91影视

The region inside the cardioid $r=3鈭�3\sin 鈦�胃$and outside the cardioid $r=1+\sin 鈦�胃$

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.

Draw a cardioid diagram

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

Given

Cardiooids are $r=3鈭�3\sin 鈦�胃\text{and}r=1+\sin 鈦�胃$

Determining the value of $胃$

The size of the area inside the cardioids $r=3鈭�3\sin 鈦�胃$and outside the cardioid $r=1+\sin 鈦�胃$can be written as

$A=\underset{\frac{蟺}{6}}{\overset{\frac{5蟺}{6}}{鈭�}}\underset{3+\sin 胃}{\overset{3鈭�\sin 胃}{鈭�}}rdrd胃$

Integrate first with regard to r

$A={鈭�}_{蟺/6}^{5蟺/6}鈥�{\left[\frac{r^2}{2}\right]}_{1+\sin 鈦�胃}^{3鈭�3\sin 鈦�胃}d胃$

Plotting the limits,

$$\begin{gathered} \mathrm{A}={鈭�}_{\mathrm{蟺}/6}^{5\mathrm{蟺}/6}鈥�\left[\frac{(3鈭�3\sin 鈦�\mathrm{胃}{)}^2鈭�(1+\sin 鈦�\mathrm{胃}{)}^2}{2}\right]\mathrm{诲胃} \\ \mathrm{A}={鈭�}_{\mathrm{蟺}/6}^{5\mathrm{蟺}/6}鈥�\left[\frac{\left(9鈭�6\sin 鈦�\mathrm{胃}+9{\sin }^2鈦�\mathrm{胃}\right)鈭�\left(1+2\sin 鈦�\mathrm{胃}+{\sin }^2鈦�\mathrm{胃}\right)}{2}\right]\mathrm{诲胃} \\ \mathrm{A}={鈭�}_{\mathrm{蟺}/6}^{5\mathrm{蟺}/6}鈥�\left[4鈭�4\sin 鈦�\mathrm{胃}+4{\sin }^2鈦�\mathrm{胃}\right]\mathrm{诲胃} \\ \mathrm{A}={鈭�}_{\mathrm{蟺}/6}^{5\mathrm{蟺}/6}鈥�[4鈭�4\sin 鈦�\mathrm{胃}+2(1鈭�\cos 鈦�2\mathrm{胃}\left)\right]\mathrm{诲胃}\left[\cos 鈦�2\mathrm{胃}=1鈭�2{\sin }^2鈦�\mathrm{胃}\right] \end{gathered}$$

Integrate in relation to $胃$

$$\begin{gathered} \mathrm{A}={\left[4\mathrm{胃}+4\cos 鈦�\mathrm{胃}+2\left(\mathrm{胃}鈭�\frac{1}{2}\sin 鈦�2\mathrm{胃}\right)\right]}_{\mathrm{蟺}/6}^{5\mathrm{蟺}/6} \\ \mathrm{A}=\left[\begin{array}{l}\left\{4\left(\frac{5\mathrm{蟺}}{6}\right)+4\cos 鈦�\left(\frac{5\mathrm{蟺}}{6}\right)+2\left(\frac{5\mathrm{蟺}}{6}鈭�\frac{1}{2}\sin 鈦�\left(\frac{5\mathrm{蟺}}{3}\right)\right)\right\}\\ 鈭�\left\{4\left(\frac{\mathrm{蟺}}{6}\right)+4\cos 鈦�\left(\frac{\mathrm{蟺}}{6}\right)+2\left(\frac{\mathrm{蟺}}{6}鈭�\frac{1}{2}\sin 鈦�\left(\frac{\mathrm{蟺}}{3}\right)\right)\right\}\end{array}\right] \\ \mathrm{A}=\left[\left\{\frac{{\displaystyle 10\mathrm{蟺}}}{{\displaystyle 3}}鈭�2\sqrt{3}+\frac{{\displaystyle 5\mathrm{蟺}}}{{\displaystyle 3}}+\frac{{\displaystyle \sqrt{3}}}{{\displaystyle 2}}\right\}鈭�\left\{\frac{{\displaystyle 2\mathrm{蟺}}}{{\displaystyle 3}}+2\sqrt{3}+\frac{{\displaystyle \mathrm{蟺}}}{{\displaystyle 3}}鈭�\frac{{\displaystyle \sqrt{3}}}{{\displaystyle 2}}\right\}\right]\left(\mathrm{simplify}\right) \\ \mathrm{A}=4\mathrm{蟺}鈭�3\sqrt{3} \\ \mathrm{A}={\left[\mathrm{胃}+\sqrt{2}\sin 鈦�\mathrm{胃}+\frac{1}{4}\sin 鈦�2\mathrm{胃}\right]}_{鈭�3\mathrm{蟺}/4}^{3\mathrm{蟺}/4}鈭�{\left[\mathrm{胃}+\sqrt{2}\sin 鈦�\mathrm{胃}+\frac{1}{4}\sin 鈦�2\mathrm{胃}\right]}_{鈭�\mathrm{蟺}/4}^{\mathrm{蟺}/4} \end{gathered}$$

Plotting the limits,

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