# The formula of an ellipse is  (4/3)πr1r2r3 , how is this equation found from a known area?

mlehuzzah | Certified Educator

The volume of an ellipsoid is `(4)/(3) pi r_1 r_2 r_3 `

The area of an ellipse is `pi r_1 r_2 `

We can use the area of an ellipse to get the volume of an ellipsoid.

Suppose our ellipsoid is given by:

`((x)/(r_1))^2 + ((y)/(r_2))^2 +((z)/(r_3))^2 =1`

We use the "slicing" method of calculus to get our answer.

Make slices perpendicular to the x-axis. (It doesn't actually matter which axis, but that is the one I will do.) So, for each slice, the x is fixed, and we want to look at the area of the resulting cross section. We have slices for all of the x values between `-r_1` and `r_1`

If we fix a slice, (and so fix the x) then the cross section of the shape has the equation:

`((y)/(r_2))^2 +((z)/(r_3))^2 =1 - ((x)/(r_1))^2`

We are now thinking of y and z as variables, but x is a fixed number, which is why it is on the right hand side.

This is not the form we want our equation in. Instead, we would like to scale the whole equation so that the number 1 is on the right hand side.

So:

`((y)/(r_2))^2 +((z)/(r_3))^2 = (r_1^2 - x^2 )/(r_1^2)`

` (y^2)/(r_2^2 ((r_1^2-x^2)/(r_1^2))) + (z^2)/(r_3^2 ((r_1^2-x^2)/(r_1^2))) = 1`

` ((y)/(r_2 ((sqrt(r_1^2-x^2))/(r_1))))^2 + ((z)/(r_3 ((sqrt(r_1^2-x^2))/(r_1))))^2 = 1`

So the "radii" of the ellipse that is the cross section is:
`r_2 ((sqrt(r_1^2-x^2))/(r_1))`, `r_3 ((sqrt(r_1^2-x^2))/(r_1))`

Thus, what we want to integrate is:

`int _(-r_1)^(r_1) pi r_2 ((sqrt(r_1^2-x^2))/(r_1)) r_3 ((sqrt(r_1^2-x^2))/(r_1)) dx`

`= (pi r_2 r_3)/(r_1^2) int_(-r_1)^(r_1) (r_1^2-x^2) dx`

`= (pi r_2 r_3)/(r_1^2) [ (r_1^2 x - (1)/(3) x^3)|_(-r_1)^(r_1) ]`

`= (pi r_2 r_3)/(r_1^2) [ (r_1^3 - (1)/(3) (r_1)^3) - (-r_1^3 + (1)/(3) (r_1)^3)] `

`= (pi r_2 r_3)/(r_1^2) [ (4)/(3)(r_1^3)]`

`= (pi r_2 r_3 r_1)(4)/(3)`