Sweep out a tiny angle \(d\theta\) at radius \(r\). The sliver you trace isn't a rectangle — it's a circular sector, a pie slice. A full circle of radius \(r\) has area \(\pi r^2\) and spans \(2\pi\); a slice of angle \(d\theta\) is that fraction of it:
Add up all the slices from the starting angle \(\alpha\) to the ending angle \(\beta\) and the sum becomes an integral — the polar area formula:
Watch the slices fill the region below. Each thin cyan wedge is one \(\tfrac12 r^2\,d\theta\) sector. Pick a curve, then drag the α and β handles to set the limits — the area updates live, so you can feel how the bounds decide the answer (drag them to the full region and the total snaps to the exact value). Switch to the two-curve mode to see where limits come from when two curves meet.
Polar area is \(\tfrac12\int r^2\,d\theta\) — not \(\int y\,dx\) (that's rectangular) and not \(\int r\,d\theta\) (that forgets the sector is a slice, not an arc). Miss the ½ or the square on r and the whole problem is wrong.
Setting up \(\tfrac12\int r^2\,d\theta\) is the easy half. The half that decides whether you get the problem right is the limits of integration — the start and end angles. There's no single rule; instead, identify which of three situations you're in.
A single petal or loop begins and ends at the pole, where the radius is zero. So set \(f(\theta)=0\) and solve; two consecutive solutions are your bounds. For the rose \(r=2\cos(2\theta)\), one petal runs between the angles where it hits the pole:
A curve traced exactly once as the angle goes around — a cardioid, a single-loop limaçon — is swept out over \(0\le\theta\le 2\pi\). But check that it isn't retraced: a circle like \(r=2\cos\theta\) completes in just \(0\) to \(\pi\) (the negative-\(r\) half retraces it), so using \(2\pi\) would double-count the area.
When the region is bounded by two different curves, the limits are the angles where they intersect. Set the radii equal and solve — these crossing angles become \(\alpha\) and \(\beta\). This is the step most often missed on the exam:
Before integrating, draw the region. A quick sketch tells you which case you're in, whether the curve retraces, which curve is outer, and roughly where the bounds should land — so you can sanity-check the angles you solve for. Skipping the sketch is how a setup that looks right silently uses the wrong limits.
The mechanics: square the radius, integrate, multiply by a half. The real work is choosing the limits — the angles where the region begins and ends. For a closed loop that's often where \(r=0\) (the curve touches the pole) or a full \(0\) to \(2\pi\) for a curve traced once.
For a single rose petal, the limits come straight from setting \(r=0\). The petal of \(r=2\cos(2\theta)\) lives between the two angles where the radius vanishes:
Setting up \(\tfrac12\int r^2\,d\theta\) is easy; the points lost are almost always on the limits of integration. Sketch the curve, find where the loop or petal starts and ends (frequently the \(r=0\) angles), and integrate over exactly that span — not blindly from \(0\) to \(2\pi\), which would trace some curves more than once.
For the region trapped between an outer and an inner polar curve, subtract the inner sector area from the outer — the polar version of "top minus bottom":
But before you can integrate, you need the limits — and now they come from where the two curves intersect. Set the radii equal and solve. That's the step that decides the whole problem, and it's where most points are lost.
Switch the visual at the top to "Two curves: where do they cross?" — it shows both pictures at once. On the left, the polar curves \(r_1=3\sin\theta\) and \(r_2=1+\sin\theta\) and the points where they physically meet. On the right, those same radii graphed against \(\theta\): solving \(r_1=r_2\) is literally where the two graphs cross. Both views mark the limits, \(\theta=\tfrac{\pi}{6}\) and \(\tfrac{5\pi}{6}\).
The integrand is \(r_{\text{outer}}^2-r_{\text{inner}}^2\), not \((r_{\text{outer}}-r_{\text{inner}})^2\). Square each radius first, then subtract. And make sure you've correctly identified which curve is outer over the interval — if they swap, the region splits into pieces.
Find the area enclosed by one petal of the rose \(r=2\cos(2\theta)\).
Find the limits. One petal runs between consecutive \(r=0\) angles: \(2\cos(2\theta)=0\Rightarrow 2\theta=\pm\tfrac{\pi}{2}\Rightarrow\theta=\pm\tfrac{\pi}{4}\).
Set up and integrate. Using \(\cos^2 u=\tfrac{1+\cos 2u}{2}\), this evaluates to π/2
Find the total area enclosed by the cardioid \(r=1+\cos\theta\).
Limits. The cardioid is traced once over \(0\) to \(2\pi\), so those are the limits.
Integrate. Expanding \((1+\cos\theta)^2\) and integrating term by term gives 3π/2
Find the area inside \(r=3\sin\theta\) and outside \(r=1+\sin\theta\).
Find the intersection (the limits). Set them equal: \(3\sin\theta=1+\sin\theta\Rightarrow 2\sin\theta=1\Rightarrow\sin\theta=\tfrac12\), so \(\theta=\tfrac{\pi}{6}\) and \(\tfrac{5\pi}{6}\).
Outer minus inner. Over that span \(r=3\sin\theta\) is outer.
Evaluate. The integral works out to a clean π
Five questions on the formula, the limits, and the two-curve setup.