# How do you find the area of the region bounded by the polar curve r^2=4cos(2theta) ?

Sep 3, 2014

The first thing to remember that an integral is a way to add up an infinite number of areas. For rectangular coordinates ($y = f \left(x\right)$), these areas are always rectangles.

${\int}_{a}^{b} f \left(x\right) \mathrm{dx}$

literally means "let's find the area of an infinite numbers of rectangles between $x = a$ and $x = b$, where $f \left(x\right)$ equals the height of each rectangle.

Polar coordinates, though it seems more complicated, follows the same general pattern. The big difference is that we're not dealing with rectangles. We are dealing with sectors of a circle. Also known as pizza slices. The area of a single pizza slice of a circle is $A = \frac{1}{2} {r}^{2} \theta$

(remember this particular area formula only works if $\theta$ is in radians!)

So the area of an infinite number of "pizza slices" is

$\frac{1}{2} {\int}_{a}^{b} {r}^{2} d \theta$

which literally means "let's find the area of an infinite number of pizza slices between $\theta =$angle $a$ and $\theta =$ angle $b$ where r equals the radius of each pizza slice.

Now for your specific problem, we substitute $4 \cos \left(2 \theta\right)$ for ${r}^{2}$.

$\frac{1}{2} {\int}_{a}^{b} {r}^{2} d \theta = \frac{1}{2} {\int}_{a}^{b} 4 \cos \left(2 \theta\right) d \theta$

Now we have to determine a suitable $a$ and $b$.

First we remember what a lemniscate looks like. The $2 \theta$ makes this a little bit tricky. Basically, this polar graph gets through it's cycle twice as fast. That means when $\theta = \frac{\pi}{4}$, $\cos \left(2 \theta\right)$ is behaving as if $\theta = \frac{\pi}{2}$. That's why the radius shrinks to zero at $\theta = \frac{\pi}{4}$. Because $\cos \left(2 \frac{\pi}{4}\right) = \cos \left(\frac{\pi}{2}\right) = 0$.

It looks like the simplest thing to do is to have our angle go from $\theta = - \frac{\pi}{4}$ to $\theta = \frac{\pi}{4}$, which will give us the right half of the lemniscate. Then we just need to double our answer to find the entire area bound by the lemniscate.

So...

$A = 2 {\int}_{- \frac{\pi}{4}}^{\frac{\pi}{4}} 4 \cos \left(2 \theta\right) d \theta$

$A = 2 \left(\frac{1}{2}\right) {\int}_{- \frac{\pi}{4}}^{\frac{\pi}{4}} 4 \cos \left(2 \theta\right) 2 d \theta$

$A = {\int}_{- \frac{\pi}{4}}^{\frac{\pi}{4}} 4 \cos \left(2 \theta\right) 2 d \theta$

$A = 4 \sin \left(2 \theta\right) {|}_{- \frac{\pi}{4}}^{\frac{\pi}{4}} = 4 \sin \left(2 \frac{\pi}{4}\right) - 2 \sin \left(2 \left(- \frac{\pi}{4}\right)\right)$

$A = 4 \sin \left(\frac{\pi}{2}\right) - 4 \sin \left(- \frac{\pi}{2}\right) = 4 \left(1\right) - 4 \left(- 1\right) = 8$