Rotating an ellipse to create a spheroid?

[gary0000]

Homework Statement

This problem began with solving for an 2D ellipse whose axis aligned with the y-axis and the line x=-z. Then the ellipse needs to be translated, rotated, and then revolved to form a spheroid. I've already found the rotated and translated ellipse.

Rotate the following ellipse about its major axis to obtain a prolate spheroid:

(195.010)^2*((x-72.850)*cos(1.423)+(y-490.030)*sin(1.423))^2+(532.419)^2*((x-72.850)*sin(1.423)-(y-490.030)*cos(1.423))^2-(532.419)^2*(195.010)^2=0

[Each x-value corresponds to a negative z-value of equal magnitude as well, since this ellipse is in the plane formed by line x=-z and the y-axis, not the xy plane.]

Relevant Equations

General equation of an ellipse: x^2/a^2+y^2/b^2=1

(in my case b>a)

Equation of an ellipse with a rotated axis and translated center:

((x−h)cos(A)+(y−k)sin(A))^2/(a^2)+((x−h)sin(A)−(y−k)cos(A))^2/(b^2)=1

General equation of a prolate spheroid: (x^2+y^2)/a^2+z^2/c^2=1, where c > a

I was able to find the equation of an ellipse where its major axis is shifted and rotated off of the x,y, or z axis. However, I could not find anywhere an equation for a spheroid that does not have its axis or revolution along the x,y, or z axis. How might I go about deriving such an equation?For this spheroid specifically I have found that:
-It's axis of revolution passes through the origin
-Its center its located at the point (72.846,490.034,-72.846)
-It's radius is 195.010 units long
-Its semi-major axis is 532.419 units long

(I attached an image of a plot of the ellipse that needs to be revolved with its axis of revolution represented by a yellow line)

 

[danyull]

Let's say your ellipse has axes of length $A$ and $B$, and center $(h,k)$, and is rotated by angle $\theta$. This is represented by the equation $$\frac{((x-h)\cos(\theta)+(y-k)\sin(\theta))^2}{A^2}+ \frac{((x-h)\sin(\theta)-(y-k)\cos(\theta))^2}{B^2} =1.$$ If $A>B$ then I suspect the equation you're looking for is
$$\frac{((x-h)\cos(\theta)+(y-k)\sin(\theta))^2}{A^2}+ \frac{((x-h)\sin(\theta)-(y-k)\cos(\theta))^2}{B^2} + \frac{z^2}{B^2}=1,$$
simply tacking on a $z^2/B^2$ term. The denominator is $B^2$ because, since we're revolving around the semi-major axis, the new axis must have the same length as the semi-minor axis. There's none of the extra stuff in the numerator either because those only deal with translations and rotations in the $xy$-plane. Using a 3D-grapher, it'll look something like this:

(I used $h=1, k=2, \theta=\pi/6, A=2, B=1$).

 

[LCKurtz]

(I attached an image of a plot of the ellipse that needs to be revolved with its axis of revolution represented by a yellow line)

I haven't checked your numbers, but unless I misunderstand, your picture can't be correct. A prolate spheroid must be an ellipse rotated about its major axis. Your yellow line apparently doesn't align with the major axis.