Exercises — Divergence theorem (Gauss's theorem) — statement, flux-divergence connection

WHAT: Add the three "matching" partial derivatives. $\nabla \cdot \mathbf{F}=\frac{\partial (3x)}{\partial x}+\frac{\partial (-2y)}{\partial y}+\frac{\partial (5z)}{\partial z}=3+(-2)+5=6.$ WHY each term: ${\partial }_x$ only sees the first component $P=3x$ (its slope in $x$ is $3$); the middle ${\partial }_y(-2y)=-2$; the last ${\partial }_z(5z)=5$. The cross-terms (like ${\partial }_x$ of $-2y$) are zero because $-2y$ has no $x$ in it. Answer: $\nabla \cdot \mathbf{F}=6$, a constant — the field pumps out at the same rate everywhere.

Answer: (a). WHY: The theorem needs a closed surface $S=\partial V$ — a complete skin enclosing a solid $V$. (b) is a line integral of flow along a curve — that's circulation, handled by Stokes' theorem, not this one. (c) is an open surface (a bowl with no lid): it doesn't bound a solid by itself, so the theorem doesn't apply until you add a closing disk.

WHAT: Use the easy (volume) side. $\nabla \cdot \mathbf{F}=1+1+1=3$. $\Phi ={\iiint }_{V}3dV=3\cdot \text{Vol}=3\cdot \frac{4}{3}\pi R^3=4\pi R^3.$ WHY the volume side: the divergence is a constant $3$, so the triple integral is just $3\times$(volume of the ball) — no surface parameterisation at all. Answer: $\Phi =4\pi R^3$. (Check $R=1$ gives $4\pi$, matching the parent's Example 1.)

WHAT: $\nabla \cdot \mathbf{F}=3x^2+3y^2+3z^2$. Integrate over the cube. ${\iiint }_{[0,1{]}^3}(3x^2+3y^2+3z^2)dV.$ By symmetry the three terms are equal, so compute one and triple it: $3{\int }_0^1{\int }_0^1{\int }_0^1{x}^{2}dxdydz\times 3=3\cdot \left(\frac{1}{3}\cdot 1\cdot 1\right)\times 3.$ WHY: ${\int }_0^1{x}^{2}dx=\frac{1}{3}$; the $y$ and $z$ integrals each give $1$ (integrating $1$ over $[0,1]$). One term $=3\cdot \frac{1}{3}=1$; three identical terms $=3$. Answer: $\Phi =3$.

$\nabla \cdot \mathbf{F}=2+3-1=4$ (constant). $\Phi =4\cdot \frac{4}{3}\pi (1{)}^3=\frac{16\pi }{3}.$ Answer: $\frac{16\pi }{3}$.

WHAT: $\nabla \cdot \mathbf{F}=1+2+3=6$. The volume of the cylinder is $\pi r^{2}h=\pi (2{)}^2(5)=20\pi$. $\Phi =6\cdot 20\pi =120\pi .$ WHY this beats the surface way: the cylinder has three pieces of skin — a curved wall, a bottom disk, a top disk — each needing its own normal and integral. The divergence theorem collapses all three into one multiplication. Answer: $\Phi =120\pi$.

Volume side: $\nabla \cdot \mathbf{F}=1$, so ${\iiint }_{V}1dV=\text{Vol}=1$.

Surface side (look at the figure): Only faces with a nonzero $x$-normal contribute, because $\mathbf{F}=(x,0,0)$ points purely in $x$.

• Face at $x=1$ (outward normal $+\hat{x}$): $\mathbf{F}\cdot \hat{\mathbf{n}}=x=1$, area $=1$ ⟹ flux $+1$.
• Face at $x=0$ (outward normal $-\hat{x}$): $\mathbf{F}\cdot \hat{\mathbf{n}}=-x=0$ ⟹ flux $0$.
• The four side faces have normals $\pm \hat{y},\pm \hat{z}$, and $\mathbf{F}$ has no $y$ or $z$ part, so $\mathbf{F}\cdot \hat{\mathbf{n}}=0$.

Surface total $=1+0=1$. Both sides equal $1$. ✓ WHY it works: the field grows in the $x$-direction; only the front face at $x=1$ actually feels flow leaving, and it leaves at rate $1$ over unit area.

$\nabla \cdot \mathbf{F}={\partial }_x(y)+{\partial }_y(z)+{\partial }_z(x)=0+0+0=0.$ So $\Phi ={\iiint }_{V}0dV=0$ for every closed surface (with no singularity — this field is smooth everywhere). WHY: each component depends on a different variable than the derivative that would catch it. This is a solenoidal (divergence-free) field: it shuffles stuff around without creating or destroying any. Answer: $\Phi =0$.

