Divergence Theorem to find the flux of a field

Question

Compute the flux of the field $(ye^z, 16xz, \arctan(\frac{z}{\sqrt{x^2+y^2}}))$ out of region $E$ that lies between spheres $x^2+y^2+z^2= 1$ and $x^2+y^2+z^2= 16$ in first octant.

I first did $\frac{\mathrm{d}P}{\mathrm{d}x}=0, \frac{\mathrm{d}Q}{\mathrm{d}y}=0$ and $\frac{\mathrm{d}R}{\mathrm{d}z}= \frac{1}{x^2+y^2+z^2}$, but I am not sure how to set up the integral. I thought the limits would be $0$ to $2\pi$ for the outer integral and for the inner integral $1$ to $4$. Would someone be able to see if I am going the right way?

Answer 1

$\nabla \cdot (e^{yz}, 16xz, \arctan \frac {z}{\sqrt {x^2 + y^2}}) = \frac {\sqrt{x^2 + y^2}}{x^2 + y^ +z^2}$

Convert to spherical
$x = \rho\cos\theta\sin\phi\\ y = \rho\sin\theta\sin\phi\\ z = \rho\cos\phi\\ dx\ dy\ dz = \rho^2 \sin\phi\ d\rho\ d\phi\ d\theta\\ \frac {\sqrt{x^2 + y^2}}{x^2 +y^2 + z^2} = \frac {\sin \phi}{\rho}$

$\int_0^{\frac {\pi}{2}}\int_0^\frac {\pi}{2}\int_1^4 \frac {\sin\phi}{\rho} \rho^2 \sin \phi \ d\rho\ d\phi\ d\theta$

$\int_0^{\frac {\pi}{2}}\int_0^\frac {\pi}{2}\frac {15}{4} (1-\cos 2\phi) \ d\phi\ d\theta$

$\int_0^{\frac {\pi}{2}}\int_0^\frac {\pi}{2}\frac {15}{4} \frac {\pi}{2} \ d\theta$

$\frac {15}{16}\pi^2$

Answer 2

I think I sorted it out.

$R=\arctan(\frac{z}{\sqrt{x^2+y^2}})$

$\frac{dR}{dz}=\frac{1}{\sqrt{x^2+y^2}}\frac{1}{1+\frac{z^2}{x^2+y^2}}=\frac{\sqrt{x^2+y^2}}{x^2+y^2+z^2}=\frac{ \sin \phi}{\rho}$

First octant: $1\le \rho\le 4, 0\le\phi\le \pi/2, 0\le \theta \le \pi/2$

$\int \nabla \cdot \vec{E} d\tau = \int_1^4 \int_0^{\pi/2}\int_0^{\pi/2}\frac{\sin \phi}{\rho} \rho^2\sin\phi d \rho d\phi d \theta=(\frac{15}{2})\cdot \frac{\pi}{2}\cdot (\frac{\pi}{4})=\frac{15\pi^2}{16}$