I was working a tutorial and it had this proof listed below. It says that S is a closed surface and H is a region $$\int_S \frac{\textbf{r.n}}{r^2} dS\, = \int_H \frac{dH}{r^2} \,$$
My approach towards this question was that r = xi +yj + zk then if we try to find n it is equal to
$$\frac{grad{\phi}}{\mid grad{\phi}\mid} = \frac{(1,1,1)}{\sqrt{3}}$$
Therefore if we open dS, we would end up with the following
$$\int_{S'} \frac{\textbf{r.n}}{{r^2}}\frac{{dS'}}{{\textbf{n.k}}} \,$$
If we take k to be equal to (0,0,1), we would end up with the following equation,
$$\int_{S'} \frac{{\textbf{r}}.(1,1,1)}{{r^2}}\frac{{dS'}}{{1}} \,$$
Then r could be changed to (x,y,z)
$$\int_{S'} \frac{{(x,y,z)}.(1,1,1)}{{r^2}}dS' \,$$
Therefore
$$\int_{S'} \frac{{(x,y,z)}}{{r^2}}dS' \,$$
Then changing for region H
$$\iiint_H \frac{{(x,y,z)}}{{r^2}}dxdydz \,$$
is equal to
$$\int_{H} \frac{{dH}}{{r^2}} \,$$
Can somebody please check if my approach is correct?
Hint