WHAT: $\nabla \cdot \mathbf{F}=3x^2+3y^2+3z^2=3(x^2+y^2+z^2)=3r^2$ in spherical coordinates, where $r$ is distance from the origin. WHY spherical: the divergence depends only on $r^2$, so the natural volume element is a thin spherical shell of radius $r$, thickness $dr$, and area $4\pi r^2$: $dV=4\pi r^{2}dr.$

WHAT: $\nabla \cdot \mathbf{F}={\partial }_x(xz)+{\partial }_y(yz)+{\partial }_z(z^2)=z+z+2z=4z.$ WHY volume side: integrating $4z$ over the solid hemisphere is a single triple integral; the surface has a curved dome and a flat disk lid — two pieces we'd rather avoid. Use spherical shells but weighted by height: for the hemisphere, $\iiint 4zdV$. Write $z=r\cos \varphi$, $dV=r^2\sin \varphi drd\varphi d\theta$ with $r\in [0,1]$, $\varphi \in [0,\pi /2]$, $\theta \in [0,2\pi ]$: $\Phi ={\int }_0^{2\pi }{\int }_0^{\pi /2}{\int }_0^{1}4(r\cos \varphi )r^2\sin \varphi drd\varphi d\theta .$ Split the pieces: $=4\left({\int }_0^{2\pi }d\theta \right)\left({\int }_0^{\pi /2}\cos \varphi \sin \varphi d\varphi \right)\left({\int }_0^1{r}^{3}dr\right).$

• ${\int }_0^{2\pi }d\theta =2\pi$.
• ${\int }_0^{\pi /2}\cos \varphi \sin \varphi d\varphi =\frac{1}{2}$ (since $\int \sin \varphi d(\sin \varphi )=\frac{1}{2}{\sin }^2\varphi {|}_0^{\pi /2}=\frac{1}{2}$).
• ${\int }_0^1{r}^{3}dr=\frac{1}{4}$.

$\Phi =4\cdot 2\pi \cdot \frac{1}{2}\cdot \frac{1}{4}=\pi .$ Answer: $\Phi =\pi$.

Naive (wrong) path: for $r\ne 0$ one can show $\nabla \cdot \mathbf{F}=0$, tempting us to say $\Phi =0$. Why that's wrong: the field is singular at the origin — $|\mathbf{r}{|}^3\to 0$ there, so $\mathbf{F}$ and its partials are undefined at that point. The theorem requires continuous partials throughout $V$, and the origin lives inside the unit ball. The hypothesis is violated, so we cannot apply the theorem over the whole ball. Correct path — compute the surface directly. On the unit sphere $|\mathbf{r}|=1$, so $\mathbf{F}=\mathbf{r}$ and $\hat{\mathbf{n}}=\mathbf{r}$ (outward radial). Thus $\mathbf{F}\cdot \hat{\mathbf{n}}=\frac{\mathbf{r}\cdot \mathbf{r}}{|\mathbf{r}{|}^3}=\frac{|\mathbf{r}{|}^2}{|\mathbf{r}{|}^3}=\frac{1}{|\mathbf{r}|}=1.$ $\Phi ={\iint }_{S}1dS=\text{area}=4\pi .$ Answer: $\Phi =4\pi$, not $0$. This is exactly the Gauss's law behaviour — a point charge at the origin produces flux $4\pi$ regardless of sphere size.

WHAT: Let $V_{\epsilon }$ = shell $\epsilon \le |\mathbf{r}|\le 1$. Now $\mathbf{F}$ is smooth everywhere in $V_{\epsilon }$ (no origin), so the theorem applies: ${\iint }_{\partial V_{\epsilon }}\mathbf{F}\cdot \hat{\mathbf{n}}dS={\iiint }_{V_{\epsilon }}\underset{=0}{\underbrace{\nabla \cdot \mathbf{F}}}dV=0.$ WHY the boundary has two pieces: $\partial V_{\epsilon }$ consists of the outer unit sphere (outward normal $+\hat{r}$) and the inner sphere of radius $\epsilon$ with normal pointing into the hole, i.e. $-\hat{r}$ (outward from the shell means toward the origin). $\underset{4\pi }{\underbrace{{\Phi }_{\text{outer}}}}+{\Phi }_{\text{inner}}=0\Rightarrow {\Phi }_{\text{inner}}=-4\pi .$ On the inner sphere with the outward-from-origin normal $+\hat{r}$, the flux is $+4\pi$ (same computation as 5.1 with radius $\epsilon$: $\mathbf{F}\cdot \hat{r}=1/{\epsilon }^2$ times area $4\pi {\epsilon }^2=4\pi$). With the shell's outward normal $-\hat{r}$ it flips sign to $-4\pi$. Everything is consistent. Answer: shell flux $=0$; the "missing" $4\pi$ is hiding on the inner sphere around the singularity. This is the standard trick from the continuity / Gauss's-law derivations.

WHAT: We are handed the divergence directly, so go straight to the volume side. In spherical shells $\nabla \cdot \mathbf{F}=r^2$, $dV=4\pi r^{2}dr$: $\Phi ={\int }_0^2{r}^2\cdot 4\pi r^{2}dr=4\pi {\int }_0^2{r}^{4}dr=4\pi \cdot \frac{2^5}{5}=4\pi \cdot \frac{32}{5}=\frac{128\pi }{5}.$ WHY shells: the divergence depends only on distance $r$ from the centre, so slicing into constant-$r$ shells makes the integrand constant on each shell — the cleanest possible decomposition. Answer: $\Phi =\frac{128\pi }{5}$.