Evaluating Radial Powers in Cylindrical Integrals

Why the Radial Power Needs Care

In cylindrical coordinates, the volume element is $dV=r\,dz\,dr\,d\theta$. The factor $r$ is part of the volume element, not automatically part of the function being integrated. When the integrand already contains a power of $r$, both contributions must be included in the integral.

This distinction is easy to lose during a coordinate change. For example, an integrand $r^2$ contributes $r^2$, while the cylindrical volume element contributes another $r$. The radial factor in the resulting iterated integral is therefore $r^3$, not $r^2$. Keeping the two sources separate makes the setup easier to check and the radial antiderivative easier to select.

Separate the Integrand and the Volume Element

Definition: In cylindrical coordinates, if a solid $E$ is described by $0\leq\theta\leq\beta$, $a(\theta)\leq r\leq b(\theta)$, and $g(r,\theta)\leq z\leq h(r,\theta)$, then an integral of a function $f$ has the form $$ \iiint_E f(r,\theta,z)\,dV = \int_0^\beta\int_{a(\theta)}^{b(\theta)} \int_{g(r,\theta)}^{h(r,\theta)} f(r,\theta,z)\,r\,dz\,dr\,d\theta. $$ The factor $r$ displayed after $f$ is the cylindrical Jacobian factor.

The listed order $dz\,dr\,d\theta$ means that $z$ is integrated first, then $r$, then $\theta$. At each stage, the bounds for the variable being integrated are used while the other variables are held fixed. After integrating in $z$, the resulting expression may include additional powers of $r$, for example if the vertical bounds depend on $r$.

If the integrand is $f(r,\theta,z)=r^p$, the product of the integrand with the Jacobian is $r^p r=r^{p+1}$. For nonnegative integer $p$, its radial antiderivative is a power. More generally, the same power rule applies for real $p$ wherever the relevant powers are defined and the radial interval stays positive.

Rule: When the integrand contains $r^p$, include the Jacobian once, giving $r^{p+1}$. For $p\neq -2$, $$ \int r^{p+1}\,dr=\frac{r^{p+2}}{p+2}+C. $$ If $p=-2$, then the combined factor is $r^{-1}$, whose antiderivative is $\ln r+C$ on $r>0$.

Why the Exponent Changes

The change from $p$ to $p+1$ follows directly from multiplying the integrand by the volume factor. It is not a special integration trick: it is ordinary multiplication of powers before integration. The power rule then applies to the resulting exponent.

Write the integrand.
If the integrand has a factor $r^p$, record that exponent before inserting the volume element.

Insert the Jacobian.
Multiply by the additional factor $r$, so the product is $r^p r=r^{p+1}$.

Integrate in the stated order.
Use the radial antiderivative only when the radial variable is reached, and apply its actual lower and upper bounds.

For $p\neq-2$, the antiderivative can be checked by differentiation: $$ \frac{d}{dr}\left(\frac{r^{p+2}}{p+2}\right) = \frac{(p+2)r^{p+1}}{p+2} = r^{p+1}. $$ Thus, when $0\leq a\leq b$ and the integral is finite, the radial contribution is $$ \int_a^b r^{p+1}\,dr = \frac{b^{p+2}-a^{p+2}}{p+2}. $$ If $a=0$, this expression is finite for $p>-2$. At $p=-2$, the integral from zero has a logarithmic divergence; for $p<-2$, it also diverges. If $a>0$, the radial interval avoids the axis, and the logarithmic formula for $p=-2$ is well-defined.

Worked Examples

Worked Example: A Constant Integrand with a Quadratic Radial Factor

Evaluate $$ \int_0^{2\pi}\int_0^2\int_0^3 r^2\,dz\,dr\,d\theta. $$ The displayed integrand is $r^2$. The volume element supplies one additional factor of $r$, so the full radial factor is $r^3$. Integrate in the stated order, beginning with $z$: $$ \int_0^{2\pi}\int_0^2\left(\int_0^3 r^3\,dz\right)dr\,d\theta = \int_0^{2\pi}\int_0^2 3r^3\,dr\,d\theta. $$ Since $\int_0^2 r^3\,dr=[r^4/4]_0^2=16/4-0=4$, the radial integral, including the factor $3$, is $3\cdot4=12$. Therefore, $$ \int_0^{2\pi}12\,d\theta = 12(2\pi-0) = 24\pi. $$ Check the radial antiderivative: $d(r^4/4)/dr=r^3$. The evaluated radial factor is $4$, and multiplying the separate $z$-length $3$ and angular length $2\pi$ gives $4\cdot3\cdot2\pi=24\pi$, as found.

