Q. 34 - Calculus

Question

In Exercises 29–38, find an iterated integral in polar coordinates that represents the area of the given region in the polar plane and then evaluate the integral. 

The region where the two cardioids $r = 3 - 3 \sin \theta$and $r = 1 + \sin \theta$overlap 

Step-by-Step Solution

Answer

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

Step 1 : Given Information

Given equations : $r = 3 - 3 \sin \theta$and  $r = 1 + \sin \theta$

Step 2: Simplifications

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.$r=3-3\sin \theta \text{ and }r=1+\sin \theta$

Mark of $\mathrm{r}=3-3\sin \mathrm{\theta }\text{ and }\mathrm{r}=1+\sin \mathrm{\theta }$

The cardioid values are $\mathrm{r}=3-3\mathrm{sin\theta }\text{ and }\mathrm{r}=1+\mathrm{sin\theta }$

Make a solution for her$\mathrm{\theta }$

$\begin{array}{r}1+\sin \mathrm{\theta }=3-3\sin \mathrm{\theta }\\ 4\sin \mathrm{\theta }=2\\ \sin \mathrm{\theta }=\frac{1}{2}\Rightarrow ,\mathrm{\theta }=\frac{\mathrm{\pi }}{6},\frac{5\mathrm{\pi }}{6}\end{array}$

At the pole, tangent

 $\begin{array}{r}3-3\sin \mathrm{\theta }=0\\ \sin \mathrm{\theta }=1\Rightarrow \mathrm{\theta }=\frac{\mathrm{\pi }}{2}\end{array}$

and $1+\sin \mathrm{\theta }=0$

Step 3:The area of the cardioids' overlapping zone

The area of the overlapping region of the cardioids $\mathrm{r}=3-3\sin \mathrm{\theta }\text{ and }\mathrm{r}=1+\sin \mathrm{\theta }$can be represented as  $\mathrm{A}=2\left[{\int }_{-\mathrm{\pi }/2}^{\mathrm{\pi }/6} {\int }_0^{1+\sin \mathrm{\theta }} \mathrm{rdrd\theta }+{\int }_{\mathrm{\pi }/6}^{\mathrm{\pi }/2} {\int }_0^{3-3\sin \mathrm{\theta }} \mathrm{rdrd\theta }\right]$

Integrate first with regard to r.,

$\mathrm{A}=2\left[{\int }_{-\mathrm{\pi }/2}^{\mathrm{\pi }/6} {\left[\frac{{\mathrm{r}}^2}{2}\right]}_0^{1+\sin \mathrm{\theta }}\mathrm{d\theta }+{\int }_{\mathrm{\pi }/6}^{\mathrm{\pi }/2} {\left[\frac{{\mathrm{r}}^2}{2}\right]}_0^{3-3\sin \mathrm{\theta }}\mathrm{d\theta }\right]$

Set the boundaries 

$$\begin{gathered} \mathrm{A}=2\left[{\int }_{-\mathrm{\pi }/2}^{\mathrm{\pi }/6} \left[\frac{(1+\sin \mathrm{\theta }{)}^2}{2}\right]\mathrm{d\theta }+{\int }_{\mathrm{\pi }/6}^{\mathrm{\pi }/2} \left[\frac{(3-3\sin \mathrm{\theta }{)}^2}{2}\right]\mathrm{\theta }\right] \\ \mathrm{A}=2\left[{\int }_{-\mathrm{\pi }/2}^{\mathrm{\pi }/6} \left[\frac{\left(1+2\sin \mathrm{\theta }+{\sin }^2\mathrm{\theta }\right)}{2}\right]\mathrm{d\theta }+{\int }_{\mathrm{\pi }/6}^{\mathrm{\pi }/2} \left[\frac{\left(9-6\sin \mathrm{\theta }+9{\sin }^2\mathrm{\theta }\right)}{2}\right]\mathrm{d\theta }\right] \\ \mathrm{A}=2\left[{\int }_{-\mathrm{\pi }/2}^{\mathrm{\pi }/6} \left[\frac{\left(1+2\sin \mathrm{\theta }+\frac{1}{2}(1-\cos 2\mathrm{\theta })\right)}{2}\right]\mathrm{d\theta }+{\int }_{\mathrm{\pi }/6}^{\mathrm{\pi }/2} \left[\frac{\left(9-6\sin \mathrm{\theta }+\frac{9}{2}(1-\cos 2\mathrm{\theta })\right)}{2}\right]\mathrm{d\theta }\right] \end{gathered}$$

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

$$\begin{gathered} \mathrm{A}=2{\left[\frac{\left(\mathrm{\theta }-2\cos \mathrm{\theta }+\frac{1}{2}\left(\mathrm{\theta }-\frac{1}{2}\sin 2\mathrm{\theta }\right)\right)}{2}\right]}_{-\mathrm{\pi }/2}^{\mathrm{\pi }/6}+2{\left[\frac{\left(9\mathrm{\theta }+6\cos \mathrm{\theta }+\frac{9}{2}\left(\mathrm{\theta }-\frac{1}{2}\sin 2\mathrm{\theta }\right)\right)}{2}\right]}_{\mathrm{\pi }/6}^{\mathrm{\pi }/2} \\ \mathrm{A}=2{\left[\frac{\left(\frac{3}{2}\mathrm{\theta }-2\cos \mathrm{\theta }-\frac{1}{4}\sin 2\mathrm{\theta }\right)}{2}\right]}_{-\mathrm{\pi }/2}^{\mathrm{\pi }/6}+2{\left[\frac{\left(\frac{27}{2}\mathrm{\theta }+6\cos \mathrm{\theta }-\frac{9}{4}\sin 2\mathrm{\theta }\right)}{2}\right]}_{\mathrm{\pi }/6}^{\mathrm{\pi }/2} \end{gathered}$$

Set the boundaries.,

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

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