# Green's theorem: region inside 8 leaved petal

1. Jul 31, 2009

### Pete_01

1. The problem statement, all variables and given/known data
Use Green's theorem to compute the area of one petal of the 8-leafed rose defined by r=9sin(4theta)
It may be useful for recall that the area of a region D enclosed by a curve C can be expressed as A =(1/2)int xdy-ydx.

2. Relevant equations
A =(1/2)int xdy-ydx

3. The attempt at a solution
I graphed the region on my calculator, but I am not sure what to use for x and y. Would the best way be to change r into rectangular coordinates? If so what would my limit be? 0 to pi/4? perhaps?

2. Aug 1, 2009

### nickmai123

EDIT: I'm stupid, lol.

3. Aug 1, 2009

### nickmai123

Okay, so you're right with the 0 to pi/4. Now you just need to parameterize the curve. Think about finding the x and y coordinates of each of the r-vectors... Then you just integrate x(t)y'(t)dt and y(t)x'(t)dt. Hope that helps.

4. Aug 1, 2009

### Pete_01

Ok, so using the fact that x=rcostheta and y=rsintheta, would the parameter of x=9sin(4theta)cos(theta) and y=9sin(4theta)sin(theta) ?

5. Aug 1, 2009

### nickmai123

That's correct. Now you just have to figure out your limits...

6. Aug 1, 2009

### Pete_01

So limits are 0 to pi/4 correct? When I write out the integral, its huge. Is there any trick to combining the terms together neatly?

7. Aug 1, 2009

### nickmai123

Haha, yeah it's a pretty big integral. I guess this is why you wouldn't normally use Green's Theorem to find the area of a rose petal. You'd just use the polar formula...

8. Aug 1, 2009

### nickmai123

$$\frac{1}{2}(\int_{0}^{\pi/4}[{9sin(4t)cos(t)[9sin(4t)cos(t)+36cos(4t)sin(t)]]dt}-\int_{0}^{\pi/4}[{9sin(4t)sin(t)[-9sin(4t)sin(t)+36cos(4t)cos(t)]]dt})$$

EDIT: Realistically, most people would use Maple, Mathematica, or a calculator to evaluate that monster. It doesn't look friendly even if there's something that can be done to shrink it down.

9. Aug 1, 2009

### Pete_01

Awesome got it! Thanks for all the help!!!

10. Nov 17, 2009

### keltix

FTW:

This actually SIMPLIFIES to something much easier if you go through the distribution.

11. Nov 17, 2009

### nickmai123

Yeah I realized that after a few minutes but I didn't edit my post.

Also, lol @ the response 3 months later... :-)

12. Nov 18, 2009

### keltix

yea i know but i just spent like an hour going through this and i felt like i needed to share it with future WEBWORK kids.