✔ 最佳答案
Let the base be x^2/p^2+y^2/q^2=1 (z=0), vertex be (0,0,h), (p, q)=(a/2, b/2)
so, the surface can be parametrized as
r(u, t)=(x,y, z)=( p u cost, q u sint, -h u), t=0~2π, u=0~1
r_u = ( p cost, q sint, -h)
r_t = (- pu sint, qu cost, 0)
| r_u x r_t |^2= u^2 [ (pq)^2+(qh)^2+ h^2(p^2-q^2)(sint)^2 ]
area A(the base is not included)
=∫[0,2π]∫[0,1] u√[(pq)^2+(qh)^2+ h^2(p^2-q^2)(sint)^2 ] du dt
=∫[0,π]√[(pq)^2+(qh)^2+ h^2(p^2-q^2)(sint)^2 ] dt
it is an elliptic integral.
if p=q, then A=∫[0,π]√[(pq)^2+(qh)^2] dt = πp√(p^2+h^2)