Area Across a Specified Angle Interval
A polar curve can sweep out an area as its angle changes. When the interval is specified, the main task is to use the radius function on that interval—not to find every region enclosed by the curve. The radius may be positive, negative, or may change sign. In each case, the area contribution depends on the square of the radius.
The angle bounds matter: they specify which part of the curve is traced. The radius function matters because it determines how far each radial segment extends from the pole. The square makes every area contribution nonnegative, including contributions made while \(r\) is negative.
Why the Radius Is Squared
For a small angular change \(\Delta\theta\), a sector with radius \(\rho\geq0\) has area \(\frac12\rho^2\Delta\theta\), when the angle is measured in radians. For a polar curve, the radial distance from the pole is \(\rho=|r(\theta)|\). Therefore \(\rho^2=|r(\theta)|^2=r(\theta)^2\).
To see how this gives the integral, divide \([\alpha,\beta]\) into small angle intervals. On each interval, approximate the swept piece by a sector whose radius is the curve's radius at a sample angle \(\theta_i^*\). Its area is approximately \(\frac12r(\theta_i^*)^2\Delta\theta_i\). Adding these sector areas gives a Riemann sum. As the largest interval width approaches zero, the sums approach the integral of \(\frac12r(\theta)^2\), since a piecewise-continuous radius is integrable on the bounded interval.
This reasoning also explains why the sign of \(r\) does not change the area contribution. A negative radius places the point in the direction opposite to the angle \(\theta\), but its distance from the pole remains \(|r|\). Squaring that distance gives \(r^2\), not a negative quantity.
A Reliable Calculation Method
For a specified interval, use this sequence:
Write \(r=f(\theta)\), and make sure the angle interval is in radians.
Use \(f(\theta)^2\), including parentheses when the radius has multiple terms.
Evaluate \(\frac12\int_\alpha^\beta f(\theta)^2\,d\theta\).
The value is nonnegative. Check whether the interval includes negative radii or repeated sweeps.
Worked Examples
Worked Example: A Radius That Is Nonnegative on the Interval
Find the area swept by \(r=1+\cos\theta\) from \(\theta=0\) to \(\theta=\pi\).
Apply the formula and square the entire radius:
Use \(\cos^2\theta=\frac12(1+\cos 2\theta)\). Then
An antiderivative of the expression inside the integral is \(\frac32\theta+2\sin\theta+\frac14\sin 2\theta\). Therefore
Check: Differentiating \(\frac32\theta+2\sin\theta+\frac14\sin 2\theta\) gives \(\frac32+2\cos\theta+\frac12\cos 2\theta\), which equals \(1+2\cos\theta+\cos^2\theta\). The evaluated area, \(3\pi/4\), is positive.
Worked Example: A Negative Radius Throughout the Interval
Find the area swept by \(r=-2\) from \(\theta=0\) to \(\theta=\frac{\pi}{2}\).
The radius is negative, but the area formula squares it:
Since \(r=-2\), the points lie at distance \(2\) from the pole in the direction opposite \(\theta\). As \(\theta\) runs from \(0\) to \(\frac{\pi}{2}\), those directions run from \(\pi\) to \(\frac{3\pi}{2}\); the curve sweeps a quarter-circle sector of radius \(2\).
Check: The quarter-circle sector has area \(\frac14\pi(2)^2=\pi\), agreeing with the integral.
Worked Example: A Continuous Radius That Changes Sign
Find the area swept by \(r=2\cos\theta\) from \(\theta=0\) to \(\theta=\pi\), and explain what happens at the sign change.
First square the radius and integrate:
The continuous radius is positive for \(0\leq\theta<\frac{\pi}{2}\), zero at \(\frac{\pi}{2}\), and negative for \(\frac{\pi}{2}<\theta\leq\pi\). Since it is continuous and changes sign, the intermediate value theorem guarantees that it equals zero at some angle; here that angle is \(\frac{\pi}{2}\). The curve therefore passes through the pole.
Check: The integrand is \(2\cos^2\theta\), with antiderivative \(\theta+\frac12\sin 2\theta\). Its derivative is \(1+\cos 2\theta=2\cos^2\theta\), and substituting the bounds gives \(\pi\).
Negative Radii, Sign Changes, and Jumps
A negative radius is not a negative distance and does not produce negative area. In polar coordinates, \((r,\theta)\) and \((-r,\theta+\pi)\) represent the same point. Thus, where \(r<0\), the curve is represented in the direction opposite \(\theta\), while its distance from the pole is \(|r|\). The area formula still uses \(r^2\).
If \(r\) is continuous and changes sign, the intermediate value theorem guarantees that \(r=0\) somewhere in the interval. The curve then passes through the pole. This conclusion depends on continuity. A piecewise-continuous radius can instead jump from positive to negative without taking the value zero.
For instance, define \(r=1\) on \(0\leq\theta<\frac{\pi}{2}\) and \(r=-1\) on \(\frac{\pi}{2}\leq\theta\leq\pi\). This radius jumps at \(\frac{\pi}{2}\); it does not equal zero there. Its accumulated swept area is
The point on the curve jumps from one side of the pole to the opposite side at that angle. It is not correct to say that this discontinuous curve passes through the pole. The integral remains valid for the piecewise-continuous radius, with the swept sectors understood on the two parts of the interval.
Common Mistake: Keeping the Sign of the Radius
A tempting but incorrect approach is to use \(r\) instead of \(r^2\), as if a negative radius should subtract area. For \(r=-2\) on \([0,\frac{\pi}{2}]\), that error gives
This is not an area: it is negative, and it incorrectly treats radius as a signed area contribution. The correction is to square the radius before integrating:
Another issue is confusing accumulated swept area with the area of the union of all points swept. If an interval traces a region more than once, the integral counts the repeated sweep each time. To interpret an answer as the area of a region counted only once, check whether the interval retraces any portion of the curve.
Bounds and Interpretation
The angle bounds should be written in increasing order when the goal is area. If the bounds are reversed, the definite integral changes sign; that signed value represents reversed orientation, not a positive area. For an area calculation, use the interval from the smaller angle to the larger angle.
Also check whether the question asks for the area swept over a particular interval or for some enclosed region. The formula here answers the first question. Selecting bounds for a region enclosed by a curve, or comparing areas between two curves, requires identifying the relevant boundaries and intervals; the bounds are not interchangeable merely because the same radius function appears.
Check Your Understanding
Use the squared-radius formula and pay attention to the specified angle intervals.
- Set up, but do not evaluate, the swept-area integral for \(r=3\sin\theta\) on \([0,\frac{\pi}{2}]\).
- Find the area swept by \(r=-1\) on \([0,\pi]\).
- If \(r\) is continuous and changes from positive to negative on an interval, what must occur at some angle?
- Can a piecewise-continuous radius jump from positive to negative without equaling zero? Explain.
- What does the area integral count if the curve sweeps over the same region twice?
