Fine the area of the region inside the polar curve r=4sin(theta) and outside the polar curve r=2.
area under polar curve = 1/2 integral (a,b) r^2 d[tex]\Theta[/tex]
The Attempt at a Solution
I set up the integral like follows:
integral (0,pi/2) (4sin([tex]\Theta[/tex])^2 d[tex]\Theta[/tex] - integral (pi/2, pi/6) 2 d[tex]\Theta[/tex]
Then from plugging those into my calculator i got 4pi - (2/3)pi = (10/3) pi
Sorry i cant get the integral latex code to work for some reason.