Question

Evaluate the given integral by changing to polar coordinates. ∬Ry2x2+y2dA, where R is the region that lies between the circles x2+y2=a2 and x2+y2=b2 with 0

Answer 1

We begin by changing to polar coordinates. In polar coordinates, the region R is described by a ≤ r ≤ b and 0 ≤ θ ≤ 2π. The differential area element is dA = r dr dθ.

Substituting x = r cos(θ) and y = r sin(θ), we have:

y^2/(x^2 + y^2) = (r sin(θ))^2/[(r cos(θ))^2 + (r sin(θ))^2] = sin^2(θ)/(cos^2(θ) + sin^2(θ)) = sin^2(θ)

So the integrand becomes:

y^2/(x^2 + y^2) dA = (r sin(θ))^2/(r^2) r dr dθ = r^3 sin^2(θ) dr dθ

Integrating with respect to r from a to b and with respect to θ from 0 to 2π, we get:

∬R y^2/(x^2 + y^2) dA = ∫[0,2π]∫[a,b] r^3 sin^2(θ) dr dθ

Evaluating the integral with respect to r first, we get:

∫[a,b] r^3 sin^2(θ) dr = (b^4 - a^4)/4 sin^2(θ)

Substituting this back into the double integral and integrating with respect to θ, we get:

∬R y^2/(x^2 + y^2) dA = ∫[0,2π] (b^4 - a^4)/4 sin^2(θ) dθ = π(b^4 - a^4)/2

Therefore, the value of the integral is π(b^4 - a^4)/2.