AP® Calculus BC · Unit 9 · Topic 9.9
Finding the Area of the Region Bounded by Two Polar Curves
The area between two polar curves comes from subtracting two sector-area integrals, (1/2) r_out2 - (1/2) r_in2, taken only over the angles where that curve is actually the outer or inner boundary.
In rectangular coordinates, intersections come from setting two functions equal, and that's the whole story. In polar coordinates, the same point can be written with infinitely many (r, theta) pairs, since adding 2pi to theta or flipping the sign of r and adding π can land on the same spot. So two curves can cross at a location even when their r-expressions, set equal, give no solution at that angle. Before integrating, you need the true angles of intersection, checked graphically or algebraically, including the pole, where r = 0 for each curve separately.
Once you know the bounding angles, the region's area is a difference of sector integrals: (1/2) times the integral of r_outer2 minus r_inner2, over the theta-interval where that labeling holds. If the two curves swap which one is outer partway through the region, you split the integral at the crossing angle and handle each piece separately, since a single formula across the whole range would subtract the wrong curve somewhere. Symmetry often cuts the work in half: find the area of one symmetric piece and double it, rather than integrating the whole messy region at once.
This topic is the capstone of the polar unit, built directly on finding single-curve polar area and on locating polar intersections. It also pairs with the idea that arc length and area in polar form both come from converting the geometry into theta-integrals of r and its derivative, the same shift in thinking that parametric and vector-valued curves demand elsewhere in this unit.
Key terms
- pole
- The origin in polar coordinates, where r = 0 regardless of theta.
- sector integral
- The integral (1/2) times the integral of r2 d theta, which computes the area swept out by a polar curve.
- outer curve
- Whichever curve has the larger r-value, and therefore farther distance from the pole, over a given range of theta.
- symmetry argument
- Using a curve's mirror or rotational symmetry to compute one piece of a region and double or multiply it instead of integrating the whole thing.
Watch it explained
- How it worksHow do I subtract two sector integrals over exactly the right range of angles?
- Worked exampleFind the area inside r = 3 sin(theta) and outside r = 1 + sin(theta).
- Worked exampleFind the area shared by the circle r = 2 and the circle r = 4 cos(theta).
- Common mistakeWhy isn't every intersection point found by setting the two r expressions equal?
Check yourself
End of topic.