integral over Green's function of a $d$-Ball

Question

I'm studying for an exam on PDE's and I'm stuck trying to solve the following problem. This is in context of the following boundary value problem: \begin{align} \begin{cases} -\Delta u \equiv 0,&\text{in } B(0,r)\\ u \equiv g,&\text{on } \partial B(0,r) \end{cases} \end{align} where $r>0$, $B(0,r)\subset \mathbb{R}^d$ is the ball of radius $r$ and $g\in C(\partial B(0,r))$. The problem then asks to show, that the Green's function $G$ of the above BVP satisfies $$\int_{B(0,r)}G(x,y)dy = \frac{1}{2d}(r^2-|x|^2)$$ My approach was the following: I want to use the property of the greens function, namely \begin{align} \begin{cases} -\Delta G(x,y) = \delta(x-y),&\text{in } B(0,r)\\ G(x,y) = 0,&\text{on } \partial B(0,r) \end{cases} \end{align} where $\delta$ is the Dirac Delta. I also know, that the following identity holds for general functions $f$ $$\frac{\partial}{\partial s}\left(\frac{1}{|\partial B(0,s)|}\int_{\partial B(0,s)}f(y)dS(y)\right) = \frac{s}{d}\frac{1}{| B(0,s)|}\int_{ B(0,s)}\Delta f(y)dy.$$ that means we can compute \begin{align} \frac{\partial}{\partial s}\left(\frac{1}{|\partial B(0,s)|}\int_{\partial B(0,s)}G(x,y)dy\right) &= \frac{s}{d}\frac{1}{| B(0,s)|}\int_{ B(0,s)}\Delta_y G(x,y)dy\\ &=-\frac{s}{d}\frac{1}{| B(0,s)|}\int_{ B(0,s)}\delta(x-y)dy = -\frac{s}{d}\frac{1}{| B(0,s)|}\\ &=-\frac{s}{d}\frac{1}{s^{d}| B(0,1)|} = -\frac{1}{ds^{d-1}}\frac{1}{| B(0,1)|} \end{align} and thus \begin{align} \frac{1}{|\partial B(0,s)|}\int_{\partial B(0,s)}G(x,y)dy = -\frac{1}{d(d-2)s^{d-2}}\frac{1}{| B(0,1)|}+c \end{align} for some constant $c$. Now using that $|\partial B(0,s)| = ds^{d-1}|B(0,1)|$, we get \begin{align} \int_{\partial B(0,s)}G(x,y)dy = -\frac{s}{(d-2)}+cds^{d-1}|B(0,1)| \end{align} The constant can be found through the boundary condition on $G$: \begin{align} 0 = \int_{\partial B(0,r)}G(x,y)dy &= \lim_{s\to r}\int_{\partial B(0,s)}G(x,y)dy = \lim_{s\to r}\left(-\frac{s}{(d-2)}+cds^{d-1}|B(0,1)|\right)\\ &= -\frac{r}{(d-2)}+cdr^{d-1}|B(0,1)| \end{align} thus $c = \frac{1}{d(d-2)r^{d-2}|B(0,1)|}$ and so \begin{align} \int_{\partial B(0,s)}G(x,y)dy = -\frac{s}{(d-2)}+\frac{1}{(d-2)r^{d-2}}s^{d-1} \end{align} We can then compute \begin{align} \int_{B(0,r)}G(x,y)dy &= \int_0^r\int_{\partial B(0,s)}G(x,y)dy ds = \int_0^r\left(-\frac{s}{(d-2)}+\frac{1}{(d-2)r^{d-2}}s^{d-1}\right)ds\\ &= -\frac{r^2}{2(d-2)}+\frac{1}{d(d-2)r^{d-2}}r^{d} = \frac{r^2}{d-2}\left(\frac{1}{d}-\frac{1}{2}\right) = \frac{r^2}{2d} . \end{align} That is close to the answer it is supposed to be but not quite it. Can someone give me a hint on how to correct my approach and how the dependency on $x$ comes in?