了 (uo+8 式(uvo+6v)-(uo,vo) y r(uo+8u vo)-=(uo, vo r(uo, vo) Figure 3.4: Area of an infinitessimal surface patch r 6A=|ra6×r2ov=|ru×rl|66v rn×rl2=(rn×r2)·(r1×rn) Using the vector identity (a x b). (cx d)=(a c(. d)-(a d)(b c), we get uIu (36) SA=VEG-F2 dudu A eg-F2 dudu Example: For the hyperbolic paraboloid r(u, u)=(u, 0,u2-u2), let us derive an expression for the area of a region of its surface corresponding to a the circle u2+u2< l in the parametric domain d We begin by forming expressions for the derivatives of the position vector r and the first fundamental form coeffients r (0,1,-2) Using Equation (3.8), we find EG-F2=(1+42)(1+42)-16x22=1+42+42>0 √1+4a2+42duder(u0,v0+δv)−r(u0,v0) r(u0+δu,v0)−r(u0,v0) r(u0,v0+δv) r(u0+δu,v0) r(u0,v0) δA Figure 3.4: Area of an infinitessimal surface patch. r(u0, v0 + δv) − r(u0, v0) ' ∂r ∂v δv r(u0 + δu, v0) − r(u0, v0) ' ∂r ∂u δu δA = |ruδu × rvδv| = |ru × rv|δuδv |ru × rv| 2 = (ru × rv) · (ru × rv) Using the vector identity (a × b) · (c × d) = (a · c)(b · d) − (a · d)(b · c), we get |ru × rv| 2 = (ru · ru)(rv · rv) − (ru · rv) 2 (3.6) = EG − F 2 (3.7) δA = p EG − F2 δuδv, A = Z Z p EG − F2 dudv (3.8) Example: For the hyperbolic paraboloid r(u, v) = (u, v, u 2−v 2 ), let us derive an expression for the area of a region of its surface corresponding to a the circle u 2 +v 2 ≤ 1 in the parametric domain D. We begin by forming expressions for the derivatives of the position vector r and the first fundamental form coeffients. ru = (1, 0, 2u) rv = (0, 1, −2v) E = ru · ru = 1 + 4u 2 F = ru · rv = −4uv G = rv · rv = 1 + 4v 2 Using Equation (3.8), we find EG − F 2 = (1 + 4u 2 )(1 + 4v 2 ) − 16u 2 v 2 = 1 + 4u 2 + 4v 2 > 0 A = Z Z D p 1 + 4u 2 + 4v 2dudv 5