Capped Cylinders and Annular Cylinders

Introduction

A cylinder bounded by two planes has more than a curved side: it also has a top and a bottom. An annular cylinder has an additional distinction. Its boundary includes both an outer cylindrical wall and an inner cylindrical wall, and their outward normals point in opposite radial directions.

These distinctions matter whenever the Divergence Theorem is used to calculate flux. The theorem converts the total outward flux across a closed boundary into a volume integral, but the geometry determines the bounds of that integral and the direction of each boundary normal. This tutorial develops those setups for capped cylinders and annular cylinders.

Geometry and the Divergence-Theorem Setup

Use cylindrical coordinates $(r,\theta,z)$, related to Cartesian coordinates by $x=r\cos\theta$ and $y=r\sin\theta$. A solid capped cylinder of radius $R>0$, with flat ends at $z=a$ and $z=b$, where $a<b$, is

E=\{(r,\theta,z):0\le r\le R,\ 0\le\theta\le 2\pi,\ a\le z\le b\}.

Its volume element is $dV=r\,dr\,d\theta\,dz$. The factor $r$ is essential: it accounts for the change in volume when Cartesian coordinates are expressed in cylindrical coordinates.

An annular cylinder has inner radius $\rho>0$, outer radius $R>\rho$, and the same planar end levels. Its solid region is

E_{\mathrm{ann}}=\{(r,\theta,z):\rho\le r\le R,\ 0\le\theta\le 2\pi,\ a\le z\le b\}.

Divergence Theorem: If a vector field $\mathbf{F}$ has continuous first partial derivatives on an open region containing a solid $E$ and its piecewise smooth closed boundary $\partial E$, then $$ \iint_{\partial E}\mathbf{F}\cdot\mathbf{n}\,dS = \iiint_E \nabla\cdot\mathbf{F}\,dV, $$ where $\mathbf{n}$ is the outward unit normal to the solid. For the cylindrical regions here, the volume integral uses $dV=r\,dr\,d\theta\,dz$.

“Outward” is relative to the solid being considered. On the outer wall, moving outward increases $r$. On the inner wall of an annular cylinder, moving outward from the material region means moving toward its central opening, so the normal points toward decreasing $r$. At the top and bottom, outward means toward increasing and decreasing $z$, respectively.

Boundary piece	Outward unit normal	Surface area element
Outer cylindrical wall, $r=R$	$\mathbf e_r$	$R\,d\theta\,dz$
Inner cylindrical wall, $r=\rho$, if present	$-\mathbf e_r$	$\rho\,d\theta\,dz$
Top cap, $z=b$	$\mathbf k$	$r\,dr\,d\theta$
Bottom cap, $z=a$	$-\mathbf k$	$r\,dr\,d\theta$

Here $\mathbf e_r$ is the radial unit vector and $\mathbf k$ is the vertical unit vector. On an annular cap, its radial coordinate ranges from $\rho$ to $R$; on a solid cylinder cap, it ranges from $0$ to $R$.

The Basic Volume Integrals

For a sufficiently smooth vector field, the Divergence Theorem gives these setups directly. Write $\nabla\cdot\mathbf F$ for the divergence of $\mathbf F$, computed in Cartesian or cylindrical coordinates as convenient.

\text{Solid cylinder:}\qquad \iint_{\partial E}\mathbf F\cdot\mathbf n\,dS = \int_a^b\int_0^{2\pi}\int_0^R (\nabla\cdot\mathbf F)\,r\,dr\,d\theta\,dz.

\text{Annular cylinder:}\qquad \iint_{\partial E_{\mathrm{ann}}}\mathbf F\cdot\mathbf n\,dS = \int_a^b\int_0^{2\pi}\int_\rho^R (\nabla\cdot\mathbf F)\,r\,dr\,d\theta\,dz.

These equations give the total flux through the entire closed boundary. They do not by themselves give the flux through an individual wall or cap. If the goal is to identify the contribution from each piece, set up the surface integrals separately, using the normals and area elements in the table.

Why the Annular Boundary Has a Negative Inner-Wall Term

The sign on the inner wall can be justified by integrating the cylindrical-coordinate divergence formula. Suppose $\mathbf F=F_r\mathbf e_r+F_\theta\mathbf e_\theta+F_z\mathbf k$ is smooth on an annular cylinder. For $r>0$, its divergence 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}.

