Given that $\textbf{F} = \langle \sqrt{x^2+y^2+z^2}, \sqrt{x^2+y^2+z^2}, \sqrt{x^2+y^2+z^2} \rangle $ and E is the volume described by $0 \le z \le \sqrt{1-x^2-y^2}$
I'm trying to use the divergence theorem to compute the flux. $$\iint_{D} \textbf{F} \cdot \textbf{N} \: dS = \iiint_E \nabla \cdot \textbf{F}\:d\textbf{V}$$
Attempt:
$$\text{div}\textbf{F} \:=\: \nabla \cdot \textbf{F} =\frac{x}{2\sqrt{x^2+y^2+z^2}} + \frac{y}{2\sqrt{x^2+y^2+z^2}} + \frac{z}{2\sqrt{x^2+y^2+z^2}} \\ =\frac{p \sin\phi \cos\theta}{2p} + \frac{p \sin\phi \sin\theta}{2p} + \frac{p cos\phi}{2p} \\ = \frac{1}{2}(\sin\phi \cos\theta + \sin\phi \sin\theta + \cos\phi) $$
$\text{d}\textbf{V} = p^2 \sin\phi$
$$\iiint_E \nabla \cdot \textbf{F}\:d\textbf{V} \\ = \frac{1}{2} \int_0^\frac{\pi}{2} \int_0^{2\pi} \int_0^{\sqrt{1-(x^2+y^2)} =\sqrt{1-p^2}?} (\sin\phi \cos\theta + \sin\phi \sin\theta + \cos\phi) \: p^2 \sin\phi \:dz\: d\theta \: d\phi$$
This seems to be a complicated integral. Is my steps/logic correct thus far? How can I think more qualitatively to simplify the integral further? Would appreciate some guidance!
PS. The answer given was $\frac{\pi}{3}$
First, there is no $1/2$ factor in $\nabla \cdot \textbf{F}$. After fixing the integration bounds, the answer is straightforward. $$ \begin{aligned} &\iiint_E \nabla \cdot \textbf{F}\:d\textbf{V} \\ &= \int_0^\frac{\pi}{2} \int_0^{2\pi} \int_0^{1} (\sin\phi \cos\theta + \sin\phi \sin\theta + \cos\phi) \: p^2 \sin\phi \:dp\: d\theta \: d\phi \\ &= \int_0^{1} \: p^2 \:dp \int_0^\frac{\pi}{2} \int_0^{2\pi} (\sin\phi \cos\theta + \sin\phi \sin\theta + \cos\phi) \sin\phi\: d\theta \: d\phi \\ &= \frac{1}{3} \int_0^\frac{\pi}{2} \int_0^{2\pi} ( \cos\phi) \sin\phi\: d\theta \: d\phi \\ &= \frac{2\pi}{3} \int_0^\frac{\pi}{2} ( \cos\phi) \sin\phi\: \: d\phi \\ &= \frac{\pi}{3} \end{aligned} $$
As a double check, do the surface integration directly. Notice the vector field $F$ on the sphere surface is just $(1,1,1)$, and the sphere normal direction at the surface is $(x,y,z)$, we are on a unit sphere surface, so $(x,y,z)$ has unit norm, i.e., it is the normalized normal vector. Then the surface integration is
$$ \oint \vec{F} \vec{n} dS = \oint(1,1,1)\cdot(x,y,z) p^2 \sin(\theta) d\theta d\phi \\ =\oint (x+y+z) \sin\theta d\theta d\phi \\ = \int_0^\frac{\pi}{2} \int_0^{2\pi} (\sin\phi \cos\theta + \sin\phi \sin\theta + \cos\phi) \sin\phi\: d\theta \: d\phi \\ = \int_0^\frac{\pi}{2} \int_0^{2\pi} ( \cos\phi) \sin\phi\: d\theta \: d\phi \\ = \pi $$ Oops! why they are different? The reason is we have a bottom surface to consider as well! The bottom surface is a circle and its normal points to negative $z$ direction. The integration for this bottom surface is $$ \begin{aligned} \quad & \oint (\sqrt{x^2+y^2},\sqrt{x^2+y^2},\sqrt{x^2+y^2}) \cdot (0, 0, -1) dS \\ & = \oint -\sqrt{x^2+y^2} dS \text{switch to 2D polar coordinate} \\ & = -\int_0^{2\pi}\int_0^1 r \cdot r dr d\theta \\ & = -\frac{2\pi}{3} \end{aligned} $$
Sum the two parts gives $\pi/3$.