Transform to polar and evaluate ∫∫(x²+y²), 2ax−x²

I = ∫₀^{2a} ∫₀^{√(2ax−x²)} (x² + y²) dy dx

Region

y ranges from 0 to √(2ax − x²), i.e. the upper half of the circle y² = 2ax − x², or (x − a)² + y² = a². This circle has centre (a, 0), radius a, and passes through the origin. x ranges from 0 to 2a. So the region is the upper half-disc.

In polar form, x² + y² = 2ax becomes r² = 2a r cos θ, i.e. r = 2a cos θ. The half-disc above the x-axis corresponds to 0 ≤ θ ≤ π/2 and 0 ≤ r ≤ 2a cos θ.

y
a |    . - .
  |  /       \
  | /         \
  |/___________\__ x
  O     a      2a

Integral

x² + y² = r², dy dx = r dr dθ:

I = ∫₀^{π/2} ∫₀^{2a cosθ} r² · r dr dθ
  = ∫₀^{π/2} (2a cosθ)⁴/4 dθ
  = 4a⁴ ∫₀^{π/2} cos⁴θ dθ

By Wallis' formula, ∫₀^{π/2} cos⁴θ dθ = (3/4)(1/2)(π/2) = 3π/16.

I = 4a⁴ · 3π/16 = 3π a⁴/4

Answer: 3π a⁴/4.