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$$\iint xy(x^2 + y^2)^{1/2}\, \mathrm dA$$ 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?

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    $\begingroup$ Don't worry about the shape of this region; just perform the integral over the square and subtract the integral over the quarter circle. $\endgroup$
    – joriki
    Commented Mar 8, 2013 at 23:28

2 Answers 2

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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.

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  • $\begingroup$ I got 1/6(2/5(2^(5/2) -1)-1) as my final answer? $\endgroup$
    – sarah
    Commented Mar 9, 2013 at 0:12
  • $\begingroup$ That's pretty confusing notation, @mathlover . Try reducing it to simpler form. $\endgroup$ Commented Mar 9, 2013 at 0:14
  • $\begingroup$ i got (2^(7/2)-7)/30 $\endgroup$
    – sarah
    Commented Mar 9, 2013 at 0:18
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    $\begingroup$ That is exactly what I got. I actually doubted my result, but agreement is reinforcement :) $\endgroup$ Commented Mar 9, 2013 at 0:26
  • $\begingroup$ You can also write is as $\frac{8\sqrt{2}-7}{30}$ $\endgroup$ Commented Mar 9, 2013 at 0:29
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A natural thing to do is to integrate over the square, and subtract the integral over the quarter-disk.

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