Why Add a Cap?
Flux through an open surface can be difficult to calculate directly, especially when the surface is curved. If the surface is part of the boundary of a simple solid, a useful strategy is to add another surface along its boundary curve. The added piece, called a cap, completes the open surface into a closed boundary.
The Divergence Theorem applies to the completed closed boundary. Its total outward flux is the sum of the flux through the original surface and the flux through the cap. Therefore, when the cap flux is known or easy to compute, the flux through the original surface follows by subtraction.
Conditions and Orientation
Suppose a vector field \(\mathbf{F}\) is continuously differentiable on an open region containing a solid \(E\). Assume the boundary \(\partial E\) is closed and piecewise smooth, and consists of the target surface \(S\) and one or more cap pieces \(C\). The Divergence Theorem states that the total outward flux through this boundary equals the integral of the divergence over the solid.
Here, \(\mathbf{n}\) is the outward unit normal on each piece of the boundary. Since the boundary is made of \(S\) and \(C\), its flux is the sum of their fluxes. Rearranging gives the cap-completion formula.
This identity is a direct consequence of the Divergence Theorem and the additivity of surface integrals over boundary pieces. The orientations matter: both fluxes on the right use normals directed outward from the same solid \(E\). The cap’s normal is not chosen just to point in a convenient direction, nor is it necessarily the direction suggested by the way the cap is drawn.
A valid cap must meet the edge of \(S\) so that their union forms the boundary of a solid, with no gaps or unintended overlaps. The field must also be regular throughout that solid. If the field has a singularity inside \(E\), the stated theorem cannot be applied to \(E\) as written.
Worked Example: An Upper Hemisphere with a Planar Cap
Worked Example: An Upper Hemisphere with a Planar Cap
Let \(S\) be the upper hemisphere \(x^2+y^2+z^2=1\), \(z\geq 0\), with its outward orientation. Find the flux of \(\mathbf{F}=(x,y,z+2)\) through \(S\).
Step 1: Complete the surface. Add the disk \(C\) given by \(z=0\), \(x^2+y^2\leq 1\). Together, the hemisphere and disk bound the upper half of the unit ball. On the disk, the outward normal from this half-ball is \(\mathbf{n}=(0,0,-1)\).
Step 2: Compute the total flux through the closed boundary. The divergence is \(\nabla\cdot\mathbf{F}=1+1+1=3\). The upper half-ball has volume \(\frac{1}{2}\cdot\frac{4}{3}\pi=\frac{2\pi}{3}\). Thus its total outward boundary flux is
Step 3: Compute the cap flux. On \(z=0\), \(\mathbf{F}=(x,y,2)\), so \(\mathbf{F}\cdot\mathbf{n}=(x,y,2)\cdot(0,0,-1)=-2\). The disk has area \(\pi\), hence
Step 4: Subtract the cap flux.
Check by direct integration. On the unit sphere, the outward unit normal is \((x,y,z)\), so \(\mathbf{F}\cdot\mathbf{n}=x^2+y^2+z^2+2z=1+2z\). The upper hemisphere has area \(2\pi\), and \(\iint_S z\,dS=\int_0^{2\pi}\int_0^{\pi/2}\cos\phi\sin\phi\,d\phi\,d\theta=\pi\). Therefore the direct flux is \(2\pi+2\pi=4\pi\), as required.
Worked Example: Completing a Sphere with a Curved Cap
Worked Example: Completing a Sphere with a Curved Cap
Let \(S\) again be the upper unit hemisphere. This time, use the lower hemisphere \(C\), \(z\leq 0\), as the cap. Find the outward flux of \(\mathbf{F}=(x,y,z+1)\) through \(S\).
Step 1: Identify the closed solid and its boundary. The upper and lower hemispheres together form the boundary of the unit ball. On both pieces, the outward unit normal is \((x,y,z)\).
Step 2: Find the total flux through the sphere. Since \(\nabla\cdot\mathbf{F}=3\) and the unit ball has volume \(\frac{4\pi}{3}\), the Divergence Theorem gives
Step 3: Find the flux through the curved cap. On the unit sphere, \(\mathbf{F}\cdot\mathbf{n}=x^2+y^2+z(z+1)=1+z\). The lower hemisphere has area \(2\pi\), and \(\iint_C z\,dS=-\pi\), as follows from the spherical-coordinate integral \(\int_0^{2\pi}\int_{\pi/2}^{\pi}\cos\phi\sin\phi\,d\phi\,d\theta=-\pi\). Thus the cap flux is \(2\pi-\pi=\pi\).
