Comparing Flux on Closed and Open Surfaces

Why an Open Surface Needs a Boundary Around It

The divergence theorem relates flux through a closed boundary to the divergence of a vector field throughout the enclosed volume. Many surfaces of interest are open: a curved sheet, a cap, or a patch cut from a larger surface has an edge and does not enclose a volume by itself.

To use the theorem on an open patch, add one or more surfaces that fill its boundary and create a closed surface. The flux through the patch is then the total flux through that closed surface minus the flux through every added piece. The signs depend on orientation, so the added pieces cannot be ignored or assigned normals independently.

Key idea: If an open patch $S$ and added pieces $C_1,\ldots,C_m$, with compatible orientations, form the outward boundary of a solid $V$, then $$ \iint_S \mathbf{F}\cdot\mathbf{n}\,dS = \iiint_V \nabla\cdot\mathbf{F}\,dV - \sum_{j=1}^{m}\iint_{C_j}\mathbf{F}\cdot\mathbf{n}\,dS. $$

Closing Surfaces and Orientation

An open surface patch has a boundary curve or curves. A closing surface is a surface, or collection of surfaces, added so that the patch and the added pieces together form the boundary of a bounded solid. To apply the divergence theorem, the normals on all these pieces must point outward from that same solid.

For adjacent pieces, this outward orientation gives opposite directions along their shared edges when the surfaces are viewed with induced boundary orientations. This compatibility is the geometric reason the pieces fit together as one closed boundary. In a flux calculation, the essential requirement is that each surface normal agree with the outward direction from the enclosed solid.

Divergence theorem: If $V$ is a bounded solid with a piecewise smooth closed boundary $\partial V$, and $\mathbf{F}$ has continuous first partial derivatives on an open region containing $V$, then, with $\mathbf{n}$ outward, $$ \iint_{\partial V}\mathbf{F}\cdot\mathbf{n}\,dS = \iiint_V \nabla\cdot\mathbf{F}\,dV. $$

If the given patch is oriented inward relative to the solid, first reverse its orientation or include a minus sign for its flux. The formula above assumes that $S$ and all $C_j$ are oriented outward relative to $V$. If the field is singular in $V$, or the boundary and field do not satisfy the theorem's regularity conditions, the theorem cannot be applied as stated.

Why Subtracting the Added Flux Works

The boundary of the completed solid is the union of the patch and all the added pieces. Flux is additive over such pieces: the total boundary flux is the sum of the fluxes through each one. The divergence theorem gives that total as the volume integral of the divergence. Isolating the flux through the original patch gives the subtraction formula.

Complete the boundary.
Choose added surfaces so that the original patch and additions bound a solid $V$.

Orient every piece outward.
Determine the outward normal for the same solid on the patch and on each added piece.

Find the total closed flux.
Evaluate $\iiint_V\nabla\cdot\mathbf{F}\,dV$, provided the divergence theorem applies.

Remove the added fluxes.
Subtract the flux through every added piece to obtain the flux through the original patch.

Worked Examples

Worked Example: A Paraboloid Patch Closed by a Disk

Let $S$ be the part of $z=4-x^2-y^2$ above the disk $x^2+y^2\leq 1$, oriented upward. Take $\mathbf{F}=(x,y,z)$. Find the flux through $S$ using a closing surface.

The surface meets the plane $z=3$ at $x^2+y^2=1$. Add the disk $C$ in $z=3$. The enclosed solid lies above the disk and below the paraboloid. Its outward normal on $S$ points upward, as specified, and its outward normal on $C$ points downward.

The divergence is $\nabla\cdot\mathbf{F}=1+1+1=3$. In polar coordinates, the volume is

\operatorname{Vol}(V) = \int_0^{2\pi}\int_0^1\bigl((4-r^2)-3\bigr)r\,dr\,d\theta = \int_0^{2\pi}\int_0^1(1-r^2)r\,dr\,d\theta = 2\pi\left(\frac12-\frac14\right) = \frac{\pi}{2}.

Thus the total outward flux is $3(\pi/2)=3\pi/2$. On the closing disk, the outward normal is $(0,0,-1)$, so $\mathbf{F}\cdot\mathbf{n}=-3$. Its flux is $-3\pi$, since the disk has area $\pi$. Therefore,

\iint_S\mathbf{F}\cdot\mathbf{n}\,dS = \frac{3\pi}{2}-(-3\pi) = \frac{9\pi}{2}.

As a check, the upward vector area element for the graph is $(2x,2y,1)\,dx\,dy$. Its dot product with $(x,y,4-x^2-y^2)$ is $4+x^2+y^2$. Hence direct integration gives

\int_0^{2\pi}\int_0^1(4+r^2)r\,dr\,d\theta = 2\pi\left(2+\frac14\right) = \frac{9\pi}{2},

agreeing with the closed-surface calculation.

Worked Example: A Half-Cylinder with Three Added Pieces

Let $S$ be the curved half-cylinder $x^2+y^2=1,\ y\geq 0,\ 0\leq z\leq 2$, oriented outward from the half-cylinder solid. For $\mathbf{F}=(x,y,z)$, find its flux by closing the surface and accounting for every addition.

