Q. 33 - 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 inside the cardioid $r = 3 - 3 \sin \theta$ and outside the cardioid $r = 1 + \sin \theta$.

Step-by-Step Solution

Answer

The integral value is $A=\mathrm{\pi }-1$

Step 1 : Given Information

The region inside the cardioid  $r=3-3\sin \theta$ and outside the cardioid   $r=1+\sin \theta$

Step 2: Simplifications

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

Given 

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

Determining the value of $\theta$

Step 3: Determine the integral value

The size of the area inside the cardioids $r=3-3\sin \theta$and outside the cardioid  $r=1+\sin \theta$can be written as

$A=\underset{\frac{\pi }{6}}{\overset{\frac{5\pi }{6}}{\int }}\underset{3+\sin \theta }{\overset{3-\sin \theta }{\int }}rdrd\theta$

Integrate first with regard to r

$A={\int }_{\pi /6}^{5\pi /6} {\left[\frac{r^2}{2}\right]}_{1+\sin \theta }^{3-3\sin \theta }d\theta$

Plotting the limits,

$$\begin{gathered} \mathrm{A}={\int }_{\mathrm{\pi }/6}^{5\mathrm{\pi }/6} \left[\frac{(3-3\sin \mathrm{\theta }{)}^2-(1+\sin \mathrm{\theta }{)}^2}{2}\right]\mathrm{d\theta } \\ \mathrm{A}={\int }_{\mathrm{\pi }/6}^{5\mathrm{\pi }/6} \left[\frac{\left(9-6\sin \mathrm{\theta }+9{\sin }^2\mathrm{\theta }\right)-\left(1+2\sin \mathrm{\theta }+{\sin }^2\mathrm{\theta }\right)}{2}\right]\mathrm{d\theta } \\ \mathrm{A}={\int }_{\mathrm{\pi }/6}^{5\mathrm{\pi }/6} \left[4-4\sin \mathrm{\theta }+4{\sin }^2\mathrm{\theta }\right]\mathrm{d\theta } \\ \mathrm{A}={\int }_{\mathrm{\pi }/6}^{5\mathrm{\pi }/6} [4-4\sin \mathrm{\theta }+2(1-\cos 2\mathrm{\theta }\left)\right]\mathrm{d\theta }\left[\cos 2\mathrm{\theta }=1-2{\sin }^2\mathrm{\theta }\right] \end{gathered}$$

Integrate in relation to $\theta$

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

Plotting the limits,

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