Worked Example: A Negative Power on an Annular Radial Interval

Evaluate $$ \int_0^{\pi/2}\int_1^3\int_0^2 r^{-1}\,dz\,dr\,d\theta. $$ Because the radial interval begins at $1$, it stays away from the axis. Multiply the integrand factor $r^{-1}$ by the Jacobian $r$: $$ r^{-1}r=r^0=1. $$ The integral becomes $$ \int_0^{\pi/2}\int_1^3\int_0^2 1\,dz\,dr\,d\theta = \int_0^{\pi/2}\int_1^3 2\,dr\,d\theta. $$ The radial integral is $2(3-1)=4$, so $$ \int_0^{\pi/2}4\,d\theta = 4\left(\frac{\pi}{2}-0\right) = 2\pi. $$ Check the cancellation: the product $r^{-1}r$ equals $1$ throughout $1\leq r\leq3$. The remaining lengths are $2$ in $z$, $2$ in $r$, and $\pi/2$ in $\theta$; their product is $2\cdot2\cdot\pi/2=2\pi$.

Worked Example: A Radial Power with a Radial-Dependent Height

Evaluate $$ \int_0^\pi\int_0^1\int_0^{1+r} r^3\,dz\,dr\,d\theta. $$ The integrand contributes $r^3$, and the Jacobian contributes one more $r$, giving $r^4$. Integrate first with respect to $z$: $$ \int_0^\pi\int_0^1 r^4(1+r)\,dr\,d\theta. $$ Expand the product before integrating in $r$: $$ r^4(1+r)=r^4+r^5. $$ Thus the radial integral is $$ \int_0^1(r^4+r^5)\,dr = \left[\frac{r^5}{5}+\frac{r^6}{6}\right]_0^1 = \left(\frac15+\frac16\right)-0 = \frac{11}{30}. $$ The angular integral gives $$ \int_0^\pi\frac{11}{30}\,d\theta = \frac{11}{30}(\pi-0) = \frac{11\pi}{30}. $$ Check both radial antiderivatives: $d(r^5/5)/dr=r^4$ and $d(r^6/6)/dr=r^5$. At the upper bound $r=1$, their sum is $1/5+1/6=11/30$, while at $r=0$ it is zero.

Worked Example: The Logarithmic Case

Evaluate $$ \int_0^{\pi}\int_1^e\int_0^1 r^{-2}\,dz\,dr\,d\theta. $$ The integrand factor $r^{-2}$ combines with the Jacobian $r$ to give $r^{-1}$. Since $r\geq1$, the logarithmic antiderivative is defined throughout the interval. Integrating in $z$ gives $$ \int_0^\pi\int_1^e r^{-1}\,dr\,d\theta = \int_0^\pi[\ln r]_1^e\,d\theta. $$ The radial value is $\ln e-\ln 1=1-0=1$. Therefore, $$ \int_0^\pi 1\,d\theta=\pi. $$ Check the radial antiderivative: $d(\ln r)/dr=1/r=r^{-1}$ for $r>0$. Evaluating at $e$ and $1$ gives exactly $1$, so the remaining angular integral is $\pi$.

A Common Error: Counting the Jacobian Incorrectly

A frequent mistake is to integrate the integrand's radial power without multiplying by the Jacobian. For the first example, this incorrect calculation would replace the radial factor $r^2$ by $r^2$ throughout and conclude $$ \int_0^{2\pi}\int_0^2\int_0^3 r^2\,dz\,dr\,d\theta \ \text{incorrectly becomes}\ \int_0^{2\pi}\int_0^2 3r^2\,dr\,d\theta = \int_0^{2\pi}8\,d\theta = 16\pi. $$ The error is that the expression on the right has omitted the Jacobian factor $r$. The correction is to form $r^2\cdot r=r^3$ before integrating. Then the radial integral is $3\int_0^2r^3\,dr=3\cdot4=12$, and the full value is $12\cdot2\pi=24\pi$.

The opposite error is to count the Jacobian twice. The factor $r$ is included exactly once in the cylindrical volume element. A reliable setup labels the integrand and the volume element separately before combining them.

What to Check Before Evaluating

Write the integrand's radial power as it appears in the problem.
Multiply by exactly one Jacobian factor $r$.
Keep the specified integration order and use the bounds for each variable.
Check whether the radial interval reaches $r=0$, especially for negative powers.
Differentiate the proposed radial antiderivative to confirm its integrand.

These checks matter because a correct radial antiderivative cannot repair an incorrectly formed radial factor. Once the integrand and Jacobian are combined properly, the remaining work is ordinary iterated integration with careful bounds.

Check Your Understanding

For each question, identify the radial factor after including the cylindrical Jacobian before evaluating or deciding whether the integral is finite.

For an integrand $r^4$, what radial power appears after multiplication by the Jacobian?
What is $\int_1^2 r^2 r\,dr$?
For an integrand $r^{-2}$, what is the combined radial factor, and which antiderivative applies on $r>0$?
Does the radial integral for an integrand $r^{-2}$ remain finite when its radial interval is $0\leq r\leq1$? Explain using the combined factor.
A student integrates an integrand $r$ using radial factor $r$. What factor was omitted, and what is the correct radial factor?