Gauss divergence theorem

Let $V$ be a solid region in space bounded by a piecewise-smooth closed surface $S = \partial V$, oriented with outward unit normal $\hat{\mathbf n}$. If $\vec F$ has continuous first partial derivatives on an open region containing $V$, Gauss's divergence theorem states:

where: $\nabla\cdot\vec F = \dfrac{\partial F_1}{\partial x}+\dfrac{\partial F_2}{\partial y}+\dfrac{\partial F_3}{\partial z}$ is the divergence of $\vec F=(F_1,F_2,F_3)$, $dV$ is a volume element, $dS$ is a surface-area element, and $\oiint_S$ denotes integration over the entire closed surface.

Why it is useful.

• Closed surfaces made of several pieces (spheres, capped cylinders, cubes) usually force you to add up several separate surface integrals if you attack the flux directly; the divergence theorem collapses all of that bookkeeping into one triple integral over the solid.
• If $\nabla\cdot\vec F=0$ everywhere inside $V$ — a solenoidal field — the theorem instantly gives $\oiint_S\vec F\cdot\hat{\mathbf n}\,dS=0$ with no computation at all, since the volume integral of zero is zero.
• The direction also runs backwards: an awkward triple integral of a divergence can sometimes be replaced by an easier flux computation over the boundary, whichever side of the equation is simpler for the given problem.

Reading the physics. Divergence at a point measures how strongly the field is "spreading out" there — outward flux per unit volume, in the limit of a tiny box shrunk to that point. Gauss's theorem says that summing this local spreading-out over the whole solid must reproduce exactly the net flow crossing the outer boundary, because any flux leaving one small cell simply enters the neighbouring cell — those interior contributions cancel in pairs, leaving only the flux that crosses the outer skin $S$.

The goal is to confirm, by two entirely independent routes, that the theorem's two sides really do produce the same number for a concrete field and solid — the volume-integral side using $\nabla\cdot\vec F$ over the ball, and the surface-integral side using an outward-normal flux computation over the sphere — so both must be carried through in full and compared, not just one side computed and assumed to match.

Step 1. Take $\vec F(x,y,z)=x\,\hat\imath+y\,\hat\jmath+z\,\hat k$ and the solid ball $V:\ x^2+y^2+z^2\le a^2$ of radius $a$, boundary $S$ the sphere of radius $a$. Compute the divergence first: $\nabla\cdot\vec F=\dfrac{\partial x}{\partial x}+\dfrac{\partial y}{\partial y}+\dfrac{\partial z}{\partial z}=1+1+1=3$, a constant.

Step 2. Evaluate the volume-integral side. Since the divergence is the constant $3$ throughout $V$, $\displaystyle\iiint_V \nabla\cdot\vec F\,dV=3\iiint_V dV=3\cdot\left(\frac43\pi a^3\right)=4\pi a^3$, using the standard volume formula for a ball of radius $a$.

Step 3. Evaluate the surface-integral side independently. On the sphere $x^2+y^2+z^2=a^2$ the outward unit normal is $\hat{\mathbf n}=\frac1a(x,y,z)$, because the position vector is itself perpendicular to a sphere centred at the origin. So $\vec F\cdot\hat{\mathbf n}=\frac1a(x^2+y^2+z^2)=\frac1a\cdot a^2=a$, a constant value everywhere on $S$.

Step 4. Integrate that constant over the sphere's area $4\pi a^2$: $\displaystyle\oiint_S\vec F\cdot\hat{\mathbf n}\,dS=a\cdot4\pi a^2=4\pi a^3$. Both sides equal $4\pi a^3$ exactly, confirming the theorem on this solid and field.

Problem. Use Gauss's divergence theorem to find the total outward flux of $\vec F(x,y,z)=x^2\,\hat\imath+y^2\,\hat\jmath+z^2\,\hat k$ through the surface of the unit cube $V:\ 0\le x\le1,\ 0\le y\le1,\ 0\le z\le1$ (lengths in metres).

Step 1. Compute the divergence: $\nabla\cdot\vec F=\dfrac{\partial(x^2)}{\partial x}+\dfrac{\partial(y^2)}{\partial y}+\dfrac{\partial(z^2)}{\partial z}=2x+2y+2z$.

Step 2. Convert the flux into the triple integral guaranteed by the theorem: $\displaystyle\iiint_V(2x+2y+2z)\,dV=\int_0^1\int_0^1\int_0^1(2x+2y+2z)\,dx\,dy\,dz$, avoiding six separate face computations.

Step 3. By symmetry of the cube, $\iiint_V 2x\,dV=\iiint_V 2y\,dV=\iiint_V 2z\,dV$, so evaluate one term and triple it. For the $2x$ term, $\int_0^1\int_0^1\int_0^1 2x\,dx\,dy\,dz=\left(\int_0^1 2x\,dx\right)(1)(1)=[x^2]_0^1=1$, since the $y$ and $z$ integrations each contribute a factor of $1$.

Step 4. Add the three equal contributions: $1+1+1=3$.

The total outward flux through the surface of the cube is $\mathbf{3}$ (SI flux units). A direct six-face computation would take considerably longer than this single triple integral.

Competitive stretch

If instead $\vec F=x^2\hat\imath+y^2\hat\jmath+z^2\hat k$ were evaluated over the same cube but you were only asked for the flux through the single face $x=1$ (not the closed surface), the divergence theorem would not apply directly — it needs the entire closed boundary. That flux must be computed as a standalone surface integral, $\int_0^1\int_0^1 (1)^2\,dy\,dz=1$, illustrating that the theorem's shortcut is only available once the full closed surface is in play.

• Applying the theorem to a surface that is not closed — for example, only the curved part of a hemisphere without its flat disc base — Gauss's theorem needs the entire boundary $\partial V$, including any flat caps.
• Using an inward-pointing normal instead of the required outward normal, which flips the sign of the surface-integral side and gives $-4\pi a^3$ instead of $4\pi a^3$ in a sphere-type check.
• Forgetting that $\vec F$ must have continuous partial derivatives throughout the closed solid $V$, not merely on its boundary — a field singular at an interior point (such as $1/r^2$ at the origin of a ball containing the origin) cannot be plugged in directly.
• Computing $\nabla\cdot\vec F$ incorrectly by dotting components with the wrong variable — divergence is specifically $\partial F_1/\partial x+\partial F_2/\partial y+\partial F_3/\partial z$, matching each component to its own coordinate, never mixing them.

Ask the tutor: Want to try the divergence theorem on a cylinder with caps, or check your own flux computation step by step?