Flux Through a Closed Cylinder by Face Decomposition

Why Decompose a Closed Cylinder?

A closed cylinder has three distinct parts: its curved lateral wall, its top disk, and its bottom disk. The outward normal points in a different direction on each part. Computing the contributions separately makes both the geometry and the signs explicit.

For a vector field \(\mathbf{F}=(P,Q,R)\), flux measures the component of the field passing through an oriented surface. On a closed surface, the outward flux is the sum of the fluxes through its pieces. The seams where the wall meets the disks do not affect the surface integrals: they are boundary curves, not surface patches.

Setup: Consider a right circular cylinder of radius \(a>0\), centered on the \(z\)-axis, with bottom at \(z=z_0\) and top at \(z=z_1\), where \(z_0<z_1\). Assume \(\mathbf{F}\) is continuous on the cylinder. Its total outward flux is the sum of the lateral-wall, top-cap, and bottom-cap fluxes.

Parametrizing the Three Pieces

On the lateral wall, the radius is fixed at \(a\), while the angle and height vary. A parametrization is

\mathbf{r}(\theta,z)=(a\cos\theta,a\sin\theta,z), \qquad 0\leq\theta\leq 2\pi,\quad z_0\leq z\leq z_1.

The outward direction is radially away from the \(z\)-axis. The cross product of the tangent vectors gives the outward-oriented area vector:

\mathbf{r}_{\theta}=(-a\sin\theta,a\cos\theta,0), \qquad \mathbf{r}_{z}=(0,0,1), $$ $$ \mathbf{r}_{\theta}\times\mathbf{r}_{z} =(a\cos\theta,a\sin\theta,0).

This vector points outward and has magnitude \(a\), so the lateral-wall flux uses the area factor \(a\,d\theta\,dz\). The top and bottom are horizontal disks. Their outward unit normals are respectively \((0,0,1)\) and \((0,0,-1)\), with area element \(dA\) on the disk of radius \(a\).

Formula: If \(\mathbf{F}=(P,Q,R)\), with each component evaluated at the indicated point, then the total outward flux is

\begin{aligned} \Phi_{\mathrm{out}} ={}& \int_{z_0}^{z_1}\int_{0}^{2\pi} a\bigl(P(a\cos\theta,a\sin\theta,z)\cos\theta +Q(a\cos\theta,a\sin\theta,z)\sin\theta\bigr) \,d\theta\,dz\\ &+\iint_{x^2+y^2\leq a^2}R(x,y,z_1)\,dA -\iint_{x^2+y^2\leq a^2}R(x,y,z_0)\,dA. \end{aligned}

The lateral term uses only \(P\) and \(Q\), because the outward normal to the wall has no \(z\)-component. The top term uses \(R\) with a plus sign; the bottom term uses \(R\) with a minus sign.

Why the Formula Works

The flux through an oriented parametrized surface \(\mathbf{r}(u,v)\) is the double integral of \(\mathbf{F}(\mathbf{r}(u,v))\cdot(\mathbf{r}_u\times\mathbf{r}_v)\). This formula takes the field’s component in the oriented normal direction and multiplies by the area scaling factor. Applying it to the wall gives the first term above. On a horizontal disk, the normal is vertical, so the dot product keeps only \(R\).

Orient the wall.
Use \(\mathbf{r}_{\theta}\times\mathbf{r}_{z}=(a\cos\theta,a\sin\theta,0)\), which points radially outward.

Orient the top disk.
Its outward normal points upward, so its contribution is the integral of \(R(x,y,z_1)\) over the disk.

Orient the bottom disk.
Its outward normal points downward, so its contribution is the negative of the integral of \(R(x,y,z_0)\) over the disk.

Add the pieces.
Flux is additive across the three patches, so their signed contributions give the total outward flux.

The sign on the bottom follows directly from the dot product: \(\mathbf{F}\cdot(0,0,-1)=-R\). Thus the minus sign is not an optional correction; it is part of the outward orientation.

Worked Examples

Worked Example: A Radial Field with a Vertical Component

Find the outward flux of \(\mathbf{F}(x,y,z)=(x,y,z)\) through the cylinder of radius \(2\), from \(z=0\) to \(z=3\).

On the wall, \(x=2\cos\theta\) and \(y=2\sin\theta\). The outward area vector is \((2\cos\theta,2\sin\theta,0)\,d\theta\,dz\). Their dot product is

(2\cos\theta,2\sin\theta,z)\cdot(2\cos\theta,2\sin\theta,0) =4\cos^2\theta+4\sin^2\theta=4.

Therefore the wall contribution is

\Phi_{\mathrm{wall}} =\int_0^3\int_0^{2\pi}4\,d\theta\,dz =4(2\pi)(3)=24\pi.

On the top disk, \(R=3\), and the disk has area \(\pi(2)^2=4\pi\). Its contribution is \(3(4\pi)=12\pi\). On the bottom disk, \(R=0\), so its contribution is zero. The total is

\Phi_{\mathrm{out}}=24\pi+12\pi+0=36\pi.

