Evaluate the double integral ∬R(y2)/x2+y2dA, where R is the region that lies between the circles x2+y2=25 and x2+y2=64, by changing to polar coordinates.

Answer 1

Hi Courtney R.,

I think the hardest part of this problem is figuring out exactly what it was supposed to say, rather than how you wrote it. I THINK it's supposed to be ((y^2)/(x^2+y^2))dA. At least, that will make some triginometric sense! Because y/(y^2+x^2)^0.5 is the definition of sin(x,y), isn't it. So your function in there to integrate is then just sin^2(P), where P is the point being considered (where dA is located). So, that you should be able to handle, and it will have for each concentric arc-slice the form of int(int(sin^2(theta)d(theta)*R*dR, because you are just integrating over theta from 0 to pi/2, and then multiplying by 4 for the whole circle (right? because your function was in x^2 and y^2 powers only, so it's symmetrical around all 4 quadrants). That leaves you only the single integration from radius 5 to radius 8 for the dR.

I've left the nice juicy calculus piece for you to do, int(sin^2(x)). If you don't know that right away, look it up. As an indefinite integral it has 3 terms, but as a definite integral the constant will cancel out, of course.

-- Cheers, --Mr. d.

Answer 2

Region R in polar system: {5 ≤ r ≤ 8; 0 ≤ θ ≤ 2π}; dA = rdrdθ;

∬R(y2)/x2+y2dA = ∫₀^2π∫₅⁸r²sin²θ/r²·rdrdθ = ∫₅⁸rdr∫₀^2π(1 - cos2θ)/2dθ = r²/4₅⁸(θ - sin2θ/2)₀^2π =

(64 - 25)/4(2π) = 39π/2