Separable Radial and Angular Integrands

When Can a Polar Integral Be Separated?

Some polar double integrals have an integrand that is a product of a function of the radius and a function of the angle. If the region also has independent radial and angular bounds, the two one-variable integrations can be performed separately. This can reduce a double integral to two simpler calculations.

The bounds matter as much as the integrand. A product in the integrand alone does not guarantee that the integral itself is a product of two integrals. First check whether the polar region is a rectangle in the variables \(r\) and \(\theta\).

Definition: A polar integrand is separable if it can be written as \(f(r)g(\theta)\), where \(f\) depends only on the radius and \(g\) depends only on the angle. On a region with bounds \(a\le r\le b\) and \(\alpha\le\theta\le\beta\), the polar area element contributes an additional factor of \(r\).

Thus, when \(f(r)g(\theta)\) is the function being integrated over the planar region, its polar integral contains \(r f(r)g(\theta)\). The factor \(r\) comes from the polar area element \(dA=r\,dr\,d\theta\); it is not optional, and it should be included exactly once.

The Separation Formula

Suppose \(0\le a\le b\), \(\alpha\le\beta\le\alpha+2\pi\), and the region is described by the independent bounds \(a\le r\le b\) and \(\alpha\le\theta\le\beta\). The angular width is at most \(2\pi\), so the polar description covers each point of the region once (apart from boundary rays of zero area); with a wider interval the same points would be traversed more than once and the formula would count them with multiplicity. Suppose also that \(f\) and \(g\) are continuous on the intervals involved. Continuity is a sufficient condition for the iterated integrals below to exist and for the order of integration to be used in the usual way.

\iint_D f(r)g(\theta)\,dA = \int_{\alpha}^{\beta}\int_a^b f(r)g(\theta)\,r\,dr\,d\theta = \left(\int_a^b r f(r)\,dr\right) \left(\int_\alpha^\beta g(\theta)\,d\theta\right).

The radial factor in this formula is \(\int_a^b r f(r)\,dr\), not merely \(\int_a^b f(r)\,dr\). The angular factor is \(\int_\alpha^\beta g(\theta)\,d\theta\). The bounds must be independent: the radial endpoints \(a,b\) are constants, and the angular endpoints \(\alpha,\beta\) are constants.

Why the Formula Works

On the rectangular parameter region \([a,b]\times[\alpha,\beta]\), the part of the integrand depending on \(r\), including the area factor, is \(r f(r)\). The remaining factor \(g(\theta)\) is constant with respect to the inner integration variable \(r\). It can therefore be taken outside that inner integral.

\begin{aligned} \int_{\alpha}^{\beta}\int_a^b f(r)g(\theta)\,r\,dr\,d\theta &=\int_{\alpha}^{\beta}g(\theta) \left(\int_a^b r f(r)\,dr\right)d\theta\\ &=\left(\int_a^b r f(r)\,dr\right) \left(\int_{\alpha}^{\beta}g(\theta)\,d\theta\right). \end{aligned}

For the second equality, the radial integral has a value that does not depend on \(\theta\), so it is a constant with respect to the outer integration. This proves the separation formula. It is a direct consequence of iterated integration and the fact that the bounds describe a rectangle in the \((r,\theta)\)-parameter plane.

The geometry of the polar region need not be a full disk or a full annulus. A sector of an annulus, for example, has constant radial endpoints and constant angular endpoints, so it also has independent bounds of this form. What matters for separation is the form of the bounds, not the name of the region.

Worked Example: A Separable Integral over an Annular Sector

Evaluate the integral of \(r^2(1+\theta)\) over the region \(1\le r\le2\), \(0\le\theta\le\frac{\pi}{2}\). The integrand separates with \(f(r)=r^2\) and \(g(\theta)=1+\theta\). Include the polar area factor \(r\):

\iint_D r^2(1+\theta)\,dA = \int_0^{\pi/2}\int_1^2 r^3(1+\theta)\,dr\,d\theta.

The bounds are independent, so apply the separation formula. The radial factor is

\int_1^2 r^3\,dr = \left[\frac{r^4}{4}\right]_1^2 = \frac{16}{4}-\frac{1}{4} = \frac{15}{4}.

The angular factor is

\int_0^{\pi/2}(1+\theta)\,d\theta = \left[\theta+\frac{\theta^2}{2}\right]_0^{\pi/2} = \frac{\pi}{2}+\frac{\pi^2}{8}.

Multiplying the factors gives

\iint_D r^2(1+\theta)\,dA = \frac{15}{4}\left(\frac{\pi}{2}+\frac{\pi^2}{8}\right) = \frac{15\pi}{8}+\frac{15\pi^2}{32}.

To check, integrate first with respect to \(r\) in the original iterated integral. The inner integral is \(\frac{15}{4}(1+\theta)\); integrating it from \(0\) to \(\frac{\pi}{2}\) gives exactly \(\frac{15}{4}\left(\frac{\pi}{2}+\frac{\pi^2}{8}\right)\), the same value.

