calculate the double integral (x^2+y^2)dxdy in the circle x^2+y^2=2x

Question

convert it to polar rdrd0. and then solve. the circle is at (1,0) not the origin.

Answer 1

I would do this way:

x² + y²=2x

(x-1)² + y²=1

Then x = 1+ rcosθ, y = rsinθ; dxdy = rdrdθ and x² + y² = (1+ rcosθ)²+sin²θ =1+r²+2rcosθ.

D= {(r, θ): 0≤r≤1, 0≤θ≤2π}

Then

∫∫_D(x² + y²)dxdy=∫∫_D(r + r³ +2r²cosθ) drdθ = 3π / 2,

which is basically the same as the previous answer by Yefim S,

Answer 2

We change to polar coordinate: x = rcosθ, y = rsinθ; dxdy = rdrdθ; D:0 ≤ r≤ 2cosθ, -π/2≤ θ ≤ π/2

∫∫(x² + y²)dxdy = ∫_-π/2^π/2∫₀^2cosθr³drdθ = ∫_-π/2^π/2r⁴/4₀^2cosθdθ = 8∫₀^π/2cos⁴θdθ = 4.7124

I use TI-84 to evaluate this integral, but you can using reducing power formula: cos²x = (1 + cos2x)/2