Sketch the region and convert to rectangular
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> Int(Int(r^3,r=0..5),theta=Pi/2..Pi);
                              Pi     5
                             /      /
                            |      |    3
                            |      |   r  dr dtheta
                            |      |
                           /      /
                              Pi    0
                             ----
                              2

Answer: theta = pi/2 is the positive y-axis, theta = pi is the
negative x-axis.

The region looks like quarter circle in the II quadrant (sort of like
the sorry picture below).

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     o0000|
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    ------+-------
          |
          |
          |
          |

So x goes from -5 to 0, y from 0 to the top of x^2+y^2 = 25 or sqrt(25-x^2)

Then for the integrand one r goes with dr dtheta to get dA and that
leave r^2 which is x^2+y^2.


> Int(Int(x^2+y^2,y=0..sqrt(25-x^2)),x=-5..0);
                                     2 1/2
                         0    (25 - x )
                        /    /
                       |    |               2    2
                       |    |              x  + y  dy dx
                       |    |
                      /    /
                        -5   0