For each positive real number $a\in\mathbb{R}_{>0}$ and continuously differentiable function $\phi\in C^{1}{\left[0,a\right]}$, define the function $\mathcal{F}$ via the double integral,
$$\mathcal{F}{\left[\phi\right]}{\left(a\right)}:=\int_{0}^{2a}\mathrm{d}x\int_{0}^{\sqrt{2ax-x^{2}}}\mathrm{d}y\,\frac{x\left(x^{2}+y^{2}\right)\phi^{\prime}{\left(y\right)}}{\sqrt{4a^{2}x^{2}-\left(x^{2}+y^{2}\right)^{2}}}.\tag{1}$$
Given $a\in\mathbb{R}_{>0}\land\phi\in C^{1}{\left[0,a\right]}$, we find by changing the order of integration
$$\begin{align}
\mathcal{F}{\left[\phi\right]}{\left(a\right)}
&=\int_{0}^{2a}\mathrm{d}x\int_{0}^{\sqrt{2ax-x^{2}}}\mathrm{d}y\,\frac{x\left(x^{2}+y^{2}\right)\phi^{\prime}{\left(y\right)}}{\sqrt{4a^{2}x^{2}-\left(x^{2}+y^{2}\right)^{2}}}\\
&=\int_{0}^{a}\mathrm{d}y\int_{a-\sqrt{a^{2}-y^{2}}}^{a+\sqrt{a^{2}-y^{2}}}\mathrm{d}x\,\frac{x\left(x^{2}+y^{2}\right)\phi^{\prime}{\left(y\right)}}{\sqrt{4a^{2}x^{2}-\left(x^{2}+y^{2}\right)^{2}}}\\
&=\int_{0}^{a}\mathrm{d}y\,\phi^{\prime}{\left(y\right)}\int_{a-a\sqrt{1-\left(\frac{y}{a}\right)^{2}}}^{a+a\sqrt{1-\left(\frac{y}{a}\right)^{2}}}\mathrm{d}x\,\frac{x\left(x^{2}+y^{2}\right)}{\sqrt{4a^{2}x^{2}-\left(x^{2}+y^{2}\right)^{2}}}.\tag{2a}\\
\end{align}$$
Rescaling the integration variables by $a$, we then find
$$\begin{align}
\mathcal{F}{\left[\phi\right]}{\left(a\right)}
&=\int_{0}^{a}\mathrm{d}y\,\phi^{\prime}{\left(y\right)}\int_{a-a\sqrt{1-\left(\frac{y}{a}\right)^{2}}}^{a+a\sqrt{1-\left(\frac{y}{a}\right)^{2}}}\mathrm{d}x\,\frac{x\left(x^{2}+y^{2}\right)}{\sqrt{4a^{2}x^{2}-\left(x^{2}+y^{2}\right)^{2}}}\\
&=\int_{0}^{1}\mathrm{d}t\,\phi^{\prime}{\left(at\right)}\int_{1-\sqrt{1-t^{2}}}^{1+\sqrt{1-t^{2}}}\mathrm{d}u\,\frac{a^{3}u\left(u^{2}+t^{2}\right)}{\sqrt{4u^{2}-\left(u^{2}+t^{2}\right)^{2}}};~~~\small{\left[\left(x,y\right)\mapsto\left(au,at\right)\right]}\\
&=a^{3}\int_{0}^{1}\mathrm{d}t\,\phi^{\prime}{\left(at\right)}\int_{1-\sqrt{1-t^{2}}}^{1+\sqrt{1-t^{2}}}\mathrm{d}u\,\frac{u\left(u^{2}+t^{2}\right)}{\sqrt{4u^{2}-\left(u^{2}+t^{2}\right)^{2}}}\\
&=a^{3}\lim_{b\to1^{-}}\int_{0}^{b}\mathrm{d}t\,\phi^{\prime}{\left(at\right)}\int_{1-\sqrt{1-t^{2}}}^{1+\sqrt{1-t^{2}}}\mathrm{d}u\,\frac{u\left(u^{2}+t^{2}\right)}{\sqrt{4u^{2}-\left(u^{2}+t^{2}\right)^{2}}}.\tag{2b}\\
\end{align}$$
Now, it's obvious from the last line above that the integration over $u$ is independent of the parameter $a$, but it also happens to be independent of the other variable $t$, as we shall see shortly.
