Volume of an ellipsoid using double integrals

[Lucas Mayr]

Homework Statement

Using double integrals, calculate the volume of the solid bound by the ellipsoid:

x²/a² + y²/b² + z²/c² = 1

2. Relevant data

must be done using double integrals

The Attempt at a Solution

i simply can't find a way to solve this by double integrals, i did with triple integral, but my teacher won't accept it, this is the last question and i can't find a solution for it, a step-by-step would be awsome.

Thanks.

 

[Simon Bridge]

If it were a solid of rotation you could do it in a single-integral right?
Basically you have to use reasoning in the place of one of the integrals ...

Find the volume between $$z=c^2\left ( 1-\frac{x^2}{a^2}-\frac{y^2}{b^2} \right )$$ and the x-y plane.

 

[mfb]

i simply can't find a way to solve this by double integrals, i did with triple integral, but my teacher won't accept it, this is the last question and i can't find a solution for it, a step-by-step would be awsome.

The first integration in the triple integral is trivial, and afterwards you get a double-integral - you can use this double-integral (ignoring the first step) to solve the problem.

 

[HallsofIvy]

Essentially the same idea: projecting the ellipsoid to the xy-plane (z= 0) gives the ellipse x^2/a^2+ y^2/b^2= 1. The two heights at each (x, y) point are z= \pm c\sqrt{1- x^2/a^2- y^2/b^2}. The difference, 2c\sqrt{1- x^2/a^2- y^2/b^2}, is the length of thin rectangle above that point and is the function to be integrated. For each x, The ellipse goes from y= -b\sqrt{1- x^2/a^2} to y= b\sqrt{1- x^2/a^2}. And, over all, x goes from -a to a. The volume of the ellipse is given by
\int_{-a}^a\int_{-b\sqrt{1- x^2/a^2}}^{b\sqrt{1- x^2/a^2}} 2c\sqrt{1- x^2/a^2- y^2/b^2} dydx.

As mfb says, this is the same as if you started with the triple integral
\int_{-a}^a\int_{-b\sqrt{1- x^2/a^2}}^{b\sqrt{1- x^2/a^2}}\int_{-c\sqrt{1- x^2/a^2- y^2/b^2}}^{c\sqrt{1- x^2/a^2- y^2/b^2}} dzdydx
and integrated once.