I am trying to use the integration by parts formula in spherical coordinates $$\int_\Omega \frac{\partial u}{\partial x_i} v d \Omega = \int_\Gamma u v \nu_i d\Gamma-\int_\Omega u \frac{\partial v}{\partial x_i} d \Omega, $$ with $\nu$ a unit vector normal to the boundary $\Gamma$, but I must be misunderstanding something.
Toy problem of integrating over unit ball: $$\int_0^{2 \pi} \int_0^\pi \int_0^1 r \cdot 1 (r^2 \sin \theta dr d\theta d\phi)=\pi. $$ Try integration by parts for the above with $\frac{\partial u}{\partial r}=1$ and $v=r$. The unit normal to the disk is $\nu=\frac{\partial}{\partial r}$ so $\nu_i=1$. At the boundary, $r=1$ so I obtain $$2\int_0^{2 \pi} \int_0^\pi \int_0^1 r (r^2 \sin \theta dr d\theta d\phi)=\int_{S^2} dS=4\pi.$$ This does not match what the answer should be! Where did I go wrong in trying to use the formula?
I don't understand how you do the integration by part, you're missing two powers of r that comes from the integral measure so you should write $\nu = r^3$ and how about the fully integrated part, sure for r=0 this is 0 but you still have for r=1 and so you would write $$ \int_0^{2\pi}\int_0^{\pi} sin\theta d\theta d\phi \left( \Big[~~r^4~~\Big]_0^1-\int_0^1 3r^3 dr \right) =4\pi(1-3/4 )=\pi $$