91影视

Consider the limacon is$r=1+\sqrt{2}\cos 鈦�胃$

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,$r=1+\sqrt{2}\cos 鈦�胃$

Graph of$r=1+\sqrt{2}\cos 鈦�胃$

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

For, $\mathrm{r}=0,1+\sqrt{2}\cos 鈦�\mathrm{胃}=0鈬�\cos 鈦�\mathrm{胃}=鈭�\frac{1}{\sqrt{2}}$

Pole tangents are$胃=\frac{蟺}{4}\text{and}胃=\frac{3蟺}{4}$

Area of small loop is$鈭�\frac{蟺}{4}鈮�胃鈮�\frac{蟺}{4}$

Area of large loop is$鈭�\frac{3蟺}{4}鈮�胃鈮�\frac{3蟺}{4}$

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

$\mathrm{A}={鈭�}_{鈭�3\mathrm{蟺}/4}^{3\mathrm{蟺}/4}鈥�{\left[\frac{{\mathrm{r}}^2}{2}\right]}_0^{3+\sqrt{2}\cos 鈦�\mathrm{胃}}\mathrm{诲胃}鈭�{鈭�}_{鈭�\mathrm{蟺}/4}^{\mathrm{蟺}/4}鈥�{\left[\frac{{\mathrm{r}}^2}{2}\right]}_{\mathrm{胃}}^{3+\sqrt{2}\cos 鈦�\mathrm{胃}}\mathrm{诲胃}$

Integrate for the first time with respect to r

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

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

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

Pointing the limits,

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

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