Evaluate the iterated integral by converting to polar coordinates: $${\int }_0^2{\int }_{2-x^2}^{4-x^2}{x}^2+y^2 dy dx$$

• A. $\frac{32\pi }{3}$
• B. $\frac{16\pi }{3}$
• C. $4\pi$
• D. $8\pi$

Answer: (A)

Verified step by step guidance

1. Step 1: Recognize that the integral involves a circular region. The integrand, x² + y², suggests that converting to polar coordinates will simplify the computation since x² + y² = r² in polar coordinates.
2. Step 2: Analyze the bounds of the integral. The inner integral's bounds are from y = 2x - x² to y = 4 - x², and the outer integral's bounds are from x = 0 to x = 2. These bounds describe a region in the xy-plane. Sketch the region to confirm it is part of a circle.
3. Step 3: Convert the bounds to polar coordinates. In polar coordinates, x = r cos(θ) and y = r sin(θ). The region described by the bounds corresponds to a quarter-circle in the first quadrant with radius 2. Thus, r ranges from 0 to 2, and θ ranges from 0 to π/2.
4. Step 4: Rewrite the integral in polar coordinates. The integrand x² + y² becomes r², and the differential area element dx dy becomes r dr dθ. The integral becomes: ∫₀^{π/2} ∫₀² r² * r \, dr \, dθ.
5. Step 5: Separate the integral into two parts and evaluate each. First, compute the inner integral with respect to r: ∫₀² r³ \, dr. Then, compute the outer integral with respect to θ: ∫₀^{π/2} (result of inner integral) \, dθ. Combine the results to find the final value.