Question

By changing to polar coordinates, evaluate the integral ∬[D] (x² + y²)^(5/2) dxdy, where D is the disk x² + y² ≤ 49.

Answer 1

To evaluate the integral, we change to polar coordinates and integrate over the disk x² + y² ≤ 49. Using the polar coordinate transformation, the integral becomes ∬[D] (r²)^(5/2) rdrdθ. Integrating with respect to r and θ gives the final result of 4π(7)^(5/2).

Step-by-step explanation:

To evaluate the integral ∬[D] (x² + y²)^(5/2) dxdy, where D is the disk x² + y² ≤ 49, we can change to polar coordinates. In polar coordinates, x = rcosθ and y = rsinθ, and the disk D becomes the region 0 ≤ r ≤ 7 and 0 ≤ θ ≤ 2π. The integral then becomes ∬[D] (r²)^(5/2) rdrdθ.

1. For the inner integral, integrate with respect to r from 0 to 7: ∫ r^(5/2) dr = (2/7)r^(7/2) ∣ 0 to 7 = (2/7)(7^(7/2)) - (2/7)(0) = 2(7)^(7/2)/7 = 2(7)^(5/2).
2. For the outer integral, integrate with respect to θ from 0 to 2π: ∫ 2(7)^(5/2) dθ = 2(7)^(5/2)θ ∣ 0 to 2π = 2(7)^(5/2)(2π - 0) = 4π(7)^(5/2).

Therefore, the value of the integral is 4π(7)^(5/2).