Given f(x, y) = x + y, evaluate ∬_D f(x, y) dA using polar coordinates over the region D = (x, y) ∈ ℝ² .

Question

Given f(x, y) = x + y, evaluate ∬_D f(x, y) dA using polar coordinates over the region D = x² + y² ≤ 1.

Answer 1

To evaluate the given double integral using polar coordinates, we express the function and region in polar coordinates. We find the limits of integration for r and θ and then solve the integral step-by-step. The value of the double integral over the given region is π.

Step-by-step explanation:

To evaluate the double integral ∬D f(x, y) dA using polar coordinates, we first need to express the function f(x, y) and the region D in terms of polar coordinates. In polar coordinates, x = r cos(θ) and y = r sin(θ). Therefore, f(x, y) becomes f(r, θ) = r cos(θ) + r sin(θ) = r(cos(θ) + sin(θ)). The region D can be described as D = (r, θ) .

Next, we need to find the limits of integration for r and θ. Since 0 ≤ r ≤ 1, the limits for r are 0 and 1. For θ, 0 ≤ θ ≤ 2π, the limits for θ are 0 and 2π. Therefore, the integral becomes ∬D f(x, y) dA = ∫01 ∫02π r(cos(θ) + sin(θ)) rdr dθ.

Solving the integral step-by-step, we start with the inner integral: ∫02π r(cos(θ) + sin(θ)) rdr. This integral simplifies to 2π, as the terms involving cos(θ) and sin(θ) integrate to zero over a full period of 2π.

Now we integrate 2π with respect to r from 0 to 1: ∫01 2πr dr. Evaluating this integral gives us 2π * 1/2 = π.

Therefore, the value of the double integral ∬D f(x, y) dA over the region D = (x, y) ∈ ℝ² using polar coordinates is π.