The document discusses curl and Stokes' theorem. It defines curl(F) for a vector field F(x,y,z)=f(x,y,z)i + g(x,y,z)j + h(x,y,z)k. An example calculates curl(F) for F=x2yi+yz3j+x2z2k. It states that a vector field F is conservative if and only if curl(F)=0. Another example verifies that F=(2xy+3)i+(x2-4)j+k is conservative by showing its curl is 0, and finds a potential function P(x,y,z) such that P

1. Curls and Stoke's Theorem

2. Let F(x, y, z) = f(x, y, z) i + g(x, y, z)j + h(x, y, z)k,
Curls and Stoke's Theorem

3. curl(F) =
Let F(x, y, z) = f(x, y, z) i + g(x, y, z)j + h(x, y, z)k,
i j k
d
x
d
d
y
d
dz
d
f g h
Curls and Stoke's Theorem

4. curl(F) =
Let F(x, y, z) = f(x, y, z) i + g(x, y, z)j + h(x, y, z)k,
i j k
d
x
d
d
y
d
dz
d
f g h
–
= ( d
y
d
h dz
d
g
)i – –
( d
x
d
h dz
df
)j + –
( d
x
d
g d
y
df
)k
Curls and Stoke's Theorem

5. curl(F) =
Let F(x, y, z) = f(x, y, z) i + g(x, y, z)j + h(x, y, z)k,
i j k
d
x
d
d
y
d
dz
d
f g h
–
= ( d
y
d
h dz
d
g
)i – –
( d
x
d
h dz
df
)j + –
( d
x
d
g d
y
df
)k
Example: Let F = x2yi + yz3j + x2z2k. Find curl(F).
Curls and Stoke's Theorem

6. curl(F) =
Let F(x, y, z) = f(x, y, z) i + g(x, y, z)j + h(x, y, z)k,
i j k
d
x
d
d
y
d
dz
d
f g h
–
= ( d
y
d
h dz
d
g
)i – –
( d
x
d
h dz
df
)j + –
( d
x
d
g d
y
df
)k
Example: Let F = x2yi + yz3j + x2z2k. Find curl(F).
curl(F) –
= ( d
y
dx2z2
dz )i – –
( d
x
dz )j + –
( d
x
d
y
)k
dyz3 dx2z2
dx2y dyz3 dx2y
Curls and Stoke's Theorem

7. curl(F) =
Let F(x, y, z) = f(x, y, z) i + g(x, y, z)j + h(x, y, z)k,
i j k
d
x
d
d
y
d
dz
d
f g h
–
= ( d
y
d
h dz
d
g
)i – –
( d
x
d
h dz
df
)j + –
( d
x
d
g d
y
df
)k
Example: Let F = x2yi + yz3j + x2z2k. Find curl(F).
curl(F) –
= ( d
y
dx2z2
dz )i – –
( d
x
dz )j + –
( d
x
d
y
)k
dyz3 dx2z2
dx2y dyz3 dx2y
–
= ( 0 )i – –
( )j + –
( )k
3yz2 0
2xz2
0 x2
Curls and Stoke's Theorem

8. curl(F) =
Let F(x, y, z) = f(x, y, z) i + g(x, y, z)j + h(x, y, z)k,
i j k
d
x
d
d
y
d
dz
d
f g h
–
= ( d
y
d
h dz
d
g
)i – –
( d
x
d
h dz
df
)j + –
( d
x
d
g d
y
df
)k
Example: Let F = x2yi + yz3j + x2z2k. Find curl(F).
curl(F) –
= ( d
y
dx2z2
dz )i – –
( d
x
dz )j + –
( d
x
d
y
)k
dyz3 dx2z2
dx2y dyz3 dx2y
= -3yz2i – 2xz2j – x2k
Curls and Stoke's Theorem
–
= ( 0 )i – –
( )j + –
( )k
3yz2 0
2xz2
0 x2

9. Theorem:
Let F(x, y, z) = f(x, y, z) i + g(x, y, z)j + h(x, y, z)k
where all the partial derivatives of f, g, h are
continuous. F is conservative if and only if curl(F) = 0.
Curls and Stoke's Theorem

10. Theorem:
Let F(x, y, z) = f(x, y, z) i + g(x, y, z)j + h(x, y, z)k
where all the partial derivatives of f, g, h are
continuous. F is conservative if and only if curl(F) = 0.
Curls and Stoke's Theorem
Example: a. Verify that F = (2xy + 3)i + (x2 – 4)j + k
is conservative.

