Hello, shogunhd!
Find the area enclosed by: .
It is a limacon, symmetric to the "y-axis".
We can integrate from to . . . and double.
We have: .
The function is: .
. .
The integral is: .
. .
. .
Ok so the equation is
r= 6 +4 sin theta
Its is a limacon so the area will be symmetrical from 0 to pi/2
Plugging this into the area formula I get
Integral (6+4 sin theta)^2
Integral (36 + 48 sin theta + 16 sin^2 theta)
Integral (36 + 48 sin theta + 16 [(1-cos 2 theta)/2]
Integral (36 + 48 sin theta + 8 (1 - cos 2 theta)
Integral (36 + 48 sin theta + 8 - 8 cos 2 theta)
Integral (44 + 48 sin theta - 8 cos 2 theta)
which gives me
44 theta - 48 cos theta - 8 sin 2 theta + C
I worked it out and I am still unable to arrive at the answer. Can anyone tell me what I'm doing wrong?