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.