11. Theorem:
Let F(x, y, z) = f(x, y, z) i + g(x, y, z)j + h(x, y, z)k
where all the partial derivatives of f, g, h are
continuous. F is conservative if and only if curl(F) = 0.
Curls and Stoke's Theorem
Example: a. Verify that F = (2xy + 3)i + (x2 – 4)j + k
is conservative.
We verify that curl(F) = 0.

12. Theorem:
Let F(x, y, z) = f(x, y, z) i + g(x, y, z)j + h(x, y, z)k
where all the partial derivatives of f, g, h are
continuous. F is conservative if and only if curl(F) = 0.
Curls and Stoke's Theorem
Example: a. Verify that F = (2xy + 3)i + (x2 – 4)j + k
is conservative.
–
( d
y
d
h dz
d
g
)i – –
( d
x
d
h dz
df
)j + –
( d
x
d
g d
y
df
)k = 0
We verify that curl(F) = 0. That is,

13. Theorem:
Let F(x, y, z) = f(x, y, z) i + g(x, y, z)j + h(x, y, z)k
where all the partial derivatives of f, g, h are
continuous. F is conservative if and only if curl(F) = 0.
Curls and Stoke's Theorem
Example: a. Verify that F = (2xy + 3)i + (x2 – 4)j + k
is conservative.
–
( d
y
d
h dz
d
g
)i – –
( d
x
d
h dz
df
)j + –
( d
x
d
g d
y
df
)k = 0
We verify that curl(F) = 0. That is,
–
d
y
d
h dz
d
g
= 0 – 0 = 0

14. Theorem:
Let F(x, y, z) = f(x, y, z) i + g(x, y, z)j + h(x, y, z)k
where all the partial derivatives of f, g, h are
continuous. F is conservative if and only if curl(F) = 0.
Curls and Stoke's Theorem
Example: a. Verify that F = (2xy + 3)i + (x2 – 4)j + k
is conservative.
–
( d
y
d
h dz
d
g
)i – –
( d
x
d
h dz
df
)j + –
( d
x
d
g d
y
df
)k = 0
We verify that curl(F) = 0. That is,
–
d
y
d
h dz
d
g
= 0 – 0 = 0 –
d
x
d
h dz
df
= 0 – 0 = 0

15. Theorem:
Let F(x, y, z) = f(x, y, z) i + g(x, y, z)j + h(x, y, z)k
where all the partial derivatives of f, g, h are
continuous. F is conservative if and only if curl(F) = 0.
Curls and Stoke's Theorem
Example: a. Verify that F = (2xy + 3)i + (x2 – 4)j + k
is conservative.
–
( d
y
d
h dz
d
g
)i – –
( d
x
d
h dz
df
)j + –
( d
x
d
g d
y
df
)k = 0
We verify that curl(F) = 0. That is,
–
d
y
d
h dz
d
g
= 0 – 0 = 0 –
d
x
d
h dz
df
= 0 – 0 = 0
–
d
x
d
g d
y
df
= 2x – 2x = 0

16. Theorem:
Let F(x, y, z) = f(x, y, z) i + g(x, y, z)j + h(x, y, z)k
where all the partial derivatives of f, g, h are
continuous. F is conservative if and only if curl(F) = 0.
Curls and Stoke's Theorem
Example: a. Verify that F = (2xy + 3)i + (x2 – 4)j + k
is conservative.
–
( d
y
d
h dz
d
g
)i – –
( d
x
d
h dz
df
)j + –
( d
x
d
g d
y
df
)k = 0
We verify that curl(F) = 0. That is,
–
d
y
d
h dz
d
g
= 0 – 0 = 0 –
d
x
d
h dz
df
= 0 – 0 = 0
–
d
x
d
g d
y
df
= 2x – 2x = 0
curl(F) = 0 so F is conservative.

