Let an ellipse have semimajor axis a and semiminor axis b. In other words, the ellipse runs from -a to +a along the x axis, and from -b to +b along the y axis. It has the following equation.
x2/a2 + y2/b2 = 1
y = b × sqrt(1-x2/a2)
When this ellipse is spun around the x axis, a spheroid appears, prolate if a > b and oblate if b > a. (We have a perfect sphere if a = b.)
As shown in the previous section, the surface area is 2π times the integral, from -a to a, of y×sqrt(y′2+1). Let's do the math.
y = b × sqrt(1-x2/a2) [formula for ellipse]
y = b/a × sqrt(a2-x2)
q = sqrt(a2-x2) [notational convenience]
y = b/a × q
y′ = b/a × -x/q [differentiate]
y′2+1 = ((b2-a2)x2 + a4) over a2q2
sqrt(y′2+1) = sqrt((b2-a2)x2 + a4) over aq
y×sqrt(y′2+1) = b/a2 × sqrt((b2-a2)x2 + a4)
y×sqrt(y′2+1) = b × sqrt((b2-a2)/a4×x2 + 1)
c2 = b2-a2 [find the focus on the y axis, oblate case]
e = c/b [eccentricity of the ellipse]
x = ua2/c (substitute and integrate)
∫ba2/c × sqrt(u2 + 1)
The integral of sqrt(u2+1) has already been determined. The result is shown below.
u×sqrt(u2+1) + log(u+sqrt(u2+1)) over 2
Evaluate this as u runs from 0 to c/a. Thendouble the result to get the entire spheroid. When u = 0 the expression drops to 0. So set u = c/a, and the surface area of the spheroid is:
2πba2/c × { u×sqrt(u2+1) + log(u+sqrt(u2+1)) }
2πa2/e × { u×b/a + log(u+b/a) }
2πb2 + 2πa2/e × log((b+c)/a)
2πb2 + πa2/e × log((b+c)2/(b2-c2))
2πb2 + πa2/e × log((1+e)/(1-e))
That's fine when a < b, but it doesn't work when a > b. The variable c isn't even defined. Back up to the definition of c and start again.
y×sqrt(y′2+1) = b × sqrt(-(a2-b2)/a4×x2 + 1)
c2 = a2-b2 [find the focus on the x axis, prolate case ]
e = c/a [eccentricity of the ellipse]
x = ua2/c (substitute and integrate)
∫ba2/c × sqrt(-u2 + 1)
The integral of sqrt(1-u2) has already been determined. The result is shown below.
u×sqrt(1-u2) + asin(u) over 2
As before, double, and let u run from 0 to c/a. When u = 0 the expression drops to 0. So set u = c/a, and the surface area of the spheroid is:
2πba2/c × { u×sqrt(1-u2) + asin(u) }
2πba2/c × { u×b/a + asin(u) }
2πb2 + 2πab/e × asin(e)
For both formulas, scale a and b by a common factor, and the area is squared, as it should be. Let a approach b, whence e approaches 0, and the area approaches 4πa2, as it should. In the prolate case, use the fact that e almost equals asin(e) for small e. In the oblate case, take the log of 1+2e/(1-e), which is practically 2e for small e.
If you wish, you can use these formulas to find the area of part of the spheroid, trapped between x1 and x2. Find the corresponding values u1 and u2, plug into the formula (based on u), and subtract. Remember to cut the formulas in half, since we are no longer assuming symmetry about the y axis.
Having tackled these surfaces, you'd think there would be a simple formula for the perimeter of an ellipse, but I've never seen one.