Evaluate $\int_0^1 \int_0^1 xy(x^2+y^2)^{\frac{1}{2}}\ dy\ dx$.
I started off using polar coordinate and then im stuck with defining the bound. And how would you solve it without polar coordinates
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1You need to identify the region. – André Nicolas Feb 17 '14 at 19:14
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sorry 0 to 1 for each x and y. Thanjs – user124473 Feb 17 '14 at 19:15
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Hint: We will integrate first with respect to $y$, and then with respect to $x$. So for a while $x$ is fixed.
Let $x^2+y^2=u$. Then $2y\,dy=du$ so $y\,dy=\frac{1}{2}\,du$. The inner integral is $$\int_{x^2}^{1+x^2} \frac{1}{2}x u^{1/2}\,du.$$
The outer integral (integration with respect to $x$) can also be done by substitution.
Polar coordinates also work nicely. We are integrating on the unit square. It will be useful to split into two parts, $\theta=0$ to $\frac{\pi}{4}$ annd $\theta=\frac{\pi}{4}$ to $\frac{\pi}{2}$. Use symmetry to cut the work in half. For the bounds on the first integral, note that the line $x=1$ has polar equation $r\cos\theta=1$.
André Nicolas
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