Polar Area and Arc Length
Area swept by a polar curve is \(\tfrac12\int r^2\,d\theta\), and polar arc length is \(\int\sqrt{r^2 + (dr/d\theta)^2}\,d\theta\) — both built by summing thin circular sectors rather than the rectangles of Cartesian calculus.
By the end you'll be able to set up and evaluate \(A = \tfrac12\int r^2\,d\theta\) for a region enclosed by a polar curve (finding the right \(\theta\)-bounds), and compute polar arc length with \(L = \int\sqrt{r^2 + (dr/d\theta)^2}\,d\theta\).
Predict: the petal is widest from the origin at its centerline (\(\theta = 0\), where \(r = 2\)) and pinches to \(r = 0\) at its edges. Does each \(d\theta\) slice add MORE area where \(r\) is large? Sweep \(\theta\) below to check — watch the green sector's size and the \(\tfrac12 r^2\) rate.
This is the rose \(r = 2\cos(3\theta)\) — the same kind of curve you learned to read in the previous lesson, but now we measure the area inside one petal. The sweep runs \(\theta\) from \(-\pi/6\) to \(\pi/6\) (the angles where \(r = 0\)); each tiny slice is a circular sector of area \(\tfrac12 r^2\,d\theta\), and the shaded region accumulates them. Hover or focus the point on the curve for exact values.
Swept area A ≈ 0.524 of petal total π/3 ≈ 1.047 · at θ = 0.000: r = 2.000, sector rate ½r² = 2.000
Polar area sums pie-slice wedges, \(dA = \tfrac12 r^2\,d\theta\); polar arc length sums tiny displacements, giving \(L = \int\sqrt{r^2 + (dr/d\theta)^2}\,d\theta\). Both follow from the geometry of sweeping an angle instead of marching along an axis.
In Cartesian calculus you build area from thin rectangles: width \(dx\), height \(y\), area \(y\,dx\). A polar curve isn't organized that way — it's organized around sweeping through an angle. So the natural slice is a thin pie wedge: as \(\theta\) advances by \(d\theta\), the radius line sweeps out a near-perfect circular sector of radius \(r\). Geometry says a sector of radius \(r\) and angle \(\theta\) has area \(\tfrac12 r^2\theta\), so the wedge contributes \(dA = \tfrac12 r^2\,d\theta\). That's exactly the green sector in the demo above — where \(r\) doubles, each slice contributes four times the area, because the rate goes as \(r^2\).
Summing the sectors as \(\theta\) sweeps from \(\alpha\) to \(\beta\) gives \[ A = \int_\alpha^\beta \tfrac12\big[f(\theta)\big]^2\,d\theta. \] For arc length, sum tiny straight displacements \(ds = \sqrt{dx^2 + dy^2}\) with \(\theta\) as the parameter: substituting \(x = r\cos\theta\), \(y = r\sin\theta\) and applying the product rule, \(\left(\frac{dx}{d\theta}\right)^2 + \left(\frac{dy}{d\theta}\right)^2\) simplifies — the cross terms cancel via \(\sin^2\theta + \cos^2\theta = 1\) — to \(r^2 + \left(\frac{dr}{d\theta}\right)^2\). Hence \[ L = \int_\alpha^\beta \sqrt{r^2 + \left(\frac{dr}{d\theta}\right)^2}\; d\theta. \] The practical hard part is the bounds: for petals and loops, find where \(r = 0\) first, and integrate between consecutive zeros.
