Evaluate $\int_{y=-\infty}^1\int_{x=0}^\infty 4xy\sqrt{x^2+y^2}\mathrm dx\mathrm dy$ Evaluate $$\int_{y=-\infty}^1\int_{x=0}^\infty 4xy\sqrt{x^2+y^2}\mathrm dx\mathrm dy$$
I tried to solve this for hours without any success. I tried substitution method of $u = x^2+y^2$ but doesn't seem to work.
Here is my attempt:

$$\int_{-\infty}^1\int_0^\infty (16x^2y^2)^{\frac{1}{2}}(x^2 + y^2)^\frac{1}{2} \; \mathrm{d}x \mathrm{d}{y} = \int_{-\infty}^1 \int_0^\infty (16x^3y^2 + 16x^2 y^3)^{\frac{1}{2}} \; \mathrm{d}x\mathrm{d}{y}.$$
Let
\begin{align}
u &= 16x^3 y^2 + 16x^2 y^3 \\
\mathrm{d}u &= (48x^2y^2 + 32xy^3) \; \mathrm{d}{x} \\
\mathrm{d}u &= 16xy^2(3x + 2y) \; \mathrm{d}{x} \\
\mathrm{d}u &= 4xy(12xy + 8y^2) \; \mathrm{d}{x} \\
4xy \; \mathrm{d}{x} &= \frac{\mathrm{d}{u}}{12xy + 8y^2}.
\end{align}
$$\int_{-\infty}^1 \int_0^\infty 4xy \sqrt{x^2 + y^2} \; \mathrm{d}x \mathrm{d}y = \int_{-\infty}^1 \int_0^\infty \frac{\mathrm{d}u}{(12xy + 8y^2)}.$$

Attaching image for reference


EDITS
Original Post asked integrate: 4xy sqrt(x^2+y^2) where 0<x, y<1, and added this image. .
Later edits misread 0<x,y<1 as if 0<x<1and 0<y<1
 A: With constant $y$ the substitution $u=x^2+y^2$ implies $$\int_0^1\sqrt{x^2+y^2}xdx=\frac{1}{2}\int_{y^2}^{1+y^2}u^{1/2}du=\frac{1}{3}[u^{3/2}]_{y^2}^{1+y^2}=\frac{(1+y^2)^{3/2}-y^3}{3}.$$Now we apply the integration operator $4\int_0^1 dy\,y$, giving$$\frac{4}{3}\int_0^1[y(1+y^2)^{3/2}-y^4]dy=\frac{4}{15}[(1+y^2)^{5/2}-y^5]_0^1=\frac{4}{15}(2^{5/2}-2)\approx 0.975.$$
A: You want to do this a piece at a time.  First compute the inner integral with respect to $y$ (considering $x$ to be a constant for the moment).  Do you see how to do that via the obvious $u$-substitution that you've already considered?
Once you have computed that integral (which will result in some expression that's a function of $x$), compute the outer integral with respect to $x$.
Let us know if you're still stuck.
A: You first integrate with respect to $x$ while treating $y$ as a constant. Then, you integrate the result of that with respect to $y$:
$$
\int_{0}^{1}\int_{0}^{1} 4xy\sqrt{x^2+y^2}\,dx\,dy=
\int_{0}^{1}\left(\frac{4}{2}y\int_{0}^{1}\sqrt{x^2+y^2}\frac{d}{dx}(x^2+y^2)\,dx\right)\,dy=
\int_{0}^{1}\left(2y\int_{0}^{1}\sqrt{x^2+y^2}\,d(x^2+y^2)\right)\,dy=\\
\int_{0}^{1}\left(2y\frac{2\sqrt{(x^2+y^2)^3}}{3}\bigg|_{0}^{1}\right)\,dy=
\int_{0}^{1}\left[2y\left(\frac{2\sqrt{(1^2+y^2)^3}}{3}-\frac{2\sqrt{(0^2+y^2)^3}}{3}\right)\right]\,dy=\\
\frac{4}{3}\int_{0}^{1}\left(y\sqrt{(1+y^2)^3}-y^4\right)\,dy=\\
\frac{4}{3}\int_{0}^{1}y\sqrt{(1+y^2)^3}\,dy-\frac{4}{3}\int_{0}^{1}y^4\,dy=\\
\frac{4/3}{2}\int_{0}^{1}\sqrt{(1+y^2)^3}\frac{d}{dy}(1+y^2)\,dy-\frac{4}{3}\frac{y^5}{5}\bigg|_{0}^{1}=\\
\frac{2}{3}\int_{0}^{1}(1+y^2)^{3/2}\,d(1+y^2)-\frac{4}{3}\left(\frac{1^5}{5}-\frac{0^5}{5}\right)=\\
\frac{2}{3}\frac{2(1+y^2)^{5/2}}{5}\bigg|_{0}^{1}-\frac{4}{15}=\\
\frac{2}{3}\left(\frac{2(1+1^2)^{5/2}}{5}-\frac{2(1+0^2)^{5/2}}{5}\right)-\frac{4}{15}=\\
\frac{2}{3}\left(\frac{2\sqrt{2^5}}{5}-\frac{2}{5}\right)-\frac{4}{15}=\\
\frac{4(4\sqrt{2}-1)}{15}-\frac{4}{15}=\frac{16\sqrt{2}-4-4}{15}=\frac{16(\sqrt{2}-8)}{15}=\frac{8(2\sqrt{2}-1)}{15}.
$$
Wolfram Alpha check.
A: Hint:
Don't switch to polar coordinates. Notice that
$$\frac d{dx}(x^2+y^2)^a=2ax(x^2+y^2)^{a-1}$$
and
$$\frac d{dy}\frac d{dx}2ax(x^2+y^2)^{a-1}=4a(a-1)xy(x^2+y^2)^{a-2}.$$
A: Using polar coordinates, the integrand is
$$4\rho^2\cos\theta\sin\theta\,\rho\,\rho\,d\rho\,d\theta.$$
By symmetry, we can integrate twice in the triangle between $\theta=0,\theta=\dfrac\pi4$ and $\rho\cos\theta=1$ (or $\rho=\dfrac1{\cos\theta}$).
After integration on $\rho$,
$$8\cos\theta\sin\theta\ \left.\frac{4\rho^5}{5}\right|_0^{1/\cos\theta}=\frac{32}5\frac{\sin\theta}{\cos^4\theta}.$$
Then
$$\int_0^{\pi/4}\frac{\sin\theta}{\cos^4\theta}\,d\theta=\int_0^{1/\sqrt2}\frac{dt}{(1-t^2)^2}\,d\theta.$$
This integral is soluble by decomposition in simple fractions.