A Disk with an Angular Factor

A disk described by \(0\le r\le R\) and a fixed angular interval also has independent bounds. Its integrand can still depend on \(\theta\). The separation calculation remains valid as long as the integral exists; the method does not require the angular factor to be constant.

Worked Example: An Angular Factor over a Half-Disk

Evaluate the integral of \(\theta\) over the half-disk \(0\le r\le2\), \(0\le\theta\le\pi\). Here \(f(r)=1\) and \(g(\theta)=\theta\). With the area factor included,

\iint_D \theta\,dA = \int_0^\pi\int_0^2 \theta r\,dr\,d\theta = \left(\int_0^2 r\,dr\right) \left(\int_0^\pi \theta\,d\theta\right).

Evaluate each factor:

\int_0^2 r\,dr = \left[\frac{r^2}{2}\right]_0^2 = \frac{4}{2}-0 = 2, \qquad \int_0^\pi\theta\,d\theta = \left[\frac{\theta^2}{2}\right]_0^\pi = \frac{\pi^2}{2}-0 = \frac{\pi^2}{2}.

Therefore,

\iint_D \theta\,dA = 2\cdot\frac{\pi^2}{2} = \pi^2.

For a check, integrate in the displayed order: the radial integral of \(\theta r\) from \(0\) to \(2\) is \(2\theta\), and \(\int_0^\pi 2\theta\,d\theta=\pi^2\). This confirms the result.

When the Bounds Do Not Allow Separation

If one radial endpoint depends on the angle, the parameter region is generally not a rectangle. In that situation, even when the integrand factors, the integral usually cannot be replaced by a product of one radial integral and one angular integral. The varying limit must be respected.

Worked Example: A Separable Integrand with a Varying Radial Bound

Consider the region \(0\le\theta\le1\), \(0\le r\le\theta\), and integrate the constant function \(1\) over it. The integrand is separable, with \(f(r)=1\) and \(g(\theta)=1\), but the upper radial bound is \(\theta\). The polar integral is

\int_0^1\int_0^\theta r\,dr\,d\theta.

Evaluate the inner integral using its actual upper bound:

\int_0^\theta r\,dr = \left[\frac{r^2}{2}\right]_0^\theta = \frac{\theta^2}{2}-0 = \frac{\theta^2}{2}.

Then evaluate the outer integral:

\int_0^1\frac{\theta^2}{2}\,d\theta = \left[\frac{\theta^3}{6}\right]_0^1 = \frac{1}{6}.

Check by reversing the order: the same region has \(0\le r\le1\) and \(r\le\theta\le1\). Thus its integral is

\int_0^1\int_r^1 r\,d\theta\,dr = \int_0^1 r(1-r)\,dr = \left[\frac{r^2}{2}-\frac{r^3}{3}\right]_0^1 = \frac{1}{2}-\frac{1}{3} = \frac{1}{6}.

The agreement checks the value. Simply multiplying \(\int_0^1 r\,dr=\frac12\) by \(\int_0^1 1\,d\theta=1\) would give \(\frac12\), which is wrong: those separate intervals describe the whole rectangle \(0\le r\le1,\ 0\le\theta\le1\), not the region \(r\le\theta\).

A Common Mistake: Separating the Integrand but Ignoring the Region

A common error is to see \(f(r)g(\theta)\) and immediately split the double integral into a radial integral times an angular integral. This step is justified only when the bounds are independent and the parameter region is a rectangle. In particular, a limit such as \(r\le h(\theta)\) couples the two variables through the region, even if the function itself has no such coupling.

A second error is to leave out the factor \(r\). If the original planar integrand is \(f(r)g(\theta)\), the radial part to integrate is \(r f(r)\). For example, over \(0\le r\le2\), the radial contribution from the constant function \(1\) is \(\int_0^2 r\,dr=2\), not \(\int_0^2 1\,dr=2\) by coincidence; for other bounds or radial functions, omitting the factor changes the result. Always identify whether the stated function is the planar integrand or the full transformed integrand before calculating.

Takeaway: Separate a polar integral only after checking both conditions: the integrand factors into a radial and an angular function, and the radial and angular bounds are independent. When integrating a planar function, include the area factor \(r\) in the radial integral.

Check Your Understanding

For each question, first check the bounds and the area factor before carrying out any integration.

Write the separated form for the integral of \(f(r)g(\theta)\) over \(2\le r\le4\), \(0\le\theta\le\frac{\pi}{3}\).
For a planar integrand \(r^3\theta^2\), what radial and angular factors result after including the polar area element?
Can an integral with bounds \(0\le r\le2\cos\theta\), \(0\le\theta\le\frac{\pi}{2}\) be split directly as a product of a radial integral and an angular integral? Explain why or why not.
Evaluate the separated radial factor \(\int_0^3 r\cdot r^2\,dr\).
In the varying-bound example, why does multiplying two integrals over the intervals from \(0\) to \(1\) describe the wrong region?