Evaluate the integral of f (x, y) = x over the region given by 4 ≤ x2 + y2 ≤ 16. Hint: convertto polar coordinates.Page 2 of 2

To solve this problem, we need to convert the given Cartesian coordinates to polar coordinates. In polar coordinates, x = rcos(θ) and y = rsin(θ). Also, the differential area element dxdy in Cartesian coordinates becomes rdrdθ in polar coordinates.

The region 4 ≤ x² + y² ≤ 16 in polar coordinates becomes 2 ≤ r ≤ 4, because r = sqrt(x² + y²).

The integral of f(x, y) = x over the region then becomes:

∫ (from 0 to 2π) ∫ (from 2 to 4) rcos(θ) * r dr dθ

This is because x = rcos(θ) and dxdy = rdrdθ.

Now we can solve the integral step by step:

= ∫ (from 0 to 2π) [ (1/3)r³ cos(θ) ] (from 2 to 4) dθ = ∫ (from 0 to 2π) [ (1/3)*4³ cos(θ) - (1/3)*2³ cos(θ) ] dθ = ∫ (from 0 to 2π) [ (64/3) cos(θ) - (8/3) cos(θ) ] dθ = ∫ (from 0 to 2π) (56/3) cos(θ) dθ

The integral of cos(θ) from 0 to 2π is 0, so the final answer is 0.