# 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?

• Don't worry about the shape of this region; just perform the integral over the square and subtract the integral over the quarter circle. Mar 8, 2013 at 23:28

Alternatively, you can consider the region as pairs $(x,y)$ with $x\in[0,1]$ and $y\in [\sqrt{1-x^2},1]$:

$$\int_{0}^1 \int_{\sqrt{1-x^2}}^1 xy\sqrt{x^2+y^2} dy dx$$

For any particular $x$, it is easy to compute the indefinite integral $\int y\sqrt{x^2+y^2} dy$ by substitution $u=x^2+y^2$. The perform another integral with the sam substitution.

• I got 1/6(2/5(2^(5/2) -1)-1) as my final answer? Mar 9, 2013 at 0:12
• That's pretty confusing notation, @mathlover . Try reducing it to simpler form. Mar 9, 2013 at 0:14
• i got (2^(7/2)-7)/30 Mar 9, 2013 at 0:18
• That is exactly what I got. I actually doubted my result, but agreement is reinforcement :) Mar 9, 2013 at 0:26
• You can also write is as $\frac{8\sqrt{2}-7}{30}$ Mar 9, 2013 at 0:29

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