Hello, rcmango!

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.