Calc III · §15.3 double integrals in polar coordinates

Volume under z = x²+y², inside the cylinder x²+y² = 2x

The cylinder x²+y²=2x is the circle (x−1)²+y²=1 — in polar it's r = 2cosθ. The base region D is swept by rays from the origin: for each θ in [−π/2, π/2], r runs from 0 to 2cosθ. The height is z = r², and dA = r dr dθ.

The solid — paraboloid roof over the circular base
Base region D in the xy-plane — the circle r = 2cosθ

Set up in polar and integrate

volume V = ∬D (x²+y²) dA
to polar = ∫−π/2π/2 ∫02cosθ (r²)·r dr dθ
inner ∫ r³ dr = ∫−π/2π/2 [r⁴/4]02cosθ dθ = ∫−π/2π/2 4cos⁴θ dθ
cos⁴ reduction = 4 · (3π/8) = 3π/2 ≈ 4.712

Polar turns an ugly circle into r = 2cosθ, and the Jacobian r makes r²·r = r³. 🧮