Calculate the vector surface integral Let $V=\{(x,y,z)\in \mathbb{R}:\frac{1}{4}\le x^2+y^2+z^2\le1\}$ and $\vec{F}=\frac{x\hat{i}+y\hat{j}+z\hat{k}}{(x^2+y^2+z^2)^2}$ for $(x,y,z)\in V$. Let $\hat{n}$ denote the outward unit normal vector to the boundary of $V$ and $S$ denote the part $\{(x,y,z)\in \mathbb{R}^3:x^2+y^2+z^2=\frac{1}{4}\}$ of the boundary $V$. Then find $\int\int_S\vec{F}.\hat{n}dS$.
Here is what I did.
$$\int\int_S\vec{F}.\hat{n}dS$$
$$=\int\int_S\frac{x\hat{i}+y\hat{j}+z\hat{k}}{(x^2+y^2+z^2)^2}.(2x\hat{i}+2y\hat{j}+2z\hat{k})dxdy$$
$$\int\int_S2(x^2+y^2+z^2)^{-1}dxdy$$
$$\int\int_S\frac{2}{1/4}dxdy=8\int\int_Sdxdy=8\pi$$
where $\int\int_Sdxdy$ is the surface area of a sphere of radius $1/2$.
Please advise on my solution. 
 A: Edit
Your answer will become right if you you replace $\hat{n}$ by -$\hat{n}$ because the normal unit vector on the inner surface will be directed towards origin.So the answer will be -8$\pi$
For The Calculation of integral over the full Boundary
${\vec{F}=\frac{x\hat{i}+y\hat{j}+z\hat{k}}{(x^{2}+y^{2}+z^{2})^{2}}}$ 
$\nabla.$$\vec{F}$ = $\frac{-1}{\left(x^{2}+y^{2}+z^{2}\right)^{2}}$
From Gauss Divergence Theorem ${\int\int_{S}\vec{F}.\hat{n}dS}$
= ${\int\int\int_{V}\vec{\nabla.F}dV}$=$\int\int\int_{V}\frac{1}{\left(x^{2}+y^{2}+z^{2}\right)^{2}}dxdydz$=8$\int\int\int_{V_{1}}\frac{1}{\left(x^{2}+y^{2}+z^{2}\right)^{2}}dxdydz$
{since integral is symmetrical}
Let x= u$^{\frac{1}{2}}$$\Longrightarrow dx=\frac{1}{2}u^{\frac{1}{2}-1}$
       $du$
y= v$^{\frac{1}{2}}$$\Longrightarrow dy=\frac{1}{2}v^{\frac{1}{2}-1}$  $dv$
z= w$^{\frac{1}{2}}$$\Longrightarrow dz=\frac{1}{2}w^{\frac{1}{2}-1}$  $dw$
$\int\int\int_{V}\frac{1}{\left(x^{2}+y^{2}+z^{2}\right)^{2}}dxdydz$
=$\frac{8}{8}$ $\int\int\int_{V_{1}}\frac{u^{\frac{1}{2}-1}v^{\frac{1}{2}-1}w^{\frac{1}{2}-1}}{\left(u+v+w\right)^{2}}dudvdz$
and $\frac{1}{4}\leq u+v+w\leq1$
Using Liouville’s Extension of Dirichlet’s Theorem
$\int\int\int_{V_{1}}\frac{u^{\frac{1}{2}-1}v^{\frac{1}{2}-1}w^{\frac{1}{2}-1}}{\left(u+v+w\right)^{2}}dudvdz$
=$\frac{\Gamma\frac{1}{2}.\Gamma\frac{1}{2}\Gamma\frac{1}{2}}{\Gamma\frac{3}{2}}\int\frac{1}{t^{2}}t^{\frac{1}{2}}dt$
= 2$\pi$$\left[\frac{t^{-\frac{1}{2}}}{-\frac{1}{2}}\right]_{\frac{1}{4}}^{1}$=
-4$\pi$$\left[\frac{1}{1}-\frac{1}{\sqrt{\frac{1}{4}}}\right]$=
-4$\pi$ $\boldsymbol{Ans.}$
