To convert to polar coordinates:
x=rcosθ, y=rsinθ, x²+y²=r²
Integrand f(x,y)dxdy becomes r²(rdrdθ)=r³drdθ.
Now we need to convert x²+y²=2x:
r²=2rcosθ, r=2cosθ.
This enables us to work out the double integral limits.
The limits for the inner integral (r) are [0,2cosθ].
The limits for the outer integral (θ) are [-π/2,π/2].
Evaluate inner integral r³dr=r⁴/4.
Applying the limits we get 16cos⁴θ/4=4cos⁴θ.
Evaluate outer integral 4∫cos⁴θdθ with limits [-π/2,π/2]:
cos²θ=(1+cos(2θ))/2; cos⁴θ=(1+2cos(2θ)+cos²(2θ))/4.
cos²(2θ)=(1+cos(4θ))/2; cos⁴θ=(2+4cos(2θ)+1+cos(4θ))/8.
cos⁴θ=(cos(4θ)+4cos(2θ)+3)/8.
Outer integral becomes:
½∫(cos(4θ)+4cos(2θ)+3)dθ=½(¼sin(4θ)+2sin(2θ)+3θ)[-π/2,π/2].
½(¼sin(2π)+2sinπ+3π/2-(¼sin(-2π)+2sin(-π)-3π/2))=
½(3π/2+3π/2)=3π/2.
So the double integral evaluates to 3π/2.