Problems: Extended Stokes’ Theorem Let F = (2xz + y, 2yz + 3x, x2 + y 2 + 5). Use Stokes’ theorem to compute C is the curve shown on the surface of the circular cylinder of radius 1. C F · dr, where Figure 1: Positively oriented curve around a cylinder. Answer: This is very similar to an earlier example; we can use Stokes’ theorem to calculate this integral even though we don’t have an exact description of C. We just make C into part of the boundary of a surface, as shown in the ﬁgure below. Figure 2: Curves C and C1 bound part of a cylinder. Let C1 be the unit circle in the xy-plane oriented to match C and S the portion of the cylinder between C and C1 . Then Stokes’ theorem says: F · dr = curlF · n dS. C1 −C curlF = S i j ∂ ∂x ∂ ∂y k 2xz + y 2yz + 3x x2 ∂ ∂z + y2 Since the normal vector to S is always orthogonal to k, Thus, C1 −C F · dr = C1 F · dr − C F · dr = 0 and 1 C = 2k. +5 S curlF · n = 0. F · dr = C1 F · dr. To ﬁnish, parametrize C1 by x = cos t, y = sin t, z = 0, 0 ≤ t < 2π and calculate: I I F · dr = (2xz + y)dx + (2yz + 3x)dy + (x2 + y 2 )dz C1 C 2π Z = sin t(− sin t dt) + 3 cos t(cos t dt) 0 Z = 2π −1 + 4 cos2 t dt 0 = 4 −t + (t + sin t cos t) 2 2 2π = 2π. 0