Course: Accelerated Engineering Calculus II Instructor: Michael Medvinsky 6.7 Stokes’ Theorem (13.7)
Stoke’s theorem (a sort of higher dimensional version of Green’s theorem): Let S be am oriented piecewise-­‐smooth surface that is bounded by a simple closed piecewise-­‐smooth boundary curve C with positive orientation. Let F be a vector field whole components have continuous partial derivatives on an open region in  3 that contains S. Then ∫ F ⋅ dr = 
∫∫ curlF ⋅ dS C
S
Note: The when S is a special degenerated surface, a curve in xy-­‐plane with upward orientation, then n = k and Stoke’s theorem become a vector form of Green’s theorem. One of the important uses of Stoke’s Theorem is in calculating surface integrals over “non convenient” surface using surface integral over more convenient surface with the same boundary: 
∫∫ curlF ⋅ dS = ∫ F ⋅ dr = 
∫∫ curlF ⋅ dS ∂S
S1
Ex 10.
S2
Verify Stoke’s theorem on F = yz, xz, xy over the surface S: z = x 2 + y 2 ≤ 1
∫ F ⋅ dr = ∫ F ( cost,sint,1 ) ⋅ − sint,cost,0 dt = ∫
2π
2π
0
0
sint,cost,cost sint ⋅ − sint,cost,0 dt
sin 2t
=0
0
0
2
From the other side we note that F = ∇ ( xyz ) , and therefore (since xyz has continuous second order partial derivatives) curlF = 0 , thus 
∫∫ curlF ⋅ dS = 
∫∫ curlF ⋅ ndS = 
∫∫ 0 ⋅ ( 1,0,2u × 0,1,2v ) dS = 0 as expected. 2π
2π
= ∫ − sin 2 t + cos 2 t dt = ∫ cos 2t dt =

r ( u,v )= u,v,u 2 +v 2
S
S1
6.8 The Divergence Theorem (13.8)
Recall: 
∫ F ⋅ n ds = ∫∫ divF ( x, y ) dA C
D
The Divergence Theorem: Let E be a simple solid region and let S be the boundary surface of E, given with positive (outward) orientation. Let F be a vector field whose component functions have continuous partial derivatives on an open region that contains E. Then 
∫∫ F ⋅ dS = ∫∫∫ divF dV S
1
2
Ex 11.
Let F = xy,− y 2 , z
and S defined by z = 4 − 3x 2 − 3y 2 ,1 ≤ z ≤ 4 on top,
0 ≤ z ≤ 1 on sides and z=0 at the bottom.
x 2 + y 2 = 1,

∫∫
F ⋅ dS = ∫∫∫ divF dV = ∫∫∫ y − y + 1dV = ∫
=∫
2π
S
E
0
E
∫
1
0
E
0
4r − 3r dr dθ = 2π ( 2r − r
3
2π
3
4
)
4 1
0
= π
5
2
87 1
4−3r 2
0
0
∫∫
r dz dr dθ = ∫
2π
0
∫
1
0
4−3r 2
rz 0
dr dθ =