Lecture: Section 20.4 Stokes’ Theorem We will start this topic by recalling Green’s Theorem for an area in the xy plane, and remember a key limitation. ·
·
So we could essentially find the circulation around a path C (a line integral) from a surface integral that summed the “microcirculation” over the area. Note that the direction of the curl vector for a field in the xy‐ plane is in the z‐direction. Furthermore, the normal to the enclosed area is also in the z‐direction (i.e., unit normal vectors are ). The key limitation of Green’s Theorem is that it is limited to 2‐D vector fields. Stokes’ Theorem generalizes this concept to 3‐D. There will be two modifications that will be made to convert Green’s Theorem to Stokes’ Theorem. First we will change our space from 2‐D to 3‐D. This means that we must replace the 2‐
D definition of with the 3‐D version. Second, we will replace with , a unit normal for the area of interest. Note that the direction of C is derived from the orientation of the surface using the right hand rule. So, we then get Stokes’ Theorem. ·
n
C
30
20
Z
·
S
10
0
5
0
Y
-5
0
-2
-4
2
4
6
X
n
T. King: MA3160 Page 1 Lecture 20.4 Example: Exercise 20.4.2 ̂
̂
, and C is a circle in the yz‐plane, oriented clockwise. 1
0.5
n
Z
0
-0.5
-1
-1
-0.5
0.5
0
0.5
1
-0.5
-1
X
Given the orientation above ·
x‐axis
1
0
1 and Y
·
1, since dA is positive, will be negative. Example: Exercise 20.4.5 Use Stokes’ Theorem to find the circulation of the field, , around a path C, consisting of line segments (1,0,1) to (1,0,0) to (0,0,1) to (1,0,1). Exercise 20.4.5
5
4
3
2
1
0
-1
-2
-3
-4
-5
-5
-4
-3
-2
-1 0
1
2
3
4
Note that there is no rotation in this Field. Therefore, T. King: MA3160 5
0 and circulation =0. Page 2 Lecture 20.4 Example: Problem 20.4.13 (a) Find when ̂
̂
Z
.
· (b) Find Exercise 20.4.13
4
2
0
-2
-4
4
4
2
2
0
0
-2
-2
-4 -4
X
Y
We observe that the normal to the triangular surface, determined by the right hand rule is in the positive z direction. Therefore, . .
.
̂
̂
·
̂
· .
.5 4 3
T. King: MA3160 ̂
6 Page 3 Lecture 20.4 Z
Example: Problem 20.4.27 ̂
̂ cos
0
-0.5
1
-1
0.5
0.5
0
0
-0.5
-0.5
X
Y
The problem with trying to find the surface integral here would be coming up with . In this case, it would be much better to evaluate the line integral. The boundary of the region forms a circle with r = 1. The normal to the area is in the negative z‐direction (towards surface) meaning the orientation of the circle is clockwise. 1
0.5
0
-0.5
-1
1
0.5
1
0.5
0
0
-0.5
-0.5
-1 -1
Z
X
The parameterization of the path is: x = cos(t) y = ‐sin(t) cos
̂ sin
̂ sin t ̂ cos
̂ sin
cos
T. King: MA3160 Page 4 Lecture 20.4 Note that the z component will disappear since r’(t) has no z component. .
·
sin
cos
·
sin t
cos
2 Z
Example: Problem 20.4.21 C = circle of radius 2 in x + y + z = 3 plane. Center of circle is (1, 1, 1) and oriented clockwise when viewed from origin. Exercise 20.4.21
4
3
2
1
0
-1
-2
2
-3
4
0
3
2
1
0
-2
-1
X
Y
T. King: MA3160 4
Page 5 Lecture 20.4 ̂
̂
The unit vector is equal to the normal to the plane divided by the magnitude. ̂
̂
√3
1
1 ̂
1
·
.
.
·
6
√3
1 ̂
1
1
2̂
2̂
2
2
2
6
√3
√3
√3
√3
6
6
√3
√3
· 2
2 24
√3
8√3 This problem was simplified by the fact that the enclosed area was flat (i.e., the normal vector was a constant) and the curl of the vector field was also a constant. If either the curl or the normal were position‐dependent, we would have had a significantly more complicated problem. In particular, defining the integration limits would be challenging, requiring us to project the circular area onto the xy plane. T. King: MA3160 Page 6