# Area of a rose petal using Green's Theorem

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• Feb 22nd 2010, 10:02 PM
My Little Pony
Area of a rose petal using Green's Theorem
Use Green's theorem to compute the area of one petal of the 16-leafed rose defined by r = 13sin(8t).

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)*integral(xdy - ydx).

Attempt at a solution:

I've got that the limit of theta should be from 0 to pi/8.

x = 13sin(8t)*cos(t)
y = 13sin(8t)*sin(t)

dx = [104cos(8t)*cos(t) - 13sin(8t)*sin(t)]dt
dy = [104cos(8t)*sin(t) + 13sin(8t)*cos(t)]dt

A = (1/2)*integral(13sin(8t)*cos(t)*[104cos(8t)*sin(t) + 13sin(8t)*cos(t)]dt - 13sin(8t)*sin(t)*[104cos(8t)*cos(t) - 13sin(8t)*sin(t)]dt)

I'm not sure how go about evaluating this integral.
• Feb 23rd 2010, 10:02 AM
My Little Pony
Okay, I've simplified the integral to:

A = (1/2)*integral(13sin^2(8t))

I evaluate it and get 13pi/32, as t goes from 0 to pi/8, but it says that is not correct. Anyone know if I'm simplifying the integral properly?