Surface area of a spheroid?

Parametrize the surface using R(u,v) = (a cos(u) sin(v), a sin(u) sin(v), c cos(v)), 0 ≤ u ≤ 2π, 0 ≤ v ≤ π. Then R_u = (-a sin(u) sin(v), a cos(u) sin(v), 0) and R_v = (a cos(u) cos(v), a sin(u) cos(v), -c sin(v)). Thus R_u x R_v = (-ac cos(u) sin^2(v), -ac sin(u) sin^2(v), -a^2 sin(v) cos(v)). The norm of R_u x R_v is

||R_u x R_v|| = √(a^2 c^2 cos^2(u) sin^4(v) + a^2 c^2 sin^2(u) sin^4(v) + a^4 sin^2(v) cos^2(v))

= √(a^2 c^2 sin^4(v) (cos^2(u) + sin^2(u)) + a^4 sin^2(v) cos^2(v))

= √(a^2 c^2 sin^2(v) sin^2(v) + a^4 sin^2(v) cos^2(v))

= √(a^2 sin^2(v) (c^2 sin^2(v) + a^2 cos^2(v))

= a sin(v) √(c^2 sin^2(v) + a^2 cos^2(v)).

So the surface area of A is

∫[0, π] ∫[0, 2π] ||R_u x R_v|| du dv

= ∫[0, π] ∫[0, 2π] a sin(v) √(c^2 sin^2(v) + a^2 cos^2(v)) du dv

= 2πa ∫[0, π] sin(v) √(c^2 sin^2(v) + a^2 cos^2(v)) dv.

For reference, I'll let S(a,c) represent that last integral. When c = a, the integral reduces to 2πa ∫[0, π] sin(v) a dv = 4πa^2. When c > a,

S(a,c) = 2πa ∫[0, π] sin(v) √(c^2 + (a^2 - c^2)cos^2(v)) dv

= 2πa ∫[0, π] sin(v) • c√(1 - e^2 cos^2(v)) dv, with e^2 = 1 - a^2 / c^2

= 2πac ∫[-e, e] √(1 - x^2) dx/e, setting x = e cos(v)

= 4πac/e ∫[0, e] √(1 - x^2) dx.

Let J = ∫[0, e] √(1 - x^2) dx. By integration by parts,

∫[0, e] √(1 - x^2) dx = x√(1 - x^2) (from x = 0 to e) - ∫[0, e] x d/dx(√(1 - x^2)) dx

= e√(1 - e^2) - ∫[0, e] x(-x/√(1 - x^2)) dx

= e√(1 - e^2) + ∫[0, e] x^2/√(1 - x^2) dx

= e√(1 - e^2) + ∫[0, e] (1/√(1 - x^2) - √(1 - x^2)) dx

= e√(1 - e^2) + sin^-1(x) (from x = 0 to e) - J

= e√(1 - e^2) + sin^-1(e) - J.

So 2J = e√(1 - e^2) + sin^-1(e), or,

J = (1/2)(e√(1 - e^2) + sin^-1(e)).

Consequently,

S(a,c) = (4πac/e)J = 2πa(c√(1 - e^2) + (c/e) sin^-1(e))

= 2πa(c√(a^2/c^2) + (c/e)sin^-1(e))

= 2πa(a + (c/e)sin^-1(e))

= 2πa^2 (1 + (c/ae) sin^-1(e)).

If c < a, then

S(a,c) = 2πa ∫[0, π] sin(v) √(c^2 + (a^2 - c^2) cos^2(v)) dv

= 2πa ∫[0, π] sin(v) a√(c^2/a^2 + (1 - c^2/a^2) cos^2(v)) dv

= 2πa^2 ∫[0, π] sin(v) √((1 - e^2) + e^2 cos^2(v)) dv, with e^2 = 1 - c^2/a^2

= 2πa^2 ∫[0, π] sin(v) √(1 - e^2) √(1 + [e^2/(1 - e^2)] cos^2(v)] dv

= 2πa^2√(1 - e^2) ∫[-e/√(1 - e^2), e/√(1 - e^2)] √(1 + x^2) dx √(1 - e^2)/e

(substituting x = e/√(1 - e^2) cos(v))

= 2πa^2 (1 - e^2)/e • 2∫[0, e/√(1 - e^2)] √(1 + x^2) dx.

Let K = ∫[0, e/√(1 - e^2)] √(1 + x^2) dx. Integration by parts yields

K = x√(1 + x^2) (from x = 0 to e/√(1 - e^2)) - ∫[0, e/√(1 - e^2)] x d/dx(√(1 + x^2)) dx

= e/√(1 - e^2) • √(1 + e^2/(1 - e^2)) - ∫[0, e/√(1 - e^2)] x^2/√(1 + x^2) dx

= e/√(1 - e^2) • 1/√(1 - e^2) - ∫[0, e/√(1 - e^2)] (√(1 + x^2) - 1/√(1 + x^2)) dx

= e/(1 - e^2) - K + ∫[0, e/√(1 - e^2)] dx/√(1 + x^2) dx

= e/(1 - e^2) - K + sinh^-1(x) (from x = 0 to e/√(1 - e^2))

= e/(1 - e^2) - K + sinh^-1(e/√(1 - e^2)

= e/(1 - e^2) - K + tanh^-1(e).

Thus, 2K = e/(1 - e^2) + tanh^-1(e), or,

K = (1/2)[e/(1 - e^2) + tanh^-1(e)].

Finally,

S(a,c) = 2πa^2 (1 - e^2)/e • 2K

= 2πa^2 (1 - e^2)/e • [e/(1 - e^2) + tanh^-1(e)]

= 2πa^2 (1 + [(1 - e^2)/e] tanh^-1(e)).

In summary,

S(a,c) = 4πa^2 if c = a

S(a,c) = 2πa^2 (1 + (c/ae) sin^-1(e)) if c > a

S(a,c) = 2πa^2 (1 + [(1 - e^2)/e] tanh^-1(e)) if c < a.

In particular, for the oblate spheroid (where c < a), the latter formula for S(a,c) is its surface area.

Well,

it depends on wether:

c<a (oblate spheroid like Earth): Sob = 2pia^2(1+(1-e^2)/e tanh^-1(x))

where e^2 = 1 - (c^2/a^2) excentricity

and

c>a (prolate like football or rugby): Spro=2pia^2(1+ c/ae si^n^-1e)

hope it' ll help !!

PS: see § 2 of the link