Solve the integral
$$I=\int_v(\nabla \cdot F dv)$$
Where $F=rr=xi+yj+zk$ And the volume $v$ is the volume is a sphere with the radius $R$ Placed in the origin.
How do i solve this directly and using the divergenxe theorem?
Would it be correct to use spherical coordinates where $dv$ is the jacobian and $r \in [0, R]$,$\theta \in [0, \pi]$ wouldn’t i have to use the divergence theorem in both solving it directly and one with div theorem? Since we have $\nabla \cdot F$?
If you evaluate the integral $I=\int_V \nabla\cdot F dV$ directly, then since $\nabla\cdot F=3$ you have $$ I=3\text{vol(V)}=3\cdot\frac43\pi R^3. $$
On the other hand, by the divergence theorem, $$ I = \int_{\partial V} F\cdot n \, dS\tag{1} $$ where the unit outer normal $$ n(x,y,z) = \frac{(x,y,z)}{R}\tag{2}. $$ Combining (1) and (2) you have $$ I = \frac{1}{R}\int_{\partial V} x^2+y^2+z^2 \, dS =\frac1R\int_{\partial V} R^2 \, dS =R\int_{\partial V} \, dS = R\cdot \text{area}(\partial V). $$ You can then calculating $\text{area}(\partial V)$ by using symmetries of the sphere: $$ \text{area}(\partial V) = 2\cdot \int_0^{2\pi}\int_0^R \frac{R}{\sqrt{R^2-r^2}} r\ drd\theta = 4\pi R^2.\tag{3} $$ To see how one can set up (3), one first calculates the normal vector for the parametric upper hemisphere $r(x,y) = (x,y, \sqrt{R^2-x^2-y^2})$ by $$ r_x = (1,0,\frac{-x}{z}),\quad r_y = (0,1,\frac{-y}{z}) $$ where $z = \sqrt{R^2-x^2-y^2}$. Now $$ r_x\times r_y = (\frac{x}{z},\frac{y}{z}, 1),\quad |r_x\times r_y|^2 = \frac{R^2}{z^2} . $$
So $$ \int_{S} \ dS = \iint_{D}|r_x\times r_y|\, dxdy = \iint_D\frac{R}{\sqrt{R^2-x^2-y^2}}\ dxdy. $$ Now you can use the polar coordinates.