Multiplication by the volume factor $r$ makes each term an ordinary derivative with respect to one coordinate. Integrating in $r$, $\theta$, and $z$, the radial part gives the outer value minus the inner value. The $\theta$-part vanishes because a full cylindrical field has the same value at $\theta=0$ and $\theta=2\pi$. The $z$-part gives the top value minus the bottom value.

\begin{aligned} \iiint_{E_{\mathrm{ann}}}(\nabla\cdot\mathbf F)\,dV &=\int_a^b\int_0^{2\pi}\int_\rho^R \left[ \frac{\partial}{\partial r}(rF_r) +\frac{\partial F_\theta}{\partial\theta} +r\frac{\partial F_z}{\partial z} \right]dr\,d\theta\,dz\\ &=\int_a^b\int_0^{2\pi} \bigl(RF_r(R,\theta,z)-\rho F_r(\rho,\theta,z)\bigr)\,d\theta\,dz\\ &\quad+\int_\rho^R\int_0^{2\pi} r\bigl(F_z(r,\theta,b)-F_z(r,\theta,a)\bigr)\,d\theta\,dr. \end{aligned}

The first term is exactly the sum of the outer-wall flux and inner-wall flux: the inner contribution carries a minus sign. The second term is the top-cap flux minus the bottom-cap flux. Thus the volume integral reproduces the boundary contributions with the correct outward orientations. This also explains why the inner wall does not get the same sign as the outer wall.

Worked Examples

Worked Example: A Solid Capped Cylinder

Find the total outward flux of $\mathbf F=(x,y,2z)$ through the cylinder $x^2+y^2\le4$, $0\le z\le3$. Check the answer by adding the fluxes of the side and caps.

The radius is $R=2$, and the end levels are $a=0$ and $b=3$. The field is smooth throughout the solid. Its divergence is

\nabla\cdot\mathbf F =\frac{\partial x}{\partial x} +\frac{\partial y}{\partial y} +\frac{\partial (2z)}{\partial z} =1+1+2=4.

The Divergence Theorem setup and evaluation are

\begin{aligned} \iint_{\partial E}\mathbf F\cdot\mathbf n\,dS &=\int_0^3\int_0^{2\pi}\int_0^2 4r\,dr\,d\theta\,dz\\ &=4\left[\frac{r^2}{2}\right]_0^2(2\pi)(3)\\ &=4(2)(2\pi)(3)=48\pi. \end{aligned}

To check by boundary pieces, on the side $r=2$, the radial component is $F_r=r=2$, so $\mathbf F\cdot\mathbf e_r=2$. Its area element is $2\,d\theta\,dz$, giving side flux $2\cdot2\cdot(2\pi)\cdot3=24\pi$. On the top, $F_z=2z=6$, so the flux is $6\cdot\pi(2)^2=24\pi$. On the bottom, $F_z=0$, so its flux is zero. The total is $24\pi+24\pi+0=48\pi$, agreeing with the volume integral.

Worked Example: An Annular Cylinder with Two Caps

Find the total outward flux of $\mathbf F=(x,y,z)$ from the annular cylinder $1\le\sqrt{x^2+y^2}\le2$, $0\le z\le2$.

The radii are $\rho=1$ and $R=2$. The divergence is $1+1+1=3$, and the setup is

\begin{aligned} \iint_{\partial E_{\mathrm{ann}}}\mathbf F\cdot\mathbf n\,dS &=\int_0^2\int_0^{2\pi}\int_1^2 3r\,dr\,d\theta\,dz\\ &=3\left[\frac{r^2}{2}\right]_1^2(2\pi)(2)\\ &=3\left(\frac{4-1}{2}\right)(2\pi)(2)=18\pi. \end{aligned}

On the outer wall, $F_r=2$, so the flux is $2\cdot2\cdot(2\pi)\cdot2=16\pi$. On the inner wall, the field has radial component $F_r=1$, but the outward normal is $-\mathbf e_r$. Its flux is therefore $(-1)\cdot1\cdot(2\pi)\cdot2=-4\pi$. The top cap has $F_z=2$; its annular area is $\pi(2^2-1^2)=3\pi$, giving $6\pi$. The bottom cap has $F_z=0$, giving zero. Adding the pieces gives $16\pi-4\pi+6\pi+0=18\pi$, as required.

