How to calculate $\iint\ln(x^2+y^2)$ over a part of a circle? 
Given the range $\frac{1}{\sqrt2}\le x\le 1$ and $\sqrt{1-x^2}\le y\le x$
  calculate $\iint\ln(x^2+y^2)$.

This is how the domain looks like:

We need to calculate the integral on the area in red. It seems quite hard next to impossible to pull off the calculation with the given ranges so I thought to calculate the integral on the triangle $ABC$ (let this domain be $B$) and from that to subtract the integral in on circle with the angle between $0$ and $\pi/4$ (let this domain be $C$).
So $\iint_C$ I actually was to able to pull off. But $\iint_B$ doesn't seem feasible at least to me.
$$
\iint_C\ln(x^2+y^2)=\int_0^{\pi/4}\int_0^1 \ln(r^2)r \,dr\,d\theta=\frac{1}{2}\int\bigg[ r^2\ln r^2-r^2\bigg]_0^1=\frac{1}{2}\int-1\,d\theta=-\frac{\pi}{8}
$$
Now:
$$
\iint_B \ln(x^2+y^2)\,dx \,dy=\int_0^1\int_0^x \ln(x^2+y^2)
$$
I don't see how this can be integrated, substitution method doesn't help here. Can I convert to polar coordinates again? But then:
$$
\int_0^1\int_0^{r\cos\theta} r\ln r^2
$$
which is not feasible for me.
How to tackle this?
 A: Note that
\begin{align*}
\int_{x=0}^1\int_{y=0}^x \ln(x^2+y^2)\,dxdy&=2\int_{x=0}^1\int_{y=0}^x \ln(x)\,dxdy+
\int_{x=0}^1\int_{y=0}^x \ln(1+(y/x)^2)\,dxdy\\
&=2\int_0^1\ln(x)\left(\int_0^x dy\right)\,dx+\int_{x=0}^1x\left(\int_{t=0}^1 \ln(1+t^2)\,dt\right) dx\\
&=2\int_0^1x\ln(x) \,dx+\frac{1}{2}\int_{0}^1 \ln(1+t^2)\,dt\\
&=\left[x^2\ln(x)-\frac{x^2}{2}\right]_0^1+\frac{1}{2}\left[t\ln(1+t^2)-2t+2\arctan(t)\right]_0^1\\
&=-\frac{3}{2}+\frac{\ln(2)}{2}+\frac{\pi}{4}.
\end{align*}
A: $$
\begin{align}
\iint_\Omega\log\left(x^2+y^2\right)\,\mathrm{d}x\,\mathrm{d}y
&=\int_0^{\pi/4}\int_1^{\sec(\theta)}2\log\left(r\right)\,r\,\mathrm{d}r\,\mathrm{d}\theta\tag{1}\\
&=\int_0^{\pi/4}\left[\left[r^2\log(r)\right]_1^{\sec(\theta)}-\int_1^{\sec(\theta)}r\,\mathrm{d}r\right]\mathrm{d}\theta\tag{2}\\
&=\int_0^{\pi/4}\left[\sec^2(\theta)\log(\sec(\theta))-\frac12\sec^2(\theta)+\frac12\right]\mathrm{d}\theta\tag{3}\\
&=\int_0^{\pi/4}\log(\sec(\theta))\,\mathrm{d}\tan(\theta)-\left[\frac12\tan(\theta)\right]_0^{\pi/4}+\left[\frac12\theta\right]_0^{\pi/4}\tag{4}\\
&=\frac12\int_0^1\log\left(1+u^2\right)\,\mathrm{d}u-\frac12+\frac\pi8\tag{5}\\
&=\frac12\left[u\log\left(1+u^2\right)\right]_0^1-\int_0^1\frac{u^2}{1+u^2}\,\mathrm{d}u-\frac12+\frac\pi8\tag{6}\\[3pt]
&=\frac12\log(2)-\left[u-\tan^{-1}(u)\right]_0^1-\frac12+\frac\pi8\tag{7}\\[6pt]
&=\frac12\log(2)-\frac32+\frac{3\pi}8\tag{8}
\end{align}
$$
Explanation:
$(1)$: convert to polar coordinates
$(2)$: integrate by parts $u=\log(r)$, $v=r^2$
$(3)$: evaluate the integral in $r$
$(4)$: integrate in $\theta$
$(5)$: substitute $u=\tan(\theta)$ and evaluate term on the right
$(6)$: integrate by parts
$(7)$: evaluate term on the left and integrate $\frac{u^2}{1+u^2}=1-\frac1{1+u^2}$
$(8)$: evaluate remaining terms
