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
1 · Read
Circles and sectors fight rectangular coordinates, so switch to radius r and angle theta. Substitute x = r cos theta and y = r sin theta, and describe the region with r and theta ranges. Since r squared equals x squared plus y squared, the point (3, 4) has r squared of 25 and r of 5.
The key move is the area element: dx dy becomes r dr dtheta, never plain dr dtheta. Mia is right that the extra r appears because polar grid cells get wider as r grows: a cell of fixed angular width has arc side r times angle, so its area grows with r. Forgetting that extra r is the classic error.
The unit disc x squared plus y squared <= 1 becomes simply 0 <= r <= 1 with theta sweeping 0 to 2 pi. Rings, called annuli, use a nonzero inner radius, and slices, called sectors, restrict theta. Whenever the boundary or the integrand involves x squared plus y squared, polar form is worth trying first.
Set up inside out. The inner r integral usually produces powers of r thanks to the extra factor: the integral from 0 to 2 of r dr uses antiderivative r squared over 2, giving 2. The outer theta integral often just multiplies by the swept angle. For z = x squared plus y squared, the integrand becomes r squared, so the volume over the unit disc follows the same pattern.
Substitute r and theta, carry the extra r, set round limits, and integrate inside out.
2 · Watch
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Where it sits
8 questions wait behind this lesson, each with its answer explained. Every answer feeds the sky: stars light as they are learned, and dim when it is time to come back.