# Area of a graph with polar coordinates

1. Jul 2, 2008

### glog

I'm trying to plot the graph r = sin 3t and find its area.

This is how far I've gotten:

The graph looks like a plane propellor with one propellor pointing downward, and two pointing up-left / up-right, with the length of each equal to 1. Now to get the area...

I have to figure out where the graph reaches 0... which I'm not sure how to do (I solve sin3t = 0 and get t = 0, pi, 2pi ?? )...
or...
I can just get the area from 0 to 2pi... so i'll try that:

A = intergral of (1/2 (sin 3t)^2 dt) from 0 to 2pi

the solution is:
1/2( t/2 - 1/12 sin(6t) )

If I plug in 0 or 2pi into this and solve, the last term becomes 0 and I get a weird answer... what do I do?!?

2. Jul 2, 2008

### rocomath

$$\frac 1 2\int_a^b r^2d\theta$$

$$\sin\theta=0$$

$$\theta=0, \ \pi, \ 2\pi$$

So the first limit starts and ends from $$[0,\pi]$$ for $$\sin\theta$$, but you want $$\sin 3\theta$$, so $$3\theta=...$$

Solve for $$\theta$$, that would be your first limit of integration, and just multiply by a "constant" that gets you the whole area.

Last edited: Jul 2, 2008
3. Jul 2, 2008

### glog

can you elaborate on what you wrote a bit?

sorry i'm just not sure how you got that theta = 0, pi, 2pi/3... since this last one does not equal 0 for sin(theta)

also 3*theta = inverse sin( 0, pi, etc.) = 0
therefore, 2pi/3 works here since it becomes 2pi and sin2pi = 0...

4. Jul 2, 2008

### rocomath

Sorry! I was fixing latex and brain froze a bit and accidently put down 2pi/3. Should be 2pi as you said.

Anyways, you just want the starting point (obviously 0) and the ending point which it becomes 0 again for the first petal. You only need sint=0, which is 0-pi. Now find it for sin3t.

Last edited: Jul 2, 2008
5. Jul 3, 2008

### glog

hey i got it! sweet!