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The Divergence Theorem · Tutorial 960 of 999

Flux Through Cylinders with Radial and Axial Fields

Use the Divergence Theorem and cylindrical coordinates to calculate and check total outward flux through solid cylinders.

Advanced 9 min read

What You'll Learn

  • Express a solid cylinder and its volume element in cylindrical coordinates
  • Calculate divergence for fields with radial, angular, and axial components
  • Integrate divergence over a solid cylinder to obtain total outward flux
  • Handle the cylindrical-coordinate formula at the axis of a smooth field
  • Check a divergence-theorem calculation by integrating flux over the cylinder boundary
  • Recognize how omitting the radial volume factor changes the result

Why Cylinders Suit Cylindrical Coordinates

A cylinder is naturally described by its distance from the axis, its angle around the axis, and its height. Those are exactly the variables used in cylindrical coordinates. When a vector field is given in radial and axial components, this choice of coordinates can make the divergence integral much simpler than setting up a direct flux integral over the curved side and the two end disks.

The Divergence Theorem converts the total outward flux across a closed boundary into a volume integral of divergence. For a cylinder, the main tasks are to express the divergence correctly and to use the cylindrical volume element, which includes a factor of \(r\).

Definition: For \(R>0\) and \(a<b\), the solid cylinder of radius \(R\) between heights \(a\) and \(b\) is $$ D=\{(x,y,z):x^2+y^2\le R^2,\ a\le z\le b\}. $$ Its boundary \(\partial D\) consists of the curved side and the disks at \(z=a\) and \(z=b\).

Cylindrical Components and Divergence

Write \(r=\sqrt{x^2+y^2}\), and let \(\theta\) denote the angle around the \(z\)-axis. The unit vectors \(\mathbf e_r\), \(\mathbf e_\theta\), and \(\mathbf e_z\) point, respectively, radially outward from the axis, tangentially around it, and vertically upward. A vector field can be expressed as

$$ \mathbf F=F_r\,\mathbf e_r+F_\theta\,\mathbf e_\theta+F_z\,\mathbf e_z. $$

Here each component may depend on \(r,\theta,z\). For a continuously differentiable field, the divergence in cylindrical coordinates is

$$ \nabla\cdot\mathbf F = \frac{1}{r}\frac{\partial}{\partial r}(rF_r) +\frac{1}{r}\frac{\partial F_\theta}{\partial\theta} +\frac{\partial F_z}{\partial z}. $$

The \(r\) in the radial term is essential: radial surfaces at larger \(r\) have greater area. The cylindrical volume element is \(dV=r\,dr\,d\theta\,dz\), so the same geometric factor appears in the volume integral.

Theorem: If \(\mathbf F\) is continuously differentiable on an open set containing the solid cylinder \(D\), then the total outward flux through its boundary is $$ \iint_{\partial D}\mathbf F\cdot\mathbf n\,dS = \iiint_D\nabla\cdot\mathbf F\,dV = \int_a^b\int_0^{2\pi}\int_0^R (\nabla\cdot\mathbf F)\,r\,dr\,d\theta\,dz. $$ Here \(\mathbf n\) is the outward unit normal. The smoothness condition ensures the theorem applies to the whole closed boundary.

At \(r=0\), the cylindrical basis vectors are not defined, so the displayed divergence formula is understood away from the axis. For a field that is continuously differentiable in Cartesian coordinates, its divergence is well-defined on the axis; the cylindrical expression is used by taking its limiting value where appropriate. The axis itself contributes no separate volume.

Why the Cylindrical Divergence Formula Has These Terms

A short flux-balance argument explains the structure of the formula. Consider a small cylindrical wedge near \((r,\theta,z)\), with widths \(dr\), \(d\theta\), and \(dz\). Its volume is approximately \(r\,dr\,d\theta\,dz\). The radial faces have areas approximately \(r\,d\theta\,dz\) and \((r+dr)\,d\theta\,dz\). Thus the net radial flux, to first order, is

$$ \big[(r+dr)F_r(r+dr,\theta,z)-rF_r(r,\theta,z)\big]\,d\theta\,dz \approx \frac{\partial}{\partial r}(rF_r)\,dr\,d\theta\,dz. $$

