91影视

Given equations : $r=3鈭�3\sin 胃$and $r=1+\sin 胃$

The objective of this problem is to find an iterated integral in polar coordinates that represents the area of the given region in the polar plane and then evaluate the integral.

Calculate the heart rate.$r=3鈭�3\sin 鈦�胃\text{and}r=1+\sin 鈦�胃$

Mark of $\mathrm{r}=3鈭�3\sin 鈦�\mathrm{胃}\text{and}\mathrm{r}=1+\sin 鈦�\mathrm{胃}$

The cardioid values are $\mathrm{r}=3-3\mathrm{蝉颈苍胃}\text{and}\mathrm{r}=1+\mathrm{蝉颈苍胃}$

Make a solution for her$\mathrm{胃}$

$\begin{array}{r}1+\sin 鈦�\mathrm{胃}=3鈭�3\sin 鈦�\mathrm{胃}\\ 4\sin 鈦�\mathrm{胃}=2\\ \sin 鈦�\mathrm{胃}=\frac{1}{2}鈬�,\mathrm{胃}=\frac{\mathrm{蟺}}{6},\frac{5\mathrm{蟺}}{6}\end{array}$

At the pole, tangent

$\begin{array}{r}3鈭�3\sin 鈦�\mathrm{胃}=0\\ \sin 鈦�\mathrm{胃}=1鈬�\mathrm{胃}=\frac{\mathrm{蟺}}{2}\end{array}$

and $1+\sin 鈦�\mathrm{胃}=0$

The area of the overlapping region of the cardioids $\mathrm{r}=3鈭�3\sin 鈦�\mathrm{胃}\text{and}\mathrm{r}=1+\sin 鈦�\mathrm{胃}$can be represented as $\mathrm{A}=2\left[{鈭�}_{鈭�\mathrm{蟺}/2}^{\mathrm{蟺}/6}鈥�{鈭�}_0^{1+\sin 鈦�\mathrm{胃}}鈥�\mathrm{谤诲谤诲胃}+{鈭�}_{\mathrm{蟺}/6}^{\mathrm{蟺}/2}鈥�{鈭�}_0^{3鈭�3\sin 鈦�\mathrm{胃}}鈥�\mathrm{谤诲谤诲胃}\right]$

Integrate first with regard to r.,

$\mathrm{A}=2\left[{鈭�}_{鈭�\mathrm{蟺}/2}^{\mathrm{蟺}/6}鈥�{\left[\frac{{\mathrm{r}}^2}{2}\right]}_0^{1+\sin 鈦�\mathrm{胃}}\mathrm{诲胃}+{鈭�}_{\mathrm{蟺}/6}^{\mathrm{蟺}/2}鈥�{\left[\frac{{\mathrm{r}}^2}{2}\right]}_0^{3鈭�3\sin 鈦�\mathrm{胃}}\mathrm{诲胃}\right]$

Set the boundaries

$$\begin{gathered} \mathrm{A}=2\left[{鈭�}_{鈭�\mathrm{蟺}/2}^{\mathrm{蟺}/6}鈥�\left[\frac{(1+\sin 鈦�\mathrm{胃}{)}^2}{2}\right]\mathrm{诲胃}+{鈭�}_{\mathrm{蟺}/6}^{\mathrm{蟺}/2}鈥�\left[\frac{(3鈭�3\sin 鈦�\mathrm{胃}{)}^2}{2}\right]\mathrm{胃}\right] \\ \mathrm{A}=2\left[{鈭�}_{鈭�\mathrm{蟺}/2}^{\mathrm{蟺}/6}鈥�\left[\frac{\left(1+2\sin 鈦�\mathrm{胃}+{\sin }^2鈦�\mathrm{胃}\right)}{2}\right]\mathrm{诲胃}+{鈭�}_{\mathrm{蟺}/6}^{\mathrm{蟺}/2}鈥�\left[\frac{\left(9鈭�6\sin 鈦�\mathrm{胃}+9{\sin }^2鈦�\mathrm{胃}\right)}{2}\right]\mathrm{诲胃}\right] \\ \mathrm{A}=2\left[{鈭�}_{鈭�\mathrm{蟺}/2}^{\mathrm{蟺}/6}鈥�\left[\frac{\left(1+2\sin 鈦�\mathrm{胃}+\frac{1}{2}(1鈭�\cos 鈦�2\mathrm{胃})\right)}{2}\right]\mathrm{诲胃}+{鈭�}_{\mathrm{蟺}/6}^{\mathrm{蟺}/2}鈥�\left[\frac{\left(9鈭�6\sin 鈦�\mathrm{胃}+\frac{9}{2}(1鈭�\cos 鈦�2\mathrm{胃})\right)}{2}\right]\mathrm{诲胃}\right] \end{gathered}$$

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

$$\begin{gathered} \mathrm{A}=2{\left[\frac{\left(\mathrm{胃}鈭�2\cos 鈦�\mathrm{胃}+\frac{1}{2}\left(\mathrm{胃}鈭�\frac{1}{2}\sin 鈦�2\mathrm{胃}\right)\right)}{2}\right]}_{鈭�\mathrm{蟺}/2}^{\mathrm{蟺}/6}+2{\left[\frac{\left(9\mathrm{胃}+6\cos 鈦�\mathrm{胃}+\frac{9}{2}\left(\mathrm{胃}鈭�\frac{1}{2}\sin 鈦�2\mathrm{胃}\right)\right)}{2}\right]}_{\mathrm{蟺}/6}^{\mathrm{蟺}/2} \\ \mathrm{A}=2{\left[\frac{\left(\frac{3}{2}\mathrm{胃}鈭�2\cos 鈦�\mathrm{胃}鈭�\frac{1}{4}\sin 鈦�2\mathrm{胃}\right)}{2}\right]}_{鈭�\mathrm{蟺}/2}^{\mathrm{蟺}/6}+2{\left[\frac{\left(\frac{27}{2}\mathrm{胃}+6\cos 鈦�\mathrm{胃}鈭�\frac{9}{4}\sin 鈦�2\mathrm{胃}\right)}{2}\right]}_{\mathrm{蟺}/6}^{\mathrm{蟺}/2} \end{gathered}$$

Set the boundaries.,

$$\begin{gathered} \mathrm{A}=2\left[\frac{\left(\frac{\mathrm{蟺}}{4}鈭�\sqrt{3}鈭�\frac{\sqrt{3}}{8}\right)鈭�\left(鈭�\frac{3\mathrm{蟺}}{4}\right)}{2}\right]+2\left[\frac{\left(\frac{27}{4}\mathrm{蟺}\right)鈭�\left(\frac{27}{12}\mathrm{蟺}+3\sqrt{3}鈭�\frac{9\sqrt{3}}{8}\right)}{2}\right] \\ \mathrm{A}=\frac{9\mathrm{蟺}}{2}鈭�\frac{15\sqrt{3}}{8} \end{gathered}$$

As a result, the area of the cardioids' overlapping zone is$A=\frac{9\mathrm{蟺}}{2}鈭�\frac{15\sqrt{3}}{8}$