Double Integrals in Polar Coordinates
Integrate over circular regions by switching to polar coordinates, where the area element carries an extra factor of r.
What a learner can do afterwards
- Convert a region and an integrand to polar form
- Explain where the factor of r in the area element comes from
- Evaluate an integral over a disc or an annulus
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Use radius r >= 0 and angle theta in radians. Substitute x = r cos theta and y = r sin theta. Then r squared = x squared + y squared. To find r, add the coordinate squares and take the nonnegative square root. For example, (5, 12) has radius 13. Substitute for x and y in other expressions too.
The area element dA measures a small patch in the xy-plane. In polar coordinates, dA = r dr dtheta. The factor r scales the patch area. To convert an integral, substitute into its function and multiply by this area factor once. The factor belongs to dA even when the function is the constant 1.
For constant radial bounds a to b and angular width h, the correct area is (b squared - a squared) h / 2. If r is omitted from dA, integrating 1 dr dtheta instead gives (b - a) h. Compare these two results to measure the error. For 0 <= a < b and h > 0, the omitted-factor result divided by the correct area is 2 / (a + b). The error can be an overestimate or an underestimate.
The unit disc centred at the origin has 0 <= r <= 1. Its angle covers one full turn, such as 0 <= theta <= 2 pi. Any consecutive interval of width 2 pi gives the same full disc. Shared boundary rays and the origin have zero area. A sweep of 4 pi counts the interior twice.
An origin-centred disc uses r from 0 to its radius; an origin-centred annulus uses its inner and outer radii. A sector of radius R with vertex at the origin has 0 <= r <= R and its actual angular limits. The first-quadrant quarter of an origin-centred disc uses 0 <= theta <= pi/2. A narrower sector uses its own interval, such as pi/4 <= theta <= pi/2, not the whole quadrant. Off-centre circles can need r bounds that depend on theta.
Integrating 1 over a region gives its area. First integrate r: an antiderivative is r squared / 2. With radial bounds a to b, the result is (b squared - a squared) / 2. If this result is independent of theta, multiply by the angular width, the upper angle minus the lower angle. For a full annulus centred at the origin, the area is pi (b squared - a squared).
At radius r, an angular width h radians gives arc length r h. A small polar cell has radial width dr and arc width about r dtheta, so its area is about r dr dtheta. This explains the differential dA = r dr dtheta. For a finite cell from radius a to b with angular width h, subtract two sector areas: its exact area is (b squared - a squared) h / 2.
For volume under a nonnegative height z = x squared + y squared, the height becomes r squared. Multiplying by dA gives r cubed dr dtheta. An antiderivative of r cubed is r to the fourth power / 4. Over an origin-centred disc of radius 3, integrate r from 0 to 3 to get 81/4, then theta from 0 to 2 pi to get volume 81 pi / 2. An annulus uses the same method with its nonzero inner radius.
Substitute r and theta, carry the extra r, set round limits, and integrate inside out.
Original lesson and exercises for this learning dataset.
Original readings and exercises for this learning dataset. Video: Polar coordinates by Khan Academy.
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