# 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 $1/2$.

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$ $\boldsymbol{Ans.}$