Green's theorem: region inside 8 leaved petal

[Pete_01]

Homework Statement

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.

Homework Equations

A =(1/2)int xdy-ydx

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?

Thanks in advance!

 

[nickmai123]

Homework Statement

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.

Homework Equations

A =(1/2)int xdy-ydx

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?

Thanks in advance!

EDIT: I'm stupid, lol.

 

[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.

 

[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) ?

 

[nickmai123]

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) ?

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

 

[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?

 

[nickmai123]

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?

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...

 

[nickmai123]

This should be your integral:

\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.

 

[Pete_01]

Awesome got it! Thanks for all the help!

 

[keltix]

This should be your integral:

\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.

FTW:

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

 

[nickmai123]

FTW:

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

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

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

 

[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.