17. Stoke's Theorem
Example: b. Fine a potential function P(x, y, z) where
P = F i.e. Pxi +Pyj + Pzk = (2xy + 3)i + (x2 – 4)j + k

18. Stoke's Theorem
Example: b. Fine a potential function P(x, y, z) where
P = F i.e. Pxi +Pyj + Pzk = (2xy + 3)i + (x2 – 4)j + k
We reconstruct P(x, y, z) by partial integrations.

19. Stoke's Theorem
Example: b. Fine a potential function P(x, y, z) where
P = F i.e. Pxi +Pyj + Pzk = (2xy + 3)i + (x2 – 4)j + k
We reconstruct P(x, y, z) by partial integrations.
Px= 2xy + 3

20. Stoke's Theorem
Example: b. Fine a potential function P(x, y, z) where
P = F i.e. Pxi +Pyj + Pzk = (2xy + 3)i + (x2 – 4)j + k
We reconstruct P(x, y, z) by partial integrations.
Px= 2xy + 3
Hence P = 2xy + 3 dx
∫

21. Stoke's Theorem
Example: b. Fine a potential function P(x, y, z) where
P = F i.e. Pxi +Pyj + Pzk = (2xy + 3)i + (x2 – 4)j + k
We reconstruct P(x, y, z) by partial integrations.
Px= 2xy + 3
Hence P = 2xy + 3 dx
∫ = x2y + 3x + k(y, z)

22. Stoke's Theorem
Example: b. Fine a potential function P(x, y, z) where
P = F i.e. Pxi +Pyj + Pzk = (2xy + 3)i + (x2 – 4)j + k
We reconstruct P(x, y, z) by partial integrations.
Px= 2xy + 3
Hence P = 2xy + 3 dx
∫ = x2y + 3x + k(y, z)
But Py = x2 + ky(y, z)

23. Stoke's Theorem
Example: b. Fine a potential function P(x, y, z) where
P = F i.e. Pxi +Pyj + Pzk = (2xy + 3)i + (x2 – 4)j + k
We reconstruct P(x, y, z) by partial integrations.
Px= 2xy + 3
Hence P = 2xy + 3 dx
∫ = x2y + 3x + k(y, z)
But Py = x2 + ky(y, z) = x2 – 4

24. Stoke's Theorem
Example: b. Fine a potential function P(x, y, z) where
P = F i.e. Pxi +Pyj + Pzk = (2xy + 3)i + (x2 – 4)j + k
We reconstruct P(x, y, z) by partial integrations.
Px= 2xy + 3
Hence P = 2xy + 3 dx
∫ = x2y + 3x + k(y, z)
But Py = x2 + ky(y, z) = x2 – 4  ky(y, z) = -4

25. Stoke's Theorem
Example: b. Fine a potential function P(x, y, z) where
P = F i.e. Pxi +Pyj + Pzk = (2xy + 3)i + (x2 – 4)j + k
We reconstruct P(x, y, z) by partial integrations.
Px= 2xy + 3
Hence P = 2xy + 3 dx
∫ = x2y + 3x + k(y, z)
But Py = x2 + ky(y, z) = x2 – 4  ky(y, z) = -4
ky(y, z) = -4  k(y, z) = -4 dy
∫

26. Stoke's Theorem
Example: b. Fine a potential function P(x, y, z) where
P = F i.e. Pxi +Pyj + Pzk = (2xy + 3)i + (x2 – 4)j + k
We reconstruct P(x, y, z) by partial integrations.
Px= 2xy + 3
Hence P = 2xy + 3 dx
∫ = x2y + 3x + k(y, z)
But Py = x2 + ky(y, z) = x2 – 4  ky(y, z) = -4
ky(y, z) = -4  k(y, z) = -4 dy = -4y + m(z)
∫

27. Stoke's Theorem
Example: b. Fine a potential function P(x, y, z) where
P = F i.e. Pxi +Pyj + Pzk = (2xy + 3)i + (x2 – 4)j + k
We reconstruct P(x, y, z) by partial integrations.
Px= 2xy + 3
Hence P = 2xy + 3 dx
∫ = x2y + 3x + k(y, z)
But Py = x2 + ky(y, z) = x2 – 4  ky(y, z) = -4
ky(y, z) = -4  k(y, z) = -4 dy = -4y + m(z)
∫
Therefore P = x2y + 3x – 4y + m(z).