Step 4: Subtract the cap contribution.
Check directly. On the upper hemisphere, the same integrand \(1+z\) gives flux \(2\pi+\pi=3\pi\), agreeing with the result. This example shows that a cap need not be planar: it must close the boundary, and its outward flux must be computed with the correct normal.
Worked Example: The Side of a Cylinder with Two Caps
Worked Example: The Side of a Cylinder with Two Caps
Let \(S\) be the lateral surface of the cylinder \(x^2+y^2=4\), \(0\leq z\leq 1\). Find the outward flux of \(\mathbf{F}=(x,y,z)\) through \(S\), excluding the top and bottom disks.
Step 1: Complete the boundary. Add the bottom disk at \(z=0\) and the top disk at \(z=1\). The resulting closed surface bounds a cylinder of radius \(2\) and height \(1\).
Step 2: Compute the total boundary flux. Since \(\nabla\cdot\mathbf{F}=3\) and the cylinder’s volume is \(\pi(2)^2(1)=4\pi\), the total outward flux is
Step 3: Compute the two cap fluxes. At \(z=1\), the top cap’s outward normal is \((0,0,1)\). There \(\mathbf{F}\cdot\mathbf{n}=1\), so the top flux is \(1\cdot 4\pi=4\pi\). At \(z=0\), the bottom cap’s outward normal is \((0,0,-1)\), while \(\mathbf{F}=(x,y,0)\). The bottom flux is therefore \(0\).
Step 4: Subtract both cap contributions.
Check directly. Parameterize the side by \((2\cos\theta,2\sin\theta,z)\), where \(0\leq\theta\leq2\pi\) and \(0\leq z\leq1\). The outward unit normal is \((\cos\theta,\sin\theta,0)\), and the surface-area factor is \(2\). The field’s normal component is \(2\), so the flux is \(\int_0^1\int_0^{2\pi}2\cdot2\,d\theta\,dz=8\pi\), confirming the result.
A Reliable Setup
Before calculating, make the geometric and orientation choices explicit. A short setup can prevent a sign error or an invalid use of the theorem.
State which piece’s flux is wanted and what orientation it has.
Check that the target surface and cap form the boundary of a solid.
Confirm that the field is continuously differentiable throughout the solid and that the boundary is piecewise smooth.
Integrate the divergence over the enclosed solid.
Use the normal pointing out of that same solid, including on each cap piece.
Use target flux = total boundary flux minus the sum of cap fluxes.
Common Mistake: Reversing the Cap Normal
For the upper half-ball in the first example, the cap lies at \(z=0\). The enclosed solid is above the disk, so the outward normal on the disk points downward, \((0,0,-1)\). If one incorrectly uses the upward normal, the calculated cap flux is \(+2\pi\). Subtracting that incorrect value from the total \(2\pi\) would give \(0\), not the true flux \(4\pi\).
The correction is to orient each boundary piece outward from the completed solid. With the downward normal, the cap flux is \(-2\pi\), and the target flux is \(2\pi-(-2\pi)=4\pi\). A negative cap flux is entirely possible; it records that the field points, on average, opposite the cap’s outward normal.
Why Cap Completion Is Useful
The method is valuable when the divergence is simple and the chosen solid has an easy volume description, while direct integration over the target surface would require a complicated parameterization or normal. The cap does not need to have zero flux. It only needs to be easier to handle than the original surface, and its contribution must be included with the correct sign.
There can be more than one cap piece, as with the cylinder. In that case, subtract the sum of all their outward fluxes. A useful final check is to verify the orientation on each piece and, when practical, compare the result with a direct surface integral or another independent calculation.
Check Your Understanding
For each question, identify the solid and use outward orientations consistently.
- A target surface and a disk cap enclose a solid. State the formula that relates the target flux, total divergence integral, and cap flux.
- For an upper half-ball closed by its bottom disk, which direction does the disk’s outward normal point, and why?
- If the total outward flux is \(10\) and the outward flux through a cap is \(-3\), what is the flux through the other boundary piece?
- Why must the field be continuously differentiable throughout the enclosed solid before applying the Divergence Theorem?
- A lateral cylinder is completed by a top and a bottom disk. How should their fluxes enter the calculation of the lateral flux?