Double Integral of $\sqrt{x^2+y^2}$

Question

I want to calcule the following integral, $$\int_0^{\frac{1}{\sqrt{2}}}\int_0^{\frac{1}{\sqrt{2}}}\sqrt{x^2+y^2}dxdy$$ what I have is ...

Let $x=y\tan\theta$ where $\theta\in(-\pi/2,\pi/2)$, then, $dx=y\sec^2\theta d\theta$, so, $$\int_0^{\frac{1}{\sqrt{2}}}\int_0^{\frac{1}{\sqrt{2}}}\sqrt{x^2+y^2}dxdy=\int_0^{\frac{1}{\sqrt{2}}}\int_0^{\arctan(\frac{y}{\sqrt{2}})}y^2\sec^3\theta \,d\theta\,dy$$

From here, $$\int_0^{\arctan(\frac{y}{\sqrt{2}})}y^2\sec^3\theta \,d\theta=\frac{\sqrt{2y^2+1}}{4}+\frac{1}{2}y^2\ln\left(\frac{|\sqrt{2y^2+1}+1|}{\sqrt{2}|y|}\right)$$

If I continue with the main problem, $$\int_0^{\frac{1}{\sqrt{2}}}\int_0^{\arctan(\frac{y}{\sqrt{2}})}y^2\sec^3\theta \,d\theta\,dy=\int_0^{\frac{1}{\sqrt{2}}}\left[\frac{\sqrt{2y^2+1}}{4}+\frac{1}{2}y^2\ln\left(\frac{|\sqrt{2y^2+1}+1|}{\sqrt{2}|y|}\right)\right]dy$$

$$=\int_0^{\frac{1}{\sqrt{2}}}\frac{\sqrt{2y^2+1}}{4}dy+\frac{1}{2}\int_0^{\frac{1}{\sqrt{2}}}y^2\ln\left(\frac{|\sqrt{2y^2+1}+1|}{\sqrt{2}|y|}\right)dy$$

$$=\frac{\sqrt{2}+\ln(1+\sqrt{2})}{8\sqrt{2}}+\frac{1}{2}\int_0^{\frac{1}{\sqrt{2}}}y^2\ln\left(\frac{|\sqrt{2y^2+1}+1|}{\sqrt{2}|y|}\right)dy$$

Here I´m stuck, I don´t know how to keep going with the second integral.

Answer 1

That integral is equal to$$2\int_0^{1/\sqrt2}\int_0^x\sqrt{x^2+y^2}\,\mathrm dy\,\mathrm dx.$$In polar coordinates, that is equal to$$2\int_0^{\pi/4}\int_0^{1/(\sqrt2\cos(\theta))}\rho^2\,\mathrm d\rho\,\mathrm d\theta=\frac{\sqrt{2}+\log \left(1+\sqrt{2}\right)}{6\sqrt{2}}.$$