Q. 24 - Calculus

Question

In Exercises 17-25 find a definite integral expression that represents the area of the given region in the polar plane, and then find the exact value of the expression.

The region bounded by the limaçon $r=1+k\sin \theta$, where $0<k<1$. Bonus: Explain why the area approaches $\pi$ as $k\to 0$.

Step-by-Step Solution

Answer

Ans:

 $$\begin{gathered} \underset{k\to 0}{\lim }A=\frac{1}{2}\underset{k\to 0}{\lim }\left(2\pi +k^2\pi \right) \\ =\frac{1}{2}\left(2\pi +(0{)}^2\pi \right) \\ \underset{k\to 0}{\lim }A=\pi \end{gathered}$$

Step 1. Given information:

A limacon $r=1+k\sin \theta$

Step 2. Implying formula:

Formula to find the area is $A={\int }_{\alpha }^{\beta }\frac{1}{2}\left(f\right(\theta ){)}^{2}d\theta$ or $A={\int }_{\alpha }^{\beta }\frac{1}{2}{r}^{2}d\theta$.

The limits are from 0 to $2\pi$.

$A=\frac{1}{2}{\int }_0^{2\pi }(1+k\sin \theta {)}^{2}d\theta$

Then,

$$\begin{gathered} A=\frac{1}{2}{\int }_0^{2\pi }\left(1+k^2{\sin }^2\theta +2k\sin \theta \right)d\theta \\ A=\frac{1}{2}{\int }_0^{2\pi }\left(1+k^2\frac{(1-\cos 2\theta )}{2}+2k\sin \theta \right)d\theta \\ A=\frac{1}{2}{\int }_0^{2\pi }\left(1+\frac{k^2}{2}-\frac{\cos 2\theta }{2}+2k\sin \theta \right)d\theta \end{gathered}$$

Step 3. Continue:

Thus,

$$\begin{gathered} A=\frac{1}{2}{\left(\theta +\frac{k^2}{2}\theta -\frac{\sin 2\theta }{4}+2k\cos \theta \right)}_0^{2\pi } \\ A=\frac{1}{2}\left(2\pi +\frac{k^2}{2}2\pi -\frac{\sin 2·2\pi }{4}+2k\cos 2\pi -2k\cos 2·0\right) \\ A=\frac{1}{2}\left(2\pi +\frac{k^2}{2}2\pi -0+2k-2k\right) \end{gathered}$$

Step 4. Proving:

Then,

$A=\frac{1}{2}\left(2\pi +k^2\pi \right)$

Therefore the area of the limacon is $A=\frac{1}{2}\left(2\pi +k^2\pi \right)$.

Here as $k\to 0$ the area is

$$\begin{gathered} \underset{k\to 0}{\lim }A=\frac{1}{2}\underset{k\to 0}{\lim }\left(2\pi +k^2\pi \right) \\ =\frac{1}{2}\left(2\pi +(0{)}^2\pi \right) \\ \underset{k\to 0}{\lim }A=\pi \end{gathered}$$

Thus when $k\to 0$ the area is $\pi$.

Hence it is proved.