Q. 37 - 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 bounded by the limacon$r = 1+k \sin \theta$, where$0 < k < 1$. Explain why it makes sense for the area to approach $\pi \mathrm{as} k \to 0$.

Step-by-Step Solution

Answer

The iterated integral that represents the area of the given region is $\mathit{\pi }\mathbf{+}\frac{\mathbf{\pi }{\mathbf{k}}^{\mathbf{2}}}{\mathbf{2}}$

Step 1 : Given Information

Given equation : $r = 1+k \sin \theta$

Step 2 : Graphing the strophoid and find the area bounded by the loop of the graph

First, we plot the curve $r=1+k \sin \theta$ with $k=\frac{1}{2}$ :

Step 3 : Finding an iterated integral that represents the area of the given region

The arc can be represented as 

$$\begin{gathered} A=2{\int }_{-\pi /2}^{\pi /2}\frac{1}{2}{r}^{2}d\theta \\ A={\int }_{-\pi /2}^{\pi /2}{(1+k \sin \theta )}^{2}d\theta \\ ={\int }_{-\pi /2}^{\pi /2}(1+2k \sin \theta +k^2 {\sin }^2\theta )d\theta \\ ={\int }_{-\pi /2}^{\pi /2}(1+2k \sin \theta +\frac{k^2}{2}(1-\cos 2\theta )d\theta \end{gathered}$$

Integrate in relation to $\theta$ :

$A={\left[(1+2k \sin \theta +\frac{k^2}{2}(1-\cos 2\theta )\right]}_{-\pi /2}^{\pi /2}$

Pointing the limits,

$$\begin{gathered} \mathrm{A}=\left[\begin{array}{l}\left\{\frac{\mathrm{\pi }}{2}-2\mathrm{kcos}\left(\frac{\mathrm{\pi }}{2}\right)+\frac{{\mathrm{k}}^2}{2}\left(\frac{\mathrm{\pi }}{2}-\frac{1}{2}\sin \left(\mathrm{\pi }\right)\right)\right\}-\\ \left\{-\frac{\mathrm{\pi }}{2}-2\mathrm{kcos}\left(-\frac{\mathrm{\pi }}{2}\right)+\frac{{\mathrm{k}}^2}{2}\left(-\frac{\mathrm{\pi }}{2}-\frac{1}{2}\sin (-\mathrm{\pi })\right)\right\}\end{array}\right] \\ \mathrm{A}=\left[\left\{\frac{\mathrm{\pi }}{2}+\frac{{\mathrm{k}}^2}{2}\left(\frac{\mathrm{\pi }}{2}\right)\right\}-\left\{-\frac{\mathrm{\pi }}{2}+\frac{{\mathrm{k}}^2}{2}\left(-\frac{\mathrm{\pi }}{2}\right)\right\}\right] \\ \mathrm{A}=\mathrm{\pi }+\frac{{\mathrm{\pi k}}^2}{2} \end{gathered}$$

As a result, the limacon's area is $A=\pi +\frac{\pi k^2}{2}$ where $\mathrm{k}\to 0,\mathrm{A}=\mathrm{\pi }$