Worked Example: Nonzero Flux on Both Annular Caps

For $1\le r\le2$ and $0\le z\le1$, let $\mathbf F=(xz,yz,x^2+y^2)$. Find the total outward flux and check each boundary piece.

In cylindrical components, $F_r=rz$ and $F_z=r^2$. In Cartesian coordinates the divergence is

\nabla\cdot\mathbf F =\frac{\partial(xz)}{\partial x} +\frac{\partial(yz)}{\partial y} +\frac{\partial(x^2+y^2)}{\partial z} =z+z+0=2z.

The volume-integral calculation is

\begin{aligned} \iint_{\partial E_{\mathrm{ann}}}\mathbf F\cdot\mathbf n\,dS &=\int_0^1\int_0^{2\pi}\int_1^2 2z\,r\,dr\,d\theta\,dz\\ &=\left(\int_0^1 2z\,dz\right) \left(\int_0^{2\pi}d\theta\right) \left(\int_1^2r\,dr\right)\\ &=(1)(2\pi)(3/2)=3\pi. \end{aligned}

On the outer wall, $F_r=2z$ and $dS=2\,d\theta\,dz$, so its flux is $\int_0^1\int_0^{2\pi}4z\,d\theta\,dz=4\pi$. On the inner wall, the normal reverses the radial component: $\mathbf F\cdot(-\mathbf e_r)=-z$, and $dS=d\theta\,dz$, so the flux is $\int_0^1\int_0^{2\pi}(-z)\,d\theta\,dz=-\pi$.

At the top, $F_z=r^2$, so the flux is $\int_0^{2\pi}\int_1^2r^2(r\,dr\,d\theta)=2\pi\int_1^2r^3\,dr=15\pi/2$. At the bottom, the outward normal is $-\mathbf k$, so the flux is $-15\pi/2$. The cap fluxes cancel, and the total is $4\pi-\pi+15\pi/2-15\pi/2=3\pi$, matching the volume integral.

A Common Sign Error

A frequent mistake is to use $+\mathbf e_r$ as the outward normal on both cylindrical walls of an annular solid. In the second example, that mistake would assign the inner wall flux $+4\pi$ rather than $-4\pi$, producing the incorrect total $16\pi+4\pi+6\pi=26\pi$.

The correction comes from the solid itself: at the inner wall, the annular material lies at larger $r$, so its outward direction points into the central opening, toward smaller $r$. The normal is $-\mathbf e_r$, and the inner-wall contribution must be negative in that example. A useful check is to imagine standing inside the solid and pointing away from the material, not away from the cylinder’s axis.

Why the Setup Matters

The Divergence Theorem is especially useful here because it replaces several surface integrals by one volume integral. But a setup still needs to reflect the region accurately:

For a solid capped cylinder, integrate $r$ from $0$ to $R$.
For an annular cylinder, integrate $r$ from $\rho$ to $R$.
Use the factor $r$ in the volume element.
When checking boundary pieces, include both caps and every cylindrical wall.
Use $-\mathbf e_r$ on an annular cylinder’s inner wall and $-\mathbf k$ on its bottom cap.

When the inner radius is zero, there is no inner cylindrical wall: the region is a solid cylinder, not an annular one. In either case, the end caps are part of the closed boundary and cannot be omitted when applying the theorem to the entire solid.

Check Your Understanding

Use the geometry, orientation rules, and integral setups above to answer the following.

What are the $r$-bounds and the volume element for an annular cylinder with inner radius $2$ and outer radius $5$?
Which way does the outward normal point on the inner cylindrical wall, and why?
For a cylinder with end levels $z=a$ and $z=b$, what are the outward normals on the bottom and top caps?
If the solid is a capped cylinder of radius $R$, what are the three types of boundary pieces to include in a direct flux check?
For $\mathbf F=(x,y,z)$ on an annular cylinder, how does the inner-wall contribution differ in sign from the outer-wall contribution?

Boundary piece	Outward unit normal	Surface area element
Outer cylindrical wall, \(r=R\)	\(\mathbf e_r\)	\(R\,d\theta\,dz\)
Inner cylindrical wall, \(r=\rho\), if present	\(-\mathbf e_r\)	\(\rho\,d\theta\,dz\)
Top cap, \(z=b\)	\(\mathbf k\)	\(r\,dr\,d\theta\)
Bottom cap, \(z=a\)	\(-\mathbf k\)	\(r\,dr\,d\theta\)