MHB How Do You Evaluate ∫∫∫ Over an Ellipsoid Using Change of Variables?

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To evaluate the triple integral over the ellipsoid defined by x^2/a^2 + y^2/b^2 + z^2/c^2 = 1, the transformation x = au, y = bv, z = cw is applied, simplifying the equation to u^2 + v^2 + w^2 = 1. The differential volume element dV is transformed to dV = abc du dv dw, leading to the integral abc ∫∫∫ du dv dw over the unit sphere. The integral of du dv dw over the unit sphere yields the volume of the sphere, which is (4/3)π. The discussion emphasizes that there are no bounds needed for this integration due to the nature of the transformation.
harpazo
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Evaluate ∫∫∫ over E, where E is the solid enclosed by the ellipsoid
x^2/a^2 + y^2/b^2 + z^2/c^2 = 1.
Use the transformation x = au,
y = bv, z = cw.

I decided to replace x with au, y with bv and z with cw in the ellipsoid.

After simplifying, I got
u^2 + v^2 + w^2 = 1.

What is the next step?

The answer is (4/3)πabc.

What is the correct set up leading to the answer?

A friend responded to my email by saying this:

You Are asked to evaluate ∫∫∫ dV over the domain E. dV = dx dy dz. From your change of variables, we get that dx = a du, dy = b dv & dz = c dw. Replacing makes dV = abc du dv dw. Factoring out abc gives

abc ∫∫∫ du dv dw over the domain u^2 + v^2 + w^2 = 1. But that integral is nothing more than the volume of a sphere of radius 1, which you [should] know by heart. Your answer then follows.

Question:

WHAT ARE THE BOUNDS?
 
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You are misunderstanding you friend's response. Using his suggestion, you don't need bounds because you don't need to integrate!

Please see my response to this same question at Ellipsoid: Change of Variables.
 

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