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?
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.