Trigonometric Integrands over Sectors

Why Angular Factors Change a Sector Integral

A sector has a simple angular interval, but an integrand such as \(\sin(k\theta)\) or \(\cos(k\theta)\) varies across that interval. Its integral therefore depends on the angular bounds, not just on the sector's radial size. The polar area element also contributes a factor of \(r\), which must be included even when the original integrand has no radial factor.

For a sector or annular sector with fixed radial and angular bounds, the radial and angular parts can be integrated separately when the integrand is a product of a function of \(r\) and a function of \(\theta\). The angular antiderivative then accounts for both the frequency \(k\) and the endpoints of the sector.

Definition: A polar sector with fixed radial bounds is the region described by \(a\le r\le b\) and \(\alpha\le\theta\le\beta\), where \(0\le a<b\) and the angular interval describes the chosen sector. If \(a=0\), the region is a circular sector; if \(a>0\), it is an annular sector.

The Polar Integral with an Angular Factor

In polar coordinates, the area element is \(dA=r\,dr\,d\theta\). Thus, when the integrand is \(q(r)\sin(k\theta)\), the integral over a sector with fixed bounds becomes

\iint_D q(r)\sin(k\theta)\,dA = \int_\alpha^\beta\int_a^b q(r)\sin(k\theta)\,r\,dr\,d\theta.

Here \(q\) is a continuous radial factor on \([a,b]\), and \(k\) is a positive integer. The same setup applies to a cosine factor. The additional \(r\) in the integrand comes from \(dA\), not from the function being integrated.

Because the radial bounds do not depend on \(\theta\), the radial and angular factors separate. In particular, if \(A\) and \(B\) are constants, then

\iint_D q(r)\bigl(A\sin(k\theta)+B\cos(k\theta)\bigr)\,dA = \left(\int_a^b q(r)r\,dr\right) \left(\int_\alpha^\beta \bigl(A\sin(k\theta)+B\cos(k\theta)\bigr)\,d\theta\right).

This factorization is useful for a sector with constant radial bounds. If the radial bounds depend on \(\theta\), the radial integral generally depends on \(\theta\) too, so this product formula need not apply.

Angular antiderivatives: For \(k\ge1\), \(\int\sin(k\theta)\,d\theta=-\cos(k\theta)/k\) and \(\int\cos(k\theta)\,d\theta=\sin(k\theta)/k\). The factor \(1/k\) is required by the chain rule.

For example, the definite angular integrals are

\int_\alpha^\beta\sin(k\theta)\,d\theta = \frac{\cos(k\alpha)-\cos(k\beta)}{k}, \qquad \int_\alpha^\beta\cos(k\theta)\,d\theta = \frac{\sin(k\beta)-\sin(k\alpha)}{k}.

If the angular factor is a linear combination, its definite integral is the same linear combination of these two expressions. These formulas also show why it is important to use the actual angular endpoints: different sectors can give different values even if they have the same radius.

Why the Separation Works

The factorization follows directly from the iterated integral. After including the polar area element, a product integrand has the form \(F(r)G(\theta)\), and the bounds form a rectangle in the \((r,\theta)\)-plane. Since each factor depends on only one variable, the iterated integral can be evaluated by first taking the radial integral and then the angular integral.

Include the area factor.
Replace \(dA\) by \(r\,dr\,d\theta\), so a radial factor \(q(r)\) becomes \(q(r)r\) in the integrand.

Separate the variables.
With fixed radial bounds, write the integrand as the product of a function of \(r\) and a function of \(\theta\).

Evaluate and multiply.
Compute the radial definite integral and the angular definite integral, then multiply their values.

This is a direct application of iterated integration for a continuous integrand on a bounded region with these fixed bounds. It is not a claim that any polar integral can be split this way: when a bound couples \(r\) and \(\theta\), or when the integrand is not separable, the product of two separate integrals may not represent the original integral.

Worked Sector Integrals

Worked Example: A Sine Factor on a Quarter-Disk Sector

Evaluate \(\iint_D r^2\sin(2\theta)\,dA\), where \(D\) is the sector \(0\le r\le2\), \(0\le\theta\le\pi/2\).

First include the polar area factor \(r\). The integrand in the iterated integral is \(r^3\sin(2\theta)\), so

\iint_D r^2\sin(2\theta)\,dA = \left(\int_0^2 r^3\,dr\right) \left(\int_0^{\pi/2}\sin(2\theta)\,d\theta\right).

For the radial integral, an antiderivative is \(r^4/4\), whose derivative is \(r^3\). Therefore,

\int_0^2 r^3\,dr = \left[\frac{r^4}{4}\right]_0^2 = \frac{2^4}{4}-\frac{0^4}{4} = 4.

For the angular integral, an antiderivative is \(-\cos(2\theta)/2\), whose derivative is \(\sin(2\theta)\). Thus,

\int_0^{\pi/2}\sin(2\theta)\,d\theta = \left[-\frac{\cos(2\theta)}{2}\right]_0^{\pi/2} = -\frac{\cos\pi}{2}+\frac{\cos0}{2} = \frac12+\frac12 = 1.

