# Thread: Help in setting the polar integral

1. ## Help in setting the polar integral

Q: Find the area inside the region bounded by both r = sin theta and r = sin (2 theta)

I set the integral to.........

2[1/2{(sin 2x)^2 dx - 1/2{(sin2x)^2 - (sinx)^2 dx]

Notes: For efficiency, I used x for theta. Also, the limits of integration for each integral are from 0 to pi/3.

But my book says this, which doesn't make sense -

2[1/2{(sinx)^2 dx + 1/2{(sin2x)^2 dx]

Limits of integration for first integral: from 0 to pi/3
Limits of integration for second integral: from pi/3 to pi/2

Cheers!

2. Believe it or not, your book is right. The area swept out by a radius r through angle $\theta$ is given by $\frac{1}{2}r^2\theta$ so your differential of area is $\frac{1}{2}r^2 d\theta$

For $\theta$ from 0 to $\pi/3$, $sin(2\theta)> sin(\theta)$ so the radius only extends from 0 to $sin(\theta)$. For $\theta$ from $\pi/3$ to $\pi/2$ $sin(2\theta)$ is less than $sin(\theta)$ so the radius extends only to that.
The area between the two curves in the first quadrant is $\frac{1}{2}\int_0^{\pi/3} sin^2(\theta)d\theta+ \int_{\pi/3}^{\pi/2} sin^2(2\theta)d\theta$.

Of course, there is a symmetrical part in the second quadrant so that is multiplied by 2.

What you appear to be calculating is the area outside $r= sin(\theta)$ and inside $r= sin(2\theta)$.

3. Hi

Have a look to the sketch below

Let's focus on the first quadrant
First you need to find where the 2 curves cross
$\sin(2\theta) = \sin(\theta)$
$\sin(\theta) (2 \cos(\theta) - 1) = 0$
$\theta = 0$ or $\theta = \frac{\pi}{3}$

Then you need to sweep the area
You can see that from $\theta = 0$ to $\theta = \frac{\pi}{3}$, r goes from 0 to $\sin(\theta)$ (green lines)
And from $\theta = \frac{\pi}{3}$ to $\theta = \frac{\pi}{2}$, r goes from 0 to $\sin(2\theta)$ (blue lines)

4. Unbelievable, you guys are the best! I'm still somewhat inexperienced with the polar form.