28. Stoke's Theorem
Example: b. Fine a potential function P(x, y, z) where
P = F i.e. Pxi +Pyj + Pzk = (2xy + 3)i + (x2 – 4)j + k
We reconstruct P(x, y, z) by partial integrations.
Px= 2xy + 3
Hence P = 2xy + 3 dx
∫ = x2y + 3x + k(y, z)
But Py = x2 + ky(y, z) = x2 – 4  ky(y, z) = -4
ky(y, z) = -4  k(y, z) = -4 dy = -4y + m(z)
∫
Therefore P = x2y + 3x – 4y + m(z).
But Pz = m'(z) = 1

29. Stoke's Theorem
Example: b. Fine a potential function P(x, y, z) where
P = F i.e. Pxi +Pyj + Pzk = (2xy + 3)i + (x2 – 4)j + k
We reconstruct P(x, y, z) by partial integrations.
Px= 2xy + 3
Hence P = 2xy + 3 dx
∫ = x2y + 3x + k(y, z)
But Py = x2 + ky(y, z) = x2 – 4  ky(y, z) = -4
ky(y, z) = -4  k(y, z) = -4 dy = -4y + m(z)
∫
Therefore P = x2y + 3x – 4y + m(z).
But Pz = m'(z) = 1  m(z) = z + C.

30. Stoke's Theorem
Example: b. Fine a potential function P(x, y, z) where
P = F i.e. Pxi +Pyj + Pzk = (2xy + 3)i + (x2 – 4)j + k
We reconstruct P(x, y, z) by partial integrations.
Px= 2xy + 3
Hence P = 2xy + 3 dx
∫ = x2y + 3x + k(y, z)
But Py = x2 + ky(y, z) = x2 – 4  ky(y, z) = -4
ky(y, z) = -4  k(y, z) = -4 dy = -4y + m(z)
∫
Therefore P = x2y + 3x – 4y + m(z).
But Pz = m'(z) = 1  m(z) = z + C.
So P = x2y + 3x – 4y + z + C

31. Stoke's Theorem: Given a simple closed piecewise
smooth 3D curve C and σ is a oriented surface with
C as boundary. Assume that C is parameterized in
the counter clockwise direction with respect to the
orientation N of σ, then
∫ F•dC = ∫ ∫curl(F)•N dS
C σ
Curls and Stoke's Theorem

32. Stoke's Theorem: Given a simple closed piecewise
smooth 3D curve C and σ is a oriented surface with
C as boundary. Assume that C is parameterized in
the counter clockwise direction with respect to the
orientation N of σ, then
∫ F•dC = ∫ ∫curl(F)•N dS
C σ
The counter clockwise direction of C with respect to
the orientation N of σ is called the positive direction.
Curls and Stoke's Theorem
σ
C
positively directed C
σ
C
positively directed C

33. Curls and Stoke's Theorem
If C = (x(t), y(t), z(t)) with a < t < b and σ is defined
by z = z(x, y) over the domain R,

34. Curls and Stoke's Theorem
If C = (x(t), y(t), z(t)) with a < t < b and σ is defined
by z = z(x, y) over the domain R, then Stoke's
theorem takes the form (for upward unit normal N):
∫F•<x'(t), y'(t), z'(t)>dt = ∫ ∫curl(F)•<-zx, -zy, 1> dA
t=a R
b

35. Curls and Stoke's Theorem
If C = (x(t), y(t), z(t)) with a < t < b and σ is defined
by z = z(x, y) over the domain R, then Stoke's
theorem takes the form (for upward unit normal N):
∫F•<x'(t), y'(t), z'(t)>dt = ∫ ∫curl(F)•<-zx, -zy, 1> dA
t=a C
b
These two theorems are analogus to the 2D cases.