Multiplying the two values gives \(4\cdot1=4\). This is consistent with the integrand being nonnegative throughout the sector, since \(2\theta\) ranges from \(0\) to \(\pi\).

Worked Example: A Cosine Factor on a Narrow Sector

Evaluate \(\iint_D \cos(2\theta)\,dA\), where \(0\le r\le2\) and \(0\le\theta\le\pi/6\).

The integrand has no radial factor, but the area element still contributes \(r\). Hence

\iint_D \cos(2\theta)\,dA = \left(\int_0^2 r\,dr\right) \left(\int_0^{\pi/6}\cos(2\theta)\,d\theta\right).

The radial antiderivative \(r^2/2\) has derivative \(r\), so

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

An angular antiderivative is \(\sin(2\theta)/2\), whose derivative is \(\cos(2\theta)\). Therefore,

\int_0^{\pi/6}\cos(2\theta)\,d\theta = \left[\frac{\sin(2\theta)}{2}\right]_0^{\pi/6} = \frac{\sin(\pi/3)}{2}-\frac{\sin0}{2} = \frac{\sqrt{3}}{4}.

The integral is \(2(\sqrt{3}/4)=\sqrt{3}/2\). This value is positive, as expected because \(\cos(2\theta)>0\) throughout the given angular interval.

Worked Example: Combining Sine and Cosine on an Annular Sector

Evaluate \(\iint_D r^2\bigl(2\sin(3\theta)+\cos(3\theta)\bigr)\,dA\), where \(1\le r\le2\) and \(\pi/6\le\theta\le\pi/3\).

Including \(dA=r\,dr\,d\theta\) gives the radial factor \(r^3\). Since the bounds are fixed, separate the integral:

\iint_D r^2\bigl(2\sin(3\theta)+\cos(3\theta)\bigr)\,dA = \left(\int_1^2 r^3\,dr\right) \left(\int_{\pi/6}^{\pi/3} \bigl(2\sin(3\theta)+\cos(3\theta)\bigr)\,d\theta\right).

The radial antiderivative is \(r^4/4\), with derivative \(r^3\), so

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

An angular antiderivative is \(-2\cos(3\theta)/3+\sin(3\theta)/3\). Differentiating gives \(2\sin(3\theta)+\cos(3\theta)\), as required. At \(\theta=\pi/3\), its value is \(2/3\); at \(\theta=\pi/6\), its value is \(1/3\). Thus the angular integral is \(2/3-1/3=1/3\), and the full integral is

\frac{15}{4}\cdot\frac13=\frac54.

As a check on the angular calculation, the sine contribution is \(2\cdot(1/3)=2/3\), while the cosine contribution is \(-1/3\). Their sum is \(1/3\), matching the evaluated antiderivative.

A Common Pitfall: Omitting the Polar Area Factor

A frequent error is to replace \(x\) and \(y\) by polar expressions but then integrate without the factor \(r\) in \(dA\). The first example makes the effect of this mistake explicit. If one incorrectly uses \(r^2\sin(2\theta)\,dr\,d\theta\), the radial integral becomes

\int_0^2 r^2\,dr = \left[\frac{r^3}{3}\right]_0^2 = \frac83.

Multiplying by the correct angular integral, \(1\), would give the incorrect answer \(8/3\). The area element is \(r\,dr\,d\theta\), so the correct radial power is \(r^3\), and the correct value is \(4\). The check is simple: before integrating, write the full integrand after including \(dA\).

When the Angular Integral Vanishes

Sine and cosine factors can take both positive and negative values across a sector. Their positive and negative contributions can cancel, making the angular integral zero even when the radial integral is positive. For example,

\int_0^\pi \cos\theta\,d\theta = [\sin\theta]_0^\pi = 0.

For a product integrand with fixed radial bounds, a zero angular integral makes the entire double integral zero. This conclusion comes from evaluating the antiderivative at the actual endpoints; it should not be inferred just from the presence of a trigonometric factor.

For a sector integral, a reliable order of work is: identify the radial and angular bounds, include the factor \(r\), separate the variables only when the bounds permit it, and evaluate both definite integrals at their endpoints. The angular frequency determines the \(1/k\) in the antiderivative, while the radial power is determined by both the original integrand and the polar area element.

Check Your Understanding

Use the polar area element and show the main radial and angular steps in each calculation.

Set up, but do not evaluate, the integral of \(r\sin(4\theta)\) over \(0\le r\le3\), \(0\le\theta\le\pi/8\).
Evaluate \(\int_{\pi/6}^{\pi/3}\sin(2\theta)\,d\theta\).
For \(0\le r\le1\), \(0\le\theta\le\pi/2\), evaluate the double integral of \(\cos(2\theta)\).
Explain why the product of a radial integral and an angular integral is available for fixed sector bounds.
What factor is missing if an integral in polar coordinates is written with \(dr\,d\theta\) but no \(r\) from the area element?