You could integrate from 0 to 2π, but look at your sketch.Find the area of the region bounded by: .r .= .3 + 2·sinθ
I know the formula to use is: .A .= .½ ∫ r² dθ
I have: .A .= .2 * ½ ∫ (3 + 2·sinθ)² dθ
Am i missing work for the limits? . Of course!
Why am i multiplying by 2?
. . The graph is symmetric to the "y-axis".
You can integrate from -½π to ½π and multiply by 2.