Checking each piece against the field: the wall’s radial component gives \(4\) per unit of \(\theta\)-height parameter area, the top’s vertical component is \(3\), and the bottom’s vertical component is \(0\). These reproduce \(24\pi\), \(12\pi\), and \(0\), respectively.

Worked Example: A Wall Contribution and a Top-Cap Contribution

Find the outward flux of \(\mathbf{F}(x,y,z)=(xz,yz,x+y+2z)\) through the cylinder of radius \(1\), from \(z=0\) to \(z=2\).

On the wall, \((x,y)=(\cos\theta,\sin\theta)\). The outward area vector is \((\cos\theta,\sin\theta,0)\,d\theta\,dz\). The field there is \((z\cos\theta,z\sin\theta,\cos\theta+\sin\theta+2z)\), so

\mathbf{F}\cdot(\cos\theta,\sin\theta,0) =z\cos^2\theta+z\sin^2\theta=z.

Integrating gives

\Phi_{\mathrm{wall}} =\int_0^2\int_0^{2\pi}z\,d\theta\,dz =2\pi\left[\frac{z^2}{2}\right]_0^2 =4\pi.

On the top disk, \(z=2\), so \(R=x+y+4\). The integrals of \(x\) and \(y\) over the disk centered at the origin are zero by symmetry; the constant \(4\) integrates to \(4\pi\). Thus \(\Phi_{\mathrm{top}}=4\pi\). On the bottom disk, \(z=0\), so \(R=x+y\), whose integral over the disk is zero; hence \(\Phi_{\mathrm{bottom}}=0\).

\Phi_{\mathrm{out}}=4\pi+4\pi+0=8\pi.

Substituting these contributions into the face sum gives \(8\pi\), confirming that the wall and top cap both contribute positively while the bottom contribution vanishes.

Worked Example: Signed Contributions from Both Caps

Find the outward flux of \(\mathbf{F}(x,y,z)=(x^2,y^2,z+x)\) through the cylinder of radius \(2\), from \(z=-1\) to \(z=1\).

On the wall, \(x=2\cos\theta\) and \(y=2\sin\theta\). The field’s dot product with the outward area vector \((2\cos\theta,2\sin\theta,0)\,d\theta\,dz\) is

(4\cos^2\theta,4\sin^2\theta,z+2\cos\theta) \cdot(2\cos\theta,2\sin\theta,0) =8\cos^3\theta+8\sin^3\theta.

Each cubic trigonometric term integrates to zero over a full period, so \(\Phi_{\mathrm{wall}}=0\). On the top disk, \(R=1+x\). The \(x\)-term integrates to zero by symmetry, and the constant \(1\) integrates over an area \(4\pi\). Thus \(\Phi_{\mathrm{top}}=4\pi\). On the bottom disk, \(R=-1+x\), but the outward normal points downward. The contribution is therefore

\Phi_{\mathrm{bottom}} =-\iint_{x^2+y^2\leq4}(-1+x)\,dA =-\bigl((-1)(4\pi)+0\bigr)=4\pi.

The total flux is

\Phi_{\mathrm{out}}=0+4\pi+4\pi=8\pi.

For a sign check, the bottom field has constant vertical component \(-1\), while the outward normal there is downward. Their dot product is \(1\), giving the positive bottom contribution \(4\pi\), as calculated.

A Common Sign Error

A frequent mistake is to use the upward normal on both caps. For example, take \(\mathbf{F}=(0,0,z)\) on the cylinder of radius \(1\) from \(z=1\) to \(z=3\). The wall contributes zero because the field is vertical and the wall normal is horizontal. The top contribution is \(3\pi\). At the bottom, the field has vertical component \(1\), but the outward normal is downward, so the bottom contribution is \(-\pi\). The correct total is \(3\pi-\pi=2\pi\).

If the bottom were incorrectly assigned an upward normal, its contribution would be \(+\pi\), and the resulting sum would be \(3\pi+\pi=4\pi\). The correction is to orient each face outward before integrating: upward at the top, downward at the bottom, and radially outward along the wall.

Using the Decomposition Reliably

For a new problem, first identify the radius and height interval. Then compute the wall contribution from the radial components \(P\) and \(Q\), and the cap contributions from the vertical component \(R\). Keep the bottom-cap minus sign visible until its integral has been evaluated. Finally, add the three signed numbers.

This decomposition is useful because it separates geometric directions that behave differently. It also provides a direct check on orientation: the wall normal has no vertical component, while the two cap normals point in opposite vertical directions. The method applies when the field is defined and continuous on the cylinder so that each of the three surface integrals exists.

Check Your Understanding

For each question, track the orientation of every face before combining contributions.

For a cylinder of radius \(a\) between heights \(z_0\) and \(z_1\), what is the outward area vector on the lateral wall in angle-height parameters?
Why does the lateral-wall flux depend on \(P\) and \(Q\), but not directly on \(R\)?
What sign must be used for the bottom-cap integral, and which outward normal determines it?
For \(\mathbf{F}=(0,0,2)\), what are the top and bottom contributions for a cylinder of radius \(3\)?
For a wall of radius \(1\), the lateral integrand is \(z\) over \(0\leq\theta\leq2\pi\) and \(0\leq z\leq1\). What is its flux contribution?