- #1

Kawrae

- 46

- 0

**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 :)