Integral of $f(x,y) \in W^{1,1}(B)$

Question

I have the follow problem. Let $B \subset\Bbb R^2 $ the unitary open ball, and let $$ f(x,y):= \begin{cases} 1+x^2+y^2 & \text{ if }x>0,\\ a(y-1)^2 +b(y+x^2) & \text{ if }x<0, \end{cases} \quad a,b \in\Bbb R. $$ Knowing that $f(x,y) \in W^{1,1}(B)$, calculate $$ I=\int\limits_B f(x,y) dxdy $$ So I have tried this, I've divided into the two part of the ball the integral, and used an angle. I have so, using the substitution $x=\cos \theta$ and $y=\sin \theta$: $$ I=-\int\limits_{\pi \over2}^{-\pi \over2} a(\sin \theta -1)^2+b(\sin \theta + \cos^2 \theta)\sin \theta \cos \theta d^2 \theta -\int_{-\pi \over2}^{\pi \over2} 2 \sin \theta \cos \theta d^2 \theta $$ How am I supposed to continue? And why I need to know that $f(x,y) \in W^{1,1}(B)$? Of course if this is true mean that $\Vert f(x,y)\Vert_{W^{1,1}(B)} = \Vert f(x,y)\Vert_{L^{1}(B)}+\Vert\nabla f(x,y)\Vert_{L^{1}(B)}$. But what's the point?

Answer 1

If $f$ is in $W^{1,1}(B)$, it has a unique trace on the segment $x=0$. This means that as you approach $x=0$ from the left and from the right, you should get the same value. Hence, $$1+y^2=a(y-1)^2 +by$$ for all $-1/2<y<1/2$. This implies that $a=1$ and $b=2$. You can now compute your integrals using trig formulas and integration by parts.