# Thread: Question involving polar coordinates..

1. ## Question involving polar coordinates..

Find the area of the region that is outside the graph of the first equation nd inside the graph of the second equation.
1. r=2
2. r^2= 8sin2x

(x=theta)

2. Originally Posted by clev88
Find the area of the region that is outside the graph of the first equation nd inside the graph of the second equation.
1. r=2
2. r^2= 8sin2x

(x=theta)
First, find the intersection between the circle and lemniscate:

They intersect when $(2)^2=8\sin(2\theta)\implies \tfrac{1}{2}=\sin\left(2x\right)$.

Thus, $2\theta=\arcsin\left(\tfrac{1}{2}\right)=\frac{\pi }{6},\frac{5\pi}{6}\implies \theta=\frac{\pi}{12},\frac{5\pi}{12}$

Thus, it follows that $A={\color{red}2}\left[\tfrac{1}{2}\int_{\frac{\pi}{12}}^{\frac{5\pi}{12} } 8\sin\left(2\theta\right)-4\,d\theta\right]$ (Why twice the integral? I leave that for you to figure out)

Can you continue?

3. Ya thanks!

4. Hello, clev88!

Find the area of the region that is outside $r \:=\:2$ and inside $r^2\:=\: 8\sin2\theta$
$r = 2$ is a circle with radius 2.
$r^2 = 8\sin2\theta$ is a lemniscate ("figure 8").

The graph looks like this . . .
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Due to the symmetry, we can find the area in quandrant 1 and multiply by 4.

Intersection: . $\begin{Bmatrix} r^2 \:=\:4 \\ r^2 \:=\:8\sin2\theta \end{Bmatrix}$
. . $8\sin2\theta \:=\:4 \quad\Rightarrow\quad \sin2\theta \:=\:\frac{1}{2} \quad\Rightarrow\quad 2\theta \:=\:\frac{\pi}{6} \quad\Rightarrow\quad \theta \:=\:\frac{\pi}{12}$

Then: . $A \;=\;4 \times \tfrac{1}{2}\int^{\frac{\pi}{12}}_0\bigg[8\sin2\theta - 4\bigg]\,d\theta$

Got it?

5. Originally Posted by clev88
Find the area of the region that is outside the graph of the first equation nd inside the graph of the second equation.
1. r=2
2. r^2= 8sin2x

(x=theta)
Also see this picture

### area of region of r=2 and r^2=8sin2@

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