36. Curls and Stoke's Theorem
If C = (x(t), y(t), z(t)) with a < t < b and σ is defined
by z = z(x, y) over the domain R, then Stoke's
theorem takes the form (for upward unit normal N):
∫F•<x'(t), y'(t), z'(t)>dt = ∫ ∫curl(F)•<-zx, -zy, 1> dA
t=a C
b
The fact curl(F) = 0 implies F is conservative is
analogus to the 2D case where the mixed partials
are equal.
These two theorems are analogus to the 2D cases.

37. Curls and Stoke's Theorem
If C = (x(t), y(t), z(t)) with a < t < b and σ is defined
by z = z(x, y) over the domain R, then Stoke's
theorem takes the form (for upward unit normal N):
∫F•<x'(t), y'(t), z'(t)>dt = ∫ ∫curl(F)•<-zx, -zy, 1> dA
t=a C
b
The fact curl(F) = 0 implies F is conservative is
analogus to the 2D case where the mixed partials
are equal.
These two theorems are analogus to the 2D cases.
Stoke's Theorem changes line integral to surface
integrals v.s. Green's Theorem changes line integral
to area integral.

38. Stoke's Theorem
Example: Verify Stoke's Theorem if σ is the portion
of the paraboloid z = 4 – x2 – y2 for z > 0 with
F = 2zi + 3xj + 5yk

39. Stoke's Theorem
Example: Verify Stoke's Theorem if σ is the portion
of the paraboloid z = 4 – x2 – y2 for z > 0 with
F = 2zi + 3xj + 5yk
N=<2x, 2y, 1>
R
σ
C
z = 4 – x2 – y2

40. Stoke's Theorem
Example: Verify Stoke's Theorem if σ is the portion
of the paraboloid z = 4 – x2 – y2 for z > 0 with
F = 2zi + 3xj + 5yk
R
σ
C
z = 4 – x2 – y2
For the surface integral over σ:

41. Stoke's Theorem
Example: Verify Stoke's Theorem if σ is the portion
of the paraboloid z = 4 – x2 – y2 for z > 0 with
F = 2zi + 3xj + 5yk
–
= ( d
y
d
h dz
d
g
)i – –
( d
x
d
h dz
df
)j + –
( d
x
d
g d
y
df
)k
curl(F)
R
σ
C
z = 4 – x2 – y2
For the surface integral over σ:

42. Stoke's Theorem
Example: Verify Stoke's Theorem if σ is the portion
of the paraboloid z = 4 – x2 – y2 for z > 0 with
F = 2zi + 3xj + 5yk
–
= ( d
y
d
h dz
d
g
)i – –
( d
x
d
h dz
df
)j + –
( d
x
d
g d
y
df
)k
curl(F)
–
= ( d
y
d5y
dz
d3x
)i – –
( d
x
d5y
dz
d2z
)j + –
( d
x
d3x
d
y
d2z
)k
N=<2x, 2y, 1>
R
σ
C
z = 4 – x2 – y2
For the surface integral over σ:

43. Stoke's Theorem
Example: Verify Stoke's Theorem if σ is the portion
of the paraboloid z = 4 – x2 – y2 for z > 0 with
F = 2zi + 3xj + 5yk
–
= ( d
y
d
h dz
d
g
)i – –
( d
x
d
h dz
df
)j + –
( d
x
d
g d
y
df
)k
curl(F)
–
= ( d
y
d5y
dz
d3x
)i – –
( d
x
d5y
dz
d2z
)j + –
( d
x
d3x
d
y
d2z
)k
= 5i + 2j + 3k
R
σ
C
z = 4 – x2 – y2
For the surface integral over σ:

44. Stoke's Theorem
Example: Verify Stoke's Theorem if σ is the portion
of the paraboloid z = 4 – x2 – y2 for z > 0 with
F = 2zi + 3xj + 5yk
–
= ( d
y
d
h dz
d
g
)i – –
( d
x
d
h dz
df
)j + –
( d
x
d
g d
y
df
)k
curl(F)
–
= ( d
y
d5y
dz
d3x
)i – –
( d
x
d5y
dz
d2z
)j + –
( d
x
d3x
d
y
d2z
)k
= 5i + 2j +3k
curl(F)•N dS
= curl(F)•<-zx, -zy, 1> dA
R
σ
C
z = 4 – x2 – y2
For the surface integral over σ:

