The equation of ellipse is (x/a)² + (y/b)² =1. Now let x' = x/a, and y'=y/b, so that the ellipse equation becomes
    x'² + y'² =1.
This is a circle of radius 1.

When we calculate the area of an ellipse we write A = ò dx dy over the bounding curve. The bounding curve starts out as (x/a)² + (y/b)² =1 , which makes for messy substitutions and a somewhat complicated calculation. But! When we go to the new coordinates we have A = ab ò dx' dy' over the new curve.  Since the bounding curve in the new coordinates is a unit circle, the area is obviously p which makes the area of the ellipse A = p ab .

Ellipsoid calculations go the same way, starting with (x/a)² + (y/b)² + (z/c)² = 1 .
When we change to x' = x/a, y' = y/b, and z' = z/c, the bounding surface is a unit sphere  Then the volume of an ellipsoid is easily worked out to be 4/3 p abc

Calculating integrals like I₁ = _ellipsoid ò x² dx dy dz or I₂ = _ellipsoid ò y² dx dy dz now becomes easy when we change variables to x', y', and z'. In the new coordinates

I₁ = a³bc _{unit sphere}ò x'² dx' dy' dz' and I₂ = ab³c_{unit sphere}ò y'² dx' dy' dz'.

For integrals over the unit sphere, _{unit sphere}ò x'² dx' dy' dz' = _{unit sphere}ò y'² dx' dy' dz' because x', y' and z' are all equivalent directions when integrating over the volume of a sphere of radius 1.

(Back to the ellipse for a moment.) For an ellipse, consider an integral like I₃ = ò x² dx dy . Over the ellipse when we change variables this becomes I₃ = a³b _{unit circle}ò x'² dx' dy' .

Because ò x'² dx' dy' = ò y'² dx' dy' over the unit circle, and r'² = x'² + y'², within the unit circle, we can conclude that I₃ = a³b _{unit circle}ò1/2 r'² dx' dy' . But when we integrate, we can use strips of area of length 2p r' and width dr', so then I₃ becomes

    I₃ = 1/2 a³b ₀ò¹ r'² 2p r' dr' = p /4 a³b.

Higher powers of x in an integral are a little more difficult, but still much easier over the unit circle.
Take I₄ = _ellipseò x⁴ dx dy. This integral becomes I₄ = a⁵ b _{unit circle}ò(r' cos q ')⁴ r' dr' dq '. The r' part is easy and we have I₄ = 1/6 a⁵ b ₀ò^2p(cos q ')⁴ dq ' . The angle integral is easily handled with double angle formulas or something similar ( to give 3p /4) .

For integrals over the ellipsoid, things are perhaps simpler than for the circle. When we transform to the unit sphere the integral dx' dy' dz' may be written as r' dq ' r' sin q ' dj ' dr'. If there is no j dependence in the integral, then the volume element is 2 p r'² dr' sin q ' dq ' . The substitution w' = cos q ' is quite convenient because dw' = - sin q ' dq ' . Then the w integral goes from 1 to -1, and when we flip the limits the negative sign disappears inside the integral.

Exercises:
1) Show that I_A = _ellipsoid ò x² dx dy dz = a³bc _{unit sphere}ò x'² dx' dy' dz' = 4p /15 a³bc .
2) Evaluate I_B = _ellipsoid òx⁴ dx dy dz .
3) Evaluate I_B = _ellipsoid òx² y² dx dy dz .