Find $\iint xy(x^2 + y^2)^{1/2} \, \mathrm dA$ $$\iint xy(x^2 + y^2)^{1/2}\,  \mathrm dA$$ where the region is square $[0,1] \times [0,1]$ removing its intersection with the circle of radius 1 at origin.
i got $\displaystyle{\frac{2^{7/2}-7}{30}}$ as my final answer? Anyone disagree?

A natural thing to do is to integrate over the square, and subtract the integral over the quarter-disk.
