$$\iint xy(x^2 + y^2)^{1/2}\, \mathrm dA$$ 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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6$\begingroup$ Don't worry about the shape of this region; just perform the integral over the square and subtract the integral over the quarter circle. $\endgroup$– jorikiMar 8, 2013 at 23:28
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2 Answers
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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.
answered Mar 8, 2013 at 23:38
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$\begingroup$ I got 1/6(2/5(2^(5/2) -1)-1) as my final answer? $\endgroup$– sarahMar 9, 2013 at 0:12
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$\begingroup$ That's pretty confusing notation, @mathlover . Try reducing it to simpler form. $\endgroup$ Mar 9, 2013 at 0:14
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1$\begingroup$ That is exactly what I got. I actually doubted my result, but agreement is reinforcement :) $\endgroup$ Mar 9, 2013 at 0:26
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$\begingroup$ You can also write is as $\frac{8\sqrt{2}-7}{30}$ $\endgroup$ 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.
answered Mar 8, 2013 at 23:29