45. Stoke's Theorem
Example: Verify Stoke's Theorem if σ is the portion
of the paraboloid z = 4 – x2 – y2 for z > 0 with
F = 2zi + 3xj + 5yk
–
= ( d
y
d
h dz
d
g
)i – –
( d
x
d
h dz
df
)j + –
( d
x
d
g d
y
df
)k
curl(F)
–
= ( d
y
d5y
dz
d3x
)i – –
( d
x
d5y
dz
d2z
)j + –
( d
x
d3x
d
y
d2z
)k
= 5i + 2j +3k
curl(F)•N dS
= curl(F)•<-zx, -zy, 1> dA
= <5, 2, 3><2x, 2y, 1>dA
R
σ
C
z = 4 – x2 – y2
For the surface integral over σ:

46. Stoke's Theorem
Example: Verify Stoke's Theorem if σ is the portion
of the paraboloid z = 4 – x2 – y2 for z > 0 with
F = 2zi + 3xj + 5yk
–
= ( d
y
d
h dz
d
g
)i – –
( d
x
d
h dz
df
)j + –
( d
x
d
g d
y
df
)k
curl(F)
–
= ( d
y
d5y
dz
d3x
)i – –
( d
x
d5y
dz
d2z
)j + –
( d
x
d3x
d
y
d2z
)k
= 5i + 2j +3k
curl(F)•N dS
= curl(F)•<-zx, -zy, 1> dA
= <5, 2, 3><2x, 2y, 1>dA
= (10x + 4y + 3)dA R
σ
C
z = 4 – x2 – y2
For the surface integral over σ:

47. Stoke's Theorem
R
σ
C
z = 4 – x2 – y2
∫ ∫curl(F)•<-zx, -zy, 1> dA =
R
∫ ∫
R
(10x + 4y + 3)dA

48. Stoke's Theorem
R
σ
C
z = 4 – x2 – y2
∫ ∫curl(F)•<-zx, -zy, 1> dA =
R
∫ ∫
R
(10x + 4y + 3)dA
Put it polar form where R is the circle of radius 2;

49. Stoke's Theorem
R
σ
C
z = 4 – x2 – y2
∫ ∫curl(F)•<-zx, -zy, 1> dA =
R
∫ ∫
R
(10x + 4y + 3)dA
Put it polar form where R is the circle of radius 2;
∫ ∫
r=0
(10rcos + 4rsin + 3) r drd
2
=0
2π

50. Stoke's Theorem
N=<2x, 2y, 1>
R
σ
C
z = 4 – x2 – y2
∫ ∫curl(F)•<-zx, -zy, 1> dA =
R
∫ ∫
R
(10x + 4y + 3)dA
Put it polar form where R is the circle of radius 2;
∫ ∫
r=0
(10rcos + 4rsin + 3) r drd = 12π
2
=0
2π

51. Stoke's Theorem
For the line integral:
N=<2x, 2y, 1>
R
σ
C
z = 4 – x2 – y2
∫ ∫curl(F)•<-zx, -zy, 1> dA =
R
∫ ∫
R
(10x + 4y + 3)dA
Put it polar form where R is the circle of radius 2;
∫ ∫
r=0
(10rcos + 4rsin + 3) r drd = 12π
2
=0
2π

52. Stoke's Theorem
For the line integral:
N=<2x, 2y, 1>
R
σ
C
z = 4 – x2 – y2
∫ ∫curl(F)•<-zx, -zy, 1> dA =
R
∫ ∫
R
(10x + 4y + 3)dA
Put it polar form where R is the circle of radius 2;
∫ ∫
r=0
(10rcos + 4rsin + 3) r drd = 12π
2
=0
2π
C is x2 + y2 = 4, it may be
paramaterized (in 3D) as
<2cos(t), 2sin(t), 0