Find the area of one petal of \(r = 2\cos(3\theta)\). The petal's edges are where \(r = 0\): \(\cos(3\theta) = 0\) at \(3\theta = \pm\pi/2\), i.e. \(\theta \in [-\pi/6, \pi/6]\). Then \[ A = \int_{-\pi/6}^{\pi/6} \tfrac12 (2\cos 3\theta)^2\, d\theta = \int_{-\pi/6}^{\pi/6} 2\cos^2(3\theta)\, d\theta = \int_{-\pi/6}^{\pi/6}\big(1 + \cos 6\theta\big)\,d\theta \] using \(\cos^2 u = \tfrac12(1 + \cos 2u)\). Evaluating \(\left[\theta + \tfrac{\sin 6\theta}{6} \right]_{-\pi/6}^{\pi/6}\): the sine term vanishes at both ends (\(\sin(\pm\pi) = 0\)), leaving \(A = \tfrac{\pi}{6} - \left(-\tfrac{\pi}{6}\right) = \) \(\pi/3 \approx 1.047\) — exactly the total the demo's sweep accumulates.
Use the polar arc-length formula on a curve you know the answer for: the circle \(r = 5\), \(\theta \in [0, 2\pi]\). Here \(r\) is constant, so \(dr/d\theta = 0\) and \[ L = \int_0^{2\pi} \sqrt{5^2 + 0^2}\; d\theta = \int_0^{2\pi} 5\, d\theta = \_\_\_\_ \]
Reveal the answer
\(L = 5\theta \big|_0^{2\pi} = \) \(10\pi\) — the familiar circumference \(2\pi \cdot 5\). Whenever \(dr/d\theta = 0\), the integrand collapses to \(\sqrt{r^2} = r\), so the formula reproduces circle geometry exactly — a quick sanity check to remember.
More info — why the arc-length cross terms cancel
Differentiating \(x = r\cos\theta\) and \(y = r\sin\theta\) (with \(r = f(\theta)\)) gives \(\frac{dx}{d\theta} = \frac{dr}{d\theta}\cos\theta - r\sin\theta\) and \(\frac{dy}{d\theta} = \frac{dr}{d\theta}\sin\theta + r\cos\theta\). Square both: each produces a \(\left(\frac{dr}{d\theta}\right)^2\) term, an \(r^2\) term, and a cross term \(\pm 2r\frac{dr}{d\theta}\sin\theta\cos\theta\). The cross terms carry opposite signs, so they cancel on adding, and the remaining pairs combine through \(\sin^2\theta + \cos^2\theta = 1\) into the clean \(r^2 + \left(\frac{dr}{d\theta}\right)^2\). This is the same \(ds = \sqrt{dx^2 + dy^2}\) idea behind every arc-length formula — polar coordinates just make the algebra collapse nicely. Paul's Online Math Notes (Dive deeper below) walks the companion area setups step by step.
Check your understanding
Find the area enclosed by the circle \(r = 2\sin\theta\), traced once as \(\theta\) runs from \(0\) to \(\pi\).
Why is the polar area element \(\tfrac12 r^2\,d\theta\) rather than \(r\,d\theta\) or \(y\,dx\)?
Which integral gives the arc length of the spiral \(r = e^{\theta}\) for \(0 \le \theta \le 1\)?
You recognize \(r = 4\cos\theta\) as a circle. Which circle is it, and what area does \(\tfrac12\int_{-\pi/2}^{\pi/2} r^2\,d\theta\) give?
Recap
- Polar area sums circular sectors: \(dA = \tfrac12 r^2\,d\theta\), so \(A = \int_\alpha^\beta \tfrac12 r^2\,d\theta\) — fundamentally different from Cartesian rectangles \(y\,dx\).
- Polar arc length is \(L = \int_\alpha^\beta \sqrt{r^2 + (dr/d\theta)^2}\,d\theta\), derived from \(ds = \sqrt{dx^2 + dy^2}\) via \(x = r\cos\theta\), \(y = r\sin\theta\).
- For petals and loops, find the \(\theta\)-bounds first — usually consecutive angles where \(r = 0\).
- Sanity check on \(r = a\): the arc-length formula gives \(\int_0^{2\pi} a\,d\theta = 2\pi a\), the circumference.
Dive deeper
- Paul's Online Math Notes — Area with Polar Coordinates Set up A = ∫½r² dθ for a rose petal and for regions between two polar curves.
Sources
- Calculus in Polar Coordinates — Area and Arc Length