Dividing by the wedge volume gives the radial contribution \(\frac{1}{r}\frac{\partial}{\partial r}(rF_r)\). The two angular faces have area approximately \(dr\,dz\). Their net flux is approximately \(\frac{\partial F_\theta}{\partial\theta}d\theta\,dr\,dz\); dividing by the volume gives \(\frac{1}{r}\frac{\partial F_\theta}{\partial\theta}\). Finally, the upper and lower faces give net axial flux approximately \(\frac{\partial F_z}{\partial z}r\,dr\,d\theta\,dz\), which yields \(\frac{\partial F_z}{\partial z}\) after division by volume. Adding these three contributions gives the cylindrical divergence formula.

A Reliable Calculation Procedure

For a cylinder with its axis along the \(z\)-axis, proceed by identifying the field components, calculating divergence, and then integrating over the cylinder’s coordinate ranges. Remember that the radial coordinate runs from \(0\) to \(R\), the angle runs through one complete revolution, and the height runs from \(a\) to \(b\).

1
Check the setting.
Specify the cylinder and verify that the field is continuously differentiable on and inside it.
2
Compute divergence.
Use all three cylindrical terms, including the angular term if \(F_\theta\) depends on \(\theta\).
3
Set up the volume integral.
Integrate over \(0\le r\le R\), \(0\le\theta\le2\pi\), and \(a\le z\le b\), with the factor \(r\).
4
Check the flux.
When practical, separately calculate the side and end-disk fluxes to verify the total and its sign.

Worked Examples

Worked Example: A Radial Field and a Height-Dependent Axial Field

Let \(\mathbf F=2r\,\mathbf e_r+3z\,\mathbf e_z\), and let \(D\) be the cylinder \(0\le r\le3\), \(-1\le z\le2\). Find the total outward flux.

The components are \(F_r=2r\), \(F_\theta=0\), and \(F_z=3z\). Therefore,

$$ \nabla\cdot\mathbf F = \frac{1}{r}\frac{\partial}{\partial r}(2r^2) +\frac{\partial}{\partial z}(3z) = \frac{1}{r}(4r)+3 = 7. $$

Integrate this constant over the cylinder, retaining the cylindrical factor \(r\):

$$ \begin{aligned} \iint_{\partial D}\mathbf F\cdot\mathbf n\,dS &=\int_{-1}^{2}\int_0^{2\pi}\int_0^3 7r\,dr\,d\theta\,dz\\ &=\int_{-1}^{2}\int_0^{2\pi}\frac{63}{2}\,d\theta\,dz\\ &=\int_{-1}^{2}63\pi\,dz\\ &=189\pi. \end{aligned} $$

To check directly, the side has \(r=3\), outward normal \(\mathbf e_r\), and surface element \(3\,d\theta\,dz\). Its flux is \(6\cdot3\cdot2\pi\cdot3=108\pi\). On the top disk, \(F_z=6\), so the flux is \(6\cdot9\pi=54\pi\). On the bottom disk, \(F_z=-3\) and the outward normal is \(-\mathbf e_z\), giving flux \(3\cdot9\pi=27\pi\). The total is \(108\pi+54\pi+27\pi=189\pi\), as required.

Worked Example: A Nonconstant Radial Component

Let \(\mathbf F=r^2\,\mathbf e_r+z^2\,\mathbf e_z\) on the cylinder \(0\le r\le1\), \(0\le z\le2\). The field is smooth when expressed in Cartesian coordinates: its radial part is \(r(x,y,0)\), and its axial part is \(z^2\mathbf e_z\).

The divergence is

$$ \nabla\cdot\mathbf F = \frac{1}{r}\frac{\partial}{\partial r}(r^3) +\frac{\partial}{\partial z}(z^2) = 3r+2z. $$

Integrate each term over the solid:

$$ \begin{aligned} \iiint_D(3r+2z)\,dV &=\int_0^2\int_0^{2\pi}\int_0^1(3r+2z)r\,dr\,d\theta\,dz\\ &=\int_0^2\int_0^{2\pi} \left(\int_0^1 3r^2\,dr+\int_0^1 2zr\,dr\right)d\theta\,dz\\ &=\int_0^2\int_0^{2\pi}(1+z)\,d\theta\,dz\\ &=2\pi\int_0^2(1+z)\,dz\\ &=2\pi\left(2+2\right)=8\pi. \end{aligned} $$

For a direct check, the side flux is \(F_r=1\) at \(r=1\), and its surface element is \(d\theta\,dz\). This gives \(2\pi\cdot2=4\pi\). The top disk has \(F_z=4\), giving flux \(4\pi\); the bottom disk has \(F_z=0\), giving zero. The boundary total is \(4\pi+4\pi+0=8\pi\).

