Find the area of the region outside r=9+9sin(theta), but inside r=27sin(theta)
I did the integral of .5((27sin(theta))^2-(9+9sin(theta))^2), from pi/6 to 5pi/6, but got the wrong answer. Any hints?
Find the area inside the curve r^2 = 72cos(2theta)
Do I just take the square root of the entire equation? If so, how do I integrate the square root of cosine?
Find the area inside one loop of the curve r^2=4sin(2theta)
Same question as the previous one
Hello, mistykz!
You seem to be forgetting the area formula: .
It says: .Find the area inside the curve: .
Do I just take the square root of the entire equation?
If so, how do I integrate the square root of cosine?
Look at the formula again!
So we have: .
Got it?