Find the volume of the solid under the graph of z=sqrt(16-x^2-y^2) and above the circular region x^2+y^2<=4 in the xy plane I know I must go to polar. So z=sqrt(16-r^2). Does r range from 0-2? I am not sure what theta ranges from (0-2pi)? I set up the integral as int(int r*sqrt(16-r^2), r=0..2), theta=0..2pi) but I get nowhere near any of the answer choices.