Sneaky change of variables...

Question

Evaluate the integral ∫∫_Ω[ x² −xy + y² ] dA , where Ω is the region bounded by the ellipse

x² −xy +y² = 2

Answer 1

problem is same ∫∫_R[( (1/2)x²+(3/2)y²]dxdy where R is the region bounded by xy plane curve (1/2)x²+(3/2)y²=2

the curve was rotated along with f(x,y) function.

boundaries for x are x=-√(4-3y²)to x=+√(4-3y²)

the boundaries for y are y= -2/√3 to y=+2/√3

integrating the dx piece volume = ∫[x³/6+(3/2)xy²]dy, x being from +√(4-3y²) on plus and -√(4-3y²) on negative, plugging in, volume = 2*∫[(4-3y²)^(3/2)/6+(3/2)y²(4-3y²)^(1/2)]dy

the boundaries for y are y= -2/√3 to y=+2/√3

now things become pretty standard with variable substitution

let y=(2/√3)sinθ,,,,dy=(2/√3cosθ),,, and nicely the θ boundaries are θ= -π/2 to θ=+π/2

wind up with 2*∫((2/√3)cos⁴θ+(3/2)(2/√3)(2)(4/3)sin²θcos²θ))dθ θ= -π/2 to θ=+π/2

use sin²θ=(1-cos²θ), use cos⁴θ=[(1/2)(1+cos2θ)]², use (1/4)cos²2θ=(1/8)(1+cos4θ)

notice when integrating any cos(2t) wind up with (-sin(2)t)/2 and at θ= -π/2 tθ=+π/2 sin(2t)=0

same cos(4t)

that is all the trigonomic terms after integrating go to zero

now volume=2*∫[(2√3/9)(1/4+1/8)+(24√3/9)(1/2)-(24√3/9)(1/4+1/8)]dθ θ= -π/2 to θ=+π/2

the first term is from (2√3/9)*cos⁴θ, second term from (24√3/9)cos²θ, third term is from -(24√3/9)cos⁴θ

the volume I calculate is (23√3/18)π

Answer 2

A rotation of the axes is in order here

.

The new ellipse Χ² + 3Ψ² = 4 suggests for a moment the substitution x= 2 u and y= (2/ √3) v but that would fail to eliminate the xy term.

The substitution x = 2u + (2/ √3) v and y = 2u - (2/ √3) v takes care of the aforementioned problem.

The Jacobian being ∂(x,y)/ ∂(u,v) = - 8 / √3

Then ∫∫_Ω [ x² -x ∫y + y² ]d A = ∫∫_R [4 u² + 4 v² ] d u d v where R : u² + v² ≤ 1/2

Hence ∫∫_Ω [ x² -x ∫y + y² ]d A = ∫∫_R [4 u² + 4 v² ](8 / √3) d u d v = ∫₀^2π ∫₀^1/√2 4r² (8 / √3) r dr dθ =

(32/√3) ∫₀^2π ∫₀^1/√2 r³ dr dθ = (32/4√3) ∫₀^2π [ r⁴ ] |₀^{2^.5} dr dθ (8√3) ∫₀^2π (1/4) dθ = (4 π) / √3