Define the auxiliary function $J:\left[0,1\right)\rightarrow\mathbb{R}$ via the definite integral,
$$J{\left(t\right)}:=\int_{1-\sqrt{1-t^{2}}}^{1+\sqrt{1-t^{2}}}\mathrm{d}u\,\frac{u\left(u^{2}+t^{2}\right)}{\sqrt{4u^{2}-\left(u^{2}+t^{2}\right)^{2}}}.\tag{3}$$
Supposing $t\in\left[0,1\right)$ and defining the auxiliary parameter $\sqrt{1-t^{2}}=:r\in\left(0,1\right]$, we obtain
$$\begin{align}
J{\left(t\right)}
&=\int_{1-\sqrt{1-t^{2}}}^{1+\sqrt{1-t^{2}}}\mathrm{d}u\,\frac{u\left(u^{2}+t^{2}\right)}{\sqrt{4u^{2}-\left(u^{2}+t^{2}\right)^{2}}}\\
&=\int_{1-r}^{1+r}\mathrm{d}u\,\frac{u\left(u^{2}+1-r^{2}\right)}{\sqrt{4u^{2}-\left(u^{2}+1-r^{2}\right)^{2}}};~~~\small{\left[\sqrt{1-t^{2}}=:r\right]}\\
&=\int_{1-r}^{1+r}\mathrm{d}u\,\frac{2u\left(u^{2}+1-r^{2}\right)}{2\sqrt{4r^{2}-\left(1+r^{2}-u^{2}\right)^{2}}}\\
&=\frac12\int_{1-r}^{1+r}\mathrm{d}u\,\frac{2u\left(u^{2}+1-r^{2}\right)}{\sqrt{\left[2r+\left(1+r^{2}-u^{2}\right)\right]\left[2r-\left(1+r^{2}-u^{2}\right)\right]}}\\
&=\frac12\int_{1-r}^{1+r}\mathrm{d}u\,\frac{2u\left(u^{2}+1-r^{2}\right)}{\sqrt{\left[\left(1+r\right)^{2}-u^{2}\right]\left[u^{2}-\left(1-r\right)^{2}\right]}}\\
&=\frac12\int_{\left(1-r\right)^{2}}^{\left(1+r\right)^{2}}\mathrm{d}v\,\frac{\left(v+1-r^{2}\right)}{\sqrt{\left[\left(1+r\right)^{2}-v\right]\left[v-\left(1-r\right)^{2}\right]}};~~~\small{\left[u^{2}=v\right]}\\
&=\frac12\int_{0}^{1}\mathrm{d}w\,4r\,\frac{2\left(2rw-r+1\right)}{\sqrt{\left[4r\left(1-w\right)\right]\left(4rw\right)}};~~~\small{\left[v=4rw+\left(1-r\right)^{2}\right]}\\
&=\int_{0}^{1}\mathrm{d}w\,\frac{2rw-r+1}{\sqrt{w\left(1-w\right)}}\\
&=\int_{0}^{1}\mathrm{d}w\,\frac{1-r}{\sqrt{w\left(1-w\right)}}+\int_{0}^{1}\mathrm{d}w\,\frac{2rw}{\sqrt{w\left(1-w\right)}}\\
&=\int_{0}^{1}\mathrm{d}w\,\frac{1-r}{\sqrt{w\left(1-w\right)}}+\left[2r\sqrt{w\left(1-w\right)}\right]_{w=0}^{w=1}+\int_{0}^{1}\mathrm{d}w\,\frac{r}{\sqrt{w\left(1-w\right)}};~~~\small{I.B.P.s}\\
&=\int_{0}^{1}\frac{\mathrm{d}w}{\sqrt{w\left(1-w\right)}}\\
&=\pi.\tag{4}\\
\end{align}$$
Finally, continuing with the main calculation from where we left off in the last line of $(2b)$ above, we obtain using result $(4)$ that
$$\begin{align}
\mathcal{F}{\left[\phi\right]}{\left(a\right)}
&=a^{3}\lim_{b\to1^{-}}\int_{0}^{b}\mathrm{d}t\,\phi^{\prime}{\left(at\right)}\int_{1-\sqrt{1-t^{2}}}^{1+\sqrt{1-t^{2}}}\mathrm{d}u\,\frac{u\left(u^{2}+t^{2}\right)}{\sqrt{4u^{2}-\left(u^{2}+t^{2}\right)^{2}}}\\
&=a^{3}\lim_{b\to1^{-}}\int_{0}^{b}\mathrm{d}t\,\phi^{\prime}{\left(at\right)}\,J{\left(t\right)}\\
&=a^{3}\lim_{b\to1^{-}}\int_{0}^{b}\mathrm{d}t\,\phi^{\prime}{\left(at\right)}\,\pi\\
&=\pi\,a^{3}\int_{0}^{1}\mathrm{d}t\,\phi^{\prime}{\left(at\right)}\\
&=\pi\,a^{2}\int_{0}^{a}\mathrm{d}y\,\phi^{\prime}{\left(y\right)};~~~\small{\left[at=y\right]}\\
&=\pi\,a^{2}\left[\phi{\left(a\right)}-\phi{\left(0\right)}\right].\blacksquare\\
\end{align}$$
Best Answer
Use change of order of integration $$\int_1^2\int_1^2\frac{y}{\sqrt{x^2+y^2}}dxdy$$ $$=\int_1^2\int_1^2\frac{y}{\sqrt{x^2+y^2}}dydx$$ $$=\int_1^2\left(\frac12\int_1^2\frac{d(x^2+y^2)}{\sqrt{x^2+y^2}}\right)dx$$ $$=\int_1^2\left(\sqrt{x^2+y^2}\right)_1^2dx$$ $$=\int_1^2\left(\sqrt{x^2+4}-\sqrt{x^2+1}\right)dx$$ $$=\left(\frac x2\sqrt{x^2+4}+2\ln|x+\sqrt{x^2+4}|-\frac x2\sqrt{x^2+1}-\frac12\ln|x+\sqrt{x^2+1}|\right)_1^2$$