Worked Example: A Mixed Field with an Angular Component

Let \(\mathbf F=r\,\mathbf e_r+r\,\mathbf e_\theta+z\,\mathbf e_z\) on the cylinder \(0\le r\le2\), \(-1\le z\le1\). The angular component is nonzero, but it does not depend on \(\theta\).

The divergence is

$$ \begin{aligned} \nabla\cdot\mathbf F &=\frac{1}{r}\frac{\partial}{\partial r}(r^2) +\frac{1}{r}\frac{\partial r}{\partial\theta} +\frac{\partial z}{\partial z}\\ &=2+0+1=3. \end{aligned} $$

The flux is therefore

$$ \begin{aligned} \iiint_D3\,dV &=\int_{-1}^{1}\int_0^{2\pi}\int_0^2 3r\,dr\,d\theta\,dz\\ &=6\cdot2\pi\cdot2\\ &=24\pi. \end{aligned} $$

To verify this result, the angular component is tangent to the cylindrical side and has zero dot product with its normal. The radial component on the side is \(2\), and the side area is \(2\cdot2\pi\cdot2=8\pi\), so the side flux is \(16\pi\). On the top disk, \(F_z=1\), giving \(4\pi\). On the bottom disk, \(F_z=-1\) and the outward normal points downward, also giving \(4\pi\). The sum is \(16\pi+4\pi+4\pi=24\pi\).

A Common Mistake: Dropping the Factor \(r\)

The divergence and the volume element are different parts of the setup. Even if the divergence has no \(r\) in it, cylindrical volume integration still uses \(r\,dr\,d\theta\,dz\). For example, in the first worked example the divergence is \(7\). Omitting the radial factor would give

$$ \int_{-1}^{2}\int_0^{2\pi}\int_0^3 7\,dr\,d\theta\,dz =126\pi, $$

which is not the correct flux, \(189\pi\). The mistake treats equal-width radial strips as if they had equal volume at every distance from the axis. But a strip farther out has greater circumference and therefore greater volume. The factor \(r\) accounts for that changing geometry.

Another sign check concerns the bottom disk. Its outward normal is \(-\mathbf e_z\), not \(+\mathbf e_z\). A negative value of \(F_z\) there can consequently produce positive outward flux, as in the first and third examples.

What the Calculation Reveals

Using divergence is especially useful when the field’s components make divergence simple and the cylinder is easy to describe in cylindrical coordinates. The volume calculation automatically includes flux through the entire closed boundary: both the curved side and the end disks. It does not calculate only the side flux.

For a field with only radial and axial components, the divergence reduces to \(\frac{1}{r}\frac{\partial}{\partial r}(rF_r)+\frac{\partial F_z}{\partial z}\). A nonzero angular component must still be included if it varies with angle. Keeping these terms distinct, using the correct cylindrical bounds, and checking the orientation of the disks make the calculation systematic.

Check Your Understanding

Answer each question by identifying the relevant component, formula, or boundary contribution.

  1. What is the cylindrical volume element, and why does it contain a factor of \(r\)?
  2. For \(\mathbf F=F_r\mathbf e_r+F_\theta\mathbf e_\theta+F_z\mathbf e_z\), which divergence term accounts for variation in \(F_\theta\)?
  3. For \(\mathbf F=4r\,\mathbf e_r+2z\,\mathbf e_z\) on \(0\le r\le1\), \(0\le z\le3\), set up the divergence-theorem integral. What is the divergence?
  4. Why can a nonzero angular component have no direct contribution to flux through the curved side of a cylinder?
  5. Which way does the outward normal point on the lower end disk of a solid cylinder?

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