Evaluating $\int_{0}^1\int_{0}^1 xy\sqrt{x^2+y^2}\,dy\,dx$ Calculate the iterated integral: $$\int_{0}^1\int_{0}^1 xy\sqrt{x^2+y^2}\,dy\,dx$$
I'm stumped with this problem. Should I do integration by parts with both variables or is there another way to do this? If someone could help me out, that would grand!
 A: Substitute $u=x^2$ and $v=y^2$
$$
\begin{align}
\int_0^1\int_0^1xy\sqrt{x^2+y^2}\,\mathrm{d}x\,\mathrm{d}y
&=\frac14\int_0^1\int_0^1\sqrt{u+v}\,\mathrm{d}u\,\mathrm{d}v\\
&=\frac14\int_0^1\frac23\left((v+1)^{3/2}-v^{3/2}\right)\,\mathrm{d}v\\
&=\frac16\cdot\frac25\left(2^{5/2}-1^{5/2}-1^{5/2}+0^{5/2}\right)\\
&=\frac2{15}(2\sqrt2-1)
\end{align}
$$
A: I would use substitution ($u = x^2$, $v = y^2$) in each variable separately. At first I thought of polar coordinates, but the boundaries would be messy that way.
A: In addition to the 2-variable change of variables formula suggested by Eric Auld, you could also use the one-variable substitution formula:
For the inner integral, consider the substitution $u=x^2+y^2$, $du=2y\,dy$. Then
$$
\int_0^1 xy\sqrt{x^2+y^2}\,dy=\int_{x^2}^{x^2+1}\frac{x}{2}\sqrt{u}\,du=\cdots
$$
A: Another more direct approach using
$$\int f'(x)f(x)^ndx=\frac1{n+1}f(x)^{n+1}+C\;:$$
$$\int\limits_0^1\int\limits_0^1 xy\sqrt{x^2+y^2}\,dydx=\int\limits_0^1 x\,dx\frac12\int\limits_0^1 2y\sqrt{x^2+y^2}dy=\frac12\int\limits_0^1 x\,dx\left.\frac23(x^2+y^2)^{3/2}\right|_0^1=$$
$$=\frac13\int\limits_0^1 x\left((x^2+1)^{3/2}-x^3\right)dx=\frac16\int\limits_0^12x(x^2+1)^{3/2}dx-\frac13\int\limits_0^1x^4dx=$$
$$=\left.\frac1{15}(x^2+1)^{5/2}\right|_0^1-\frac1{15}=\frac1{15}\left(4\sqrt2-1-1\right)=\frac2{15}\left(2\sqrt2-1\right)$$
