Polar coordinates: ∫∫(x²+y²) dy dx

I = ∫₀¹ ∫ₓ^{√(2x−x²)} (x² + y²) dy dx

Region

• Lower limit y = x is a straight line through the origin (θ = π/4).
• Upper limit y = √(2x − x²) gives y² = 2x − x², i.e. (x − 1)² + y² = 1: the upper half of the circle with centre (1, 0), radius 1. In polar form: r² = 2r cos θ, so r = 2 cos θ.
• The line and circle meet at (0,0) and (1,1).
• For 0 ≤ x ≤ 1 the region lies above y = x and inside the circle.

y
1 |      *(1,1)
  |    /' .
  |  /  region  .
  |/ _ _ _ _ _ _ . x
 0                 2
 (line y=x, circle r = 2cosθ)

So θ runs from π/4 (the line) to π/2 (the y-axis, where the circle meets the origin), and r runs from 0 to 2 cos θ.

Integrate

x² + y² = r², dy dx = r dr dθ:

I = ∫_{π/4}^{π/2} ∫₀^{2cosθ} r³ dr dθ
  = ∫_{π/4}^{π/2} (2cosθ)⁴/4 dθ
  = 4 ∫_{π/4}^{π/2} cos⁴θ dθ

Use cos⁴θ = (3 + 4 cos 2θ + cos 4θ)/8:

I = (1/2) [ 3θ + 2 sin 2θ + (1/4) sin 4θ ]_{π/4}^{π/2}
  = (1/2) [ (3π/2 + 0 + 0) − (3π/4 + 2 + 0) ]
  = (1/2)(3π/4 − 2)
  = 3π/8 − 1

Answer: I = 3π/8 − 1 ≈ 0.178.