# Finding Area in Polar Coordinates

1. Nov 9, 2006

### Kawrae

The problem is to find the area of r = 4cos3θ.

I know the formula for finding the area in polar coordinates is ∫ (from α to β) ½r²dθ.

I substituted into this formula the given equation and got:
A = ½ ∫ (from 0 to 2π) (4cos3θ)²dθ
= ½ ∫ (from 0 to 2π) (16cos²9θ)dθ
= 8 [(9/2)θ + (9/4)sin18θ) |(2π - 0)]
= 8 (9π + 0)
= 72π

This answer seems really high and the answer in the books gives an answer of 4π... can someone help show me where exactly I am messing up? Thanks :)

2. Nov 9, 2006

### Office_Shredder

Staff Emeritus
Something nice to remember is that cos and sin squared have an average value of 1/2 over a period of 2pi

3. Nov 9, 2006

### marcusl

Do not square 3*theta when you square the cosine function! Also you have gotten the integral (line 3 above) wrong, although it won't change your answer. Please look up the integral of cos^2 again. Finally, are you sure you are supposed to integrate from 0 to 2pi?