Find the volume of the body of revolution of the ellipse x ^ 2 / 4 + y ^ 2 / 6 = 1

Find the volume of the body of revolution of the ellipse x ^ 2 / 4 + y ^ 2 / 6 = 1

The volume of the body of revolution around the x-axis: VX = 2 ∫ (2,0) π y ^ 2 (x) DX = 4 π ∫ (2,0) (6-3x ^ 2 / 2) DX
                                       =16π；
The volume of the body of revolution around the y-axis: vy = 2 ∫ (√ 6,0) π x ^ 2 (y) dy = 4 π ∫ (√ 6,0) (4-2y ^ 2 / 3) dy
                                       =16[√(2/3)]π
It seems that the volume of an ellipse with its center at (0,0) around the X and Y axes is not equal when the major and minor axes are not equal!
Let the elliptic equation: x ^ 2 / A ^ 2 + y ^ 2 / b ^ 2 = 1, and the volume of the body of revolution around X and y are:
Vx=4πab^2/3
Vy=4πa^2b/3
When a ≠ B & nbsp;, VX ≠ vy & nbsp; & nbsp; & nbsp; & nbsp; / /: & nbsp; pay attention to this conclusion! In addition, VX / vy = B / A
When a = b = R & nbsp;, VX = vy = 4 π R ^ 3 / 3