# Computing Flux (Divergence Theorem)

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 $\frac{\pi}{3}$

• Other problem is you're conflating cylindrical and spherical coordinates. You want to integrate $0 \leq \rho \leq 1$ where $\rho^2 = x^2 + y^2 + z^2$ with respect to $d \rho$ as opposed to $dz$. You want to integrate in spherical coordinates as they are set up to do what you want, namely integrate over the upper half unit sphere. – Chris K Apr 5 '17 at 4:25
• Yes. but would $0 \le \rho \le 1$ be the right bound in this case...? hmm – misheekoh Apr 5 '17 at 4:30
• Yes, you want $1$ for an upper bound of $\rho$. Note that $\rho = \sqrt{x^2+y^2+z^2}$. In cylindrical coordinates, $r = \sqrt{x^2+y^2}$, but that is not what we want here. – Chris K Apr 5 '17 at 4:31

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 $\pi/3$.

• Can you explain to why the bound of $\rho$ is between $0 \: \text{and} \:1$ in this case..? – misheekoh Apr 5 '17 at 4:41
• @misheekoh because you are integrating over the (half) unit sphere's volume, to cover this volume the radius has to change from $0$ to $1$. – Taozi Apr 5 '17 at 4:47
• @misheekoh I added to the answer a direct evaluation of the surface flux integration. – Taozi Apr 5 '17 at 5:03
• Thank you. Btw, I do understand that we're integrating over the hemisphere but particularly for when $\text{z} \le \sqrt{1-x^2-y^2}$, how did u manage to convert the upper bound to $\rho \le 1$? – misheekoh Apr 5 '17 at 5:06
• @misheekoh Square both sides of this inequality to have $z^2 \le 1 - x^2 -y^2$, rewrite as $x^2 +y^2+z^2 \le 1$, since $x^2 +y^2+z^ = \rho^2$, this is just $\rho^2 \le 1$, or $\rho\le 1$. – Taozi Apr 5 '17 at 5:09