The solid is the half-cylinder $x^2+y^2\leq1,\ y\geq0,\ 0\leq z\leq2$. Besides $S$, its boundary has a rectangular cut face at $y=0$, a top semicircle at $z=2$, and a bottom semicircle at $z=0$. The outward normals on these additions are, respectively, $(0,-1,0)$, $(0,0,1)$, and $(0,0,-1)$.

Since $\nabla\cdot\mathbf{F}=3$ and the half-cylinder has volume $\frac12\pi(1)^2(2)=\pi$, its total outward flux is $3\pi$. On the cut face, $y=0$, so $\mathbf{F}\cdot(0,-1,0)=-y=0$; its flux is zero. On the top face, $\mathbf{F}\cdot(0,0,1)=z=2$, and the semicircle has area $\pi/2$; its flux is $2(\pi/2)=\pi$. On the bottom face, $z=0$, so the flux is zero.

Subtracting all three added fluxes gives

\iint_S\mathbf{F}\cdot\mathbf{n}\,dS = 3\pi-(0+\pi+0) = 2\pi.

To check, parametrize the curved surface by $(\cos\theta,\sin\theta,z)$, where $0\leq\theta\leq\pi$ and $0\leq z\leq2$. Its outward vector area element is $(\cos\theta,\sin\theta,0)\,d\theta\,dz$. The dot product with $\mathbf{F}$ is $\cos^2\theta+\sin^2\theta=1$, so the direct flux is

\int_0^2\int_0^\pi 1\,d\theta\,dz=2\pi.

The agreement confirms that none of the added boundary pieces was omitted.

Worked Example: Two Closures for the Same Hemisphere

Let $S$ be the upper unit hemisphere $x^2+y^2+z^2=1,\ z\geq0$, with outward orientation, and let $\mathbf{F}=(x,y,z)$. We will calculate its flux using two different closing surfaces.

First close $S$ with the unit disk $z=0$. The solid is the upper half-ball, with volume $2\pi/3$, and $\nabla\cdot\mathbf{F}=3$. Its total outward flux is $3(2\pi/3)=2\pi$. The disk's outward normal points downward, and $\mathbf{F}\cdot(0,0,-1)=-z=0$ there. The disk flux is zero, so the flux through $S$ is $2\pi-0=2\pi$.

Now close the same $S$ with the paraboloid $C\colon z=1-x^2-y^2$, over the unit disk, oriented downward. The solid lies between the hemisphere and this paraboloid. In polar coordinates its volume is

\operatorname{Vol}(V) = 2\pi\int_0^1\left(\sqrt{1-r^2}-(1-r^2)\right)r\,dr = 2\pi\left(\frac13-\frac14\right) = \frac{\pi}{6}.

The total outward flux is therefore $3(\pi/6)=\pi/2$. For the downward orientation of the graph, a vector area element is $(-2x,-2y,-1)\,dx\,dy$. Its dot product with $\mathbf{F}=(x,y,1-x^2-y^2)$ is $-1-x^2-y^2$. Thus

\iint_C\mathbf{F}\cdot\mathbf{n}\,dS = \int_0^{2\pi}\int_0^1(-1-r^2)r\,dr\,d\theta = 2\pi\left(-\frac12-\frac14\right) = -\frac{3\pi}{2}.

Subtract this added flux from the total:

\iint_S\mathbf{F}\cdot\mathbf{n}\,dS = \frac{\pi}{2}-\left(-\frac{3\pi}{2}\right) = 2\pi.

Both closures give the same flux through $S$. The enclosed volumes and the added-surface fluxes differ, but each calculation accounts for the complete closed boundary.

How the Choice of Closure Helps

A closing surface is a computational choice, not part of the original patch. One closure may produce an easy volume integral and a simple added-surface flux; another may make one of those calculations harder. The result for the patch is unchanged when each closure is oriented consistently and the theorem's assumptions hold.

In practice, list every piece of the completed boundary before calculating. For each piece, record its outward normal and its flux. This is particularly important when the closure has several caps, side faces, or internal-looking cuts that are actually part of the boundary of the chosen solid.

Common Orientation Mistake

In the second hemisphere calculation, the paraboloid must be oriented downward because it is the bottom of the solid. A common error is to use the upward vector area element $(2x,2y,1)\,dx\,dy$ instead. That gives the paraboloid flux $+3\pi/2$, then the incorrect patch flux $\pi/2-3\pi/2=-\pi$. The error is not in the divergence theorem: the chosen normal points into the solid, not outward from it. Reversing the normal changes the added flux to $-3\pi/2$, giving the correct patch flux $2\pi$.

Check before subtracting: Confirm that the original patch and every added piece use outward normals for one and the same solid. A flux value with the wrong orientation has the wrong sign in the subtraction.

Check Your Understanding

For each question, identify the solid and the orientations before setting up a flux calculation.

If a patch and two added pieces form an outward boundary, how do you express the patch flux in terms of the total divergence integral and the two added fluxes?
A horizontal disk is the bottom of a solid. Which direction does its outward normal point?
For the half-cylinder example, what would the curved-surface flux be if the top semicircle were accidentally omitted from the subtraction?
What regularity and region conditions are required to apply the divergence theorem?
Why can two different valid closing surfaces give the same flux through the original patch?