AP Calculus BC

Polar Area vs Cartesian Area

Q: Why does the polar area formula have a one half in it?

Because the slice is a circular sector, whose area is \frac{1}{2}r^2\Delta\theta. The rectangle behind the Cartesian formula carries no such factor.

Q: Do I square the difference of the two radii?

No. Square each radius first, then subtract: \frac{1}{2}\int_{\alpha}^{\beta}\left(r_{\text{out}}^2 - r_{\text{in}}^2\right)\,d\theta. Squaring the difference gives the wrong area, exactly as it does for a washer.

Q: Could I convert the region to Cartesian and integrate instead?

In principle, using x = r\cos\theta and y = r\sin\theta, but the bounds usually become far worse. On the BC exam the sector formula is always the intended route.

Use polar area when the region is swept out by an angle and the boundary is r as a function of theta, because the natural slice is a circular sector and that is where the factor of one half comes from. Use Cartesian area when the boundary is y as a function of x and the slice is a vertical strip of width dx.

Polar area

Use when: The boundary is $r = f(\theta)$ and the region is swept between two rays, as for a rose petal or a cardioid.

Cartesian area

Use when: The boundary is $y = f(x)$ and the region sits between two vertical lines.

Side by side

	Polar area	Cartesian area
Shape of one slice	A thin circular sector	A thin vertical rectangle
Formula	$\frac{1}{2}\int_{\alpha}^{\beta} r^2\,d\theta$ , with $\alpha$ to $\beta$ sweeping the region exactly once	$\int_a^b y\,dx$ when $y \ge 0$ ; otherwise $\int_a^b \lvert y \rvert\,dx$
Limits are	Angles	$x$ values
Region between two curves	$\frac{1}{2}\int_{\alpha}^{\beta}\left(r_{\text{out}}^2 - r_{\text{in}}^2\right)\,d\theta$	$\int_a^b \left(y_{\text{top}} - y_{\text{bot}}\right)\,dx$
Common trap	Dropping the $\frac{1}{2}$ , subtracting the radii before squaring, or sweeping the region twice, as $[0, 2\pi]$ does for $r = \cos(3\theta)$	Using $\int y\,dx$ on a region swept by an angle

The one half is geometry, not a correction factor. A sector of radius $r$ and angle $\Delta\theta$ takes up the fraction $\frac{\Delta\theta}{2\pi}$ of a full disc of area $\pi r^2$ , which works out to $\frac{1}{2}r^2\Delta\theta$ . Adding those slices and taking the limit produces the integral.

A = \frac{1}{2}\int_{\alpha}^{\beta} \left[f(\theta)\right]^2\,d\theta

The Cartesian formula adds rectangles instead: height $y$ , width $dx$ , no leftover factor. The two are not swappable, since a sector and a rectangle are different shapes. Applying $\int y\,dx$ to a region swept by an angle is not off by a constant, it is measuring something else entirely.

Finding the polar limits

The limits are the angles where the region starts and stops, usually found by solving $r = 0$ . For one petal of $r = \cos(3\theta)$ , $r = 0$ at $\theta = -\frac{\pi}{6}$ and $\theta = \frac{\pi}{6}$ , so that pair of angles sweeps exactly one petal.

Frequently asked questions

Why does the polar area formula have a one half in it?

Because the slice is a circular sector, whose area is $\frac{1}{2}r^2\Delta\theta$ . The rectangle behind the Cartesian formula carries no such factor.

Do I square the difference of the two radii?

No. Square each radius first, then subtract: $\frac{1}{2}\int_{\alpha}^{\beta}\left(r_{\text{out}}^2 - r_{\text{in}}^2\right)\,d\theta$ . Squaring the difference gives the wrong area, exactly as it does for a washer.

Could I convert the region to Cartesian and integrate instead?

In principle, using $x = r\cos\theta$ and $y = r\sin\theta$ , but the bounds usually become far worse. On the BC exam the sector formula is always the intended route.

In the CED: Unit 8: Applications of Integration, Unit 9: Parametric, Polar, and Vector (BC)