_{\mathrm{蟺}/4}^{3\mathrm{蟺}/4}鈥�{\left[\frac{{\mathrm{r}}^2}{2}\right]}_0^{\sqrt{3}鈭�2\sin 鈦�\mathrm{胃}}\mathrm{诲胃}鈭�{鈭�}_{5\mathrm{蟺}/4}^{7\mathrm{蟺}/4}鈥�{\left[\frac{{\mathrm{r}}^2}{2}\right]}_0^{\sqrt{3}鈭�2\sin 鈦�\mathrm{胃}}\mathrm{诲胃}\\ \mathrm{A}={鈭�}_{\mathrm{蟺}/4}^{3\mathrm{蟺}/4}鈥�\left[\frac{(\sqrt{3}鈭�2\sin 鈦�\mathrm{胃}{)}^2}{2}\right]\mathrm{诲胃}鈭�{鈭�}_{5\mathrm{蟺}/4}^{7\mathrm{蟺}/4}鈥�\left[\frac{\left.(\sqrt{3}鈭�2\sin 鈦�\mathrm{胃}{)}^2\right]}{2}\mathrm{诲胃}\right.\\ \mathrm{A}={鈭�}_{\mathrm{蟺}/4}^{3\mathrm{蟺}/4}鈥�\left[\frac{\left.\left(3鈭�4\sqrt{3}\sin 鈦�\mathrm{胃}+4{\sin }^2鈦�\mathrm{胃}\right)\right]}{2}\right]\mathrm{胃}鈭�{鈭�}_{5\mathrm{蟺}/4}^{7\mathrm{蟺}/4}鈥�\left[\frac{\left(3鈭�4\sqrt{3}\sin 鈦�\mathrm{胃}+4{\sin }^2鈦�\mathrm{胃}\right)}{2}\right]\mathrm{胃}\end{array} \\ \begin{array}{r}\mathrm{A}={鈭�}_{\mathrm{蟺}/4}^{3\mathrm{蟺}/4}鈥�\left[\frac{(3鈭�4\sqrt{3}\sin 鈦�\mathrm{胃}+2(1鈭�\cos 鈦�2\mathrm{胃}\left)\right)}{2}\right]\mathrm{诲胃}鈭�{鈭�}_{5\mathrm{蟺}/4}^{7\mathrm{蟺}/4}鈥�\left[\frac{(3鈭�4\sqrt{3}\sin 鈦�\mathrm{胃}+2(1鈭�\cos 鈦�2\mathrm{胃}\left)\right)}{2}\right]\mathrm{诲胃}\\ \mathrm{A}={\mathrm{A}}_1鈭�{\mathrm{A}}_2\end{array} \end{gathered}$$

Since,

${\mathrm{A}}_1={鈭�}_{\mathrm{蟺}/4}^{3\mathrm{蟺}/4}鈥�\left[\frac{(3鈭�4\sqrt{3}\sin 鈦�\mathrm{胃}+2(1鈭�\cos 鈦�2\mathrm{胃}\left)\right)}{2}\right]\mathrm{诲胃}$

Integrate in relation to $胃$,

localid="1652351070686" $$\begin{gathered} {\mathrm{A}}_1={\left[\frac{\left\{3\mathrm{胃}+4\sqrt{3}\cos 鈦�\mathrm{胃}+2\left(\mathrm{胃}鈭�\frac{1}{2}\sin 鈦�2\mathrm{胃}\right)\right\}}{2}\right]}_{\mathrm{蟺}/4}^{3\mathrm{蟺}/4} \\ \begin{array}{l}{\mathrm{A}}_1=\left[\begin{array}{l}\frac{\left\{9\mathrm{蟺}/4+4\sqrt{3}\cos (3\mathrm{蟺}/4)+2\left(3\mathrm{蟺}/4鈭�\frac{1}{2}\sin (3\mathrm{蟺}/2)\right)\right\}}{2}鈭�\\ \frac{\left\{3\mathrm{蟺}/4+4\sqrt{3}\cos (\mathrm{蟺}/4)+2\left(\mathrm{蟺}/4鈭�\frac{1}{2}\sin (\mathrm{蟺}/2)\right)\right\}}{2}\end{array}\right]\\ {\mathrm{A}}_1=\left[\frac{\left\{\frac{9\mathrm{蟺}}{4}鈭�\frac{4\sqrt{3}}{\sqrt{2}}+2\left(\frac{3\mathrm{蟺}}{4}+\frac{1}{2}\right)\right\}}{2}鈭�\frac{\left\{\frac{3\mathrm{蟺}}{4}+\frac{4\sqrt{3}}{\sqrt{2}}+2\left(\frac{\mathrm{蟺}}{4}鈭�\frac{1}{2}\right)\right\}}{2}\right]\\ {\mathrm{A}}_1=\frac{5\mathrm{蟺}}{4}鈭�4\sqrt{\frac{3}{2}}+1\end{array} \end{gathered}$$

localid="1652351053949" $$\begin{gathered} {\mathrm{A}}_2={鈭�}_{5\mathrm{蟺}/4}^{7\mathrm{蟺}/4}鈥�\left[\frac{(3鈭�4\sqrt{3}\sin 鈦�\mathrm{胃}+2(1鈭�\cos 鈦�2\mathrm{胃}\left)\right)}{2}\right]\mathrm{诲胃} \\ {\mathrm{A}}_2={\left[\frac{\left\{3\mathrm{胃}+4\sqrt{3}\cos 鈦�\mathrm{胃}+2\left(\mathrm{胃}鈭�\frac{1}{2}\sin 鈦�2\mathrm{胃}\right)\right\}}{2}\right]}_{5\mathrm{蟺}/4}^{7\mathrm{蟺}/4} \\ {\mathrm{A}}_2=\left[\frac{{\displaystyle \left\{21\mathrm{蟺}/4+4\sqrt{3}\cos 鈦�(7\mathrm{蟺}/4)+2\left(7\mathrm{蟺}/4鈭�\frac{{\displaystyle 1}}{{\displaystyle 2}}\sin 鈦�(7\mathrm{蟺}/2)\right)\right\}}}{{\displaystyle 2}}鈭�\frac{{\displaystyle \left\{15\mathrm{蟺}/4+4\sqrt{3}\cos 鈦�(5\mathrm{蟺}/4)+2\left(5\mathrm{蟺}/4鈭�\frac{{\displaystyle 1}}{{\displaystyle 2}}\sin 鈦�(5\mathrm{蟺}/2)\right)\right\}}}{2}\right] \\ {\mathrm{A}}_2=\left[\frac{\left\{\frac{21\mathrm{蟺}}{4}+\frac{4\sqrt{3}}{\sqrt{2}}+2\left(\frac{7\mathrm{蟺}}{4}+\frac{1}{2}\right)\right\}}{2}鈭�\frac{\left\{\frac{15\mathrm{蟺}}{4}鈭�\frac{4\sqrt{3}}{\sqrt{2}}+2\left(\frac{5\mathrm{蟺}}{4}鈭�\frac{1}{2}\right)\right\}}{2}\right] \\ {\mathrm{A}}_2=\frac{5\mathrm{蟺}}{4}+4\sqrt{\frac{3}{2}}+1 \end{gathered}$$

As a result, the region between the loops is

localid="1652351063549" $\begin{array}{r}\mathrm{A}={\mathrm{A}}_1鈭�{\mathrm{A}}_2\\ \mathrm{A}=\left(\frac{5\mathrm{蟺}}{4}鈭�4\sqrt{\frac{3}{2}}+1\right)鈭�\left(\frac{5\mathrm{蟺}}{4}+4\sqrt{\frac{3}{2}}+1\right)\\ \mathrm{A}=鈭�8\sqrt{\frac{3}{2}}\\ \mathrm{A}=鈭�4\sqrt{6}\end{array}$

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

localid="1652351079409" $\begin{array}{r}\left|\mathrm{A}\right|=4\sqrt{6}\\ \mathrm{A}={\left[\mathrm{胃}+\sqrt{2}\sin 鈦�\mathrm{胃}+\frac{1}{4}\sin 鈦�2\mathrm{胃}\right]}_{鈭�3\mathrm{蟺}/4}^{3\mathrm{蟺}/4}鈭�{\left[\mathrm{胃}+\sqrt{2}\sin 鈦�\mathrm{胃}+\frac{1}{4}\sin 鈦�2\mathrm{胃}\right]}_{鈭�\mathrm{蟺}/4}^{\mathrm{蟺}/4}\end{array}$

Plotting the limits,

localid="1652351090561" $\begin{array}{r}\mathrm{A}=\left[\left\{\frac{3\mathrm{蟺}}{4}+\sqrt{2}\sin 鈦�\left(\frac{3\mathrm{蟺}}{4}\right)+\frac{1}{4}\sin 鈦�\left(\frac{3\mathrm{蟺}}{2}\right)\right\}鈭�\left\{鈭�\frac{3\mathrm{蟺}}{4}+\sqrt{2}\sin 鈦�\left(鈭�\frac{3\mathrm{蟺}}{4}\right)+\frac{1}{4}\sin 鈦�\left(鈭�\frac{3\mathrm{蟺}}{2}\right)\right\}\right]\\ 鈭�\left[\left\{\frac{\mathrm{蟺}}{4}+\sqrt{2}\sin 鈦�\left(\frac{\mathrm{蟺}}{4}\right)+\frac{1}{4}\sin 鈦�\left(\frac{\mathrm{蟺}}{2}\right)\right\}鈭�\left\{鈭�\frac{\mathrm{蟺}}{4}+\sqrt{2}\sin 鈦�\left(鈭�\frac{\mathrm{蟺}}{4}\right)+\frac{1}{4}\sin 鈦�\left(鈭�\frac{\mathrm{蟺}}{2}\right)\right\}\right]\\ \mathrm{A}=\left[\left\{\frac{3\mathrm{蟺}}{4}+1鈭�\frac{1}{4}\right\}鈭�\left\{鈭�\frac{3\mathrm{蟺}}{4}鈭�1+\frac{1}{4}\right\}\right]鈭�\left[\left\{\frac{\mathrm{蟺}}{4}+1+\frac{1}{4}\right\}鈭�\left\{鈭�\frac{\mathrm{蟺}}{4}鈭�1鈭�\frac{1}{4}\right\}\right]\\ \mathrm{A}=\mathrm{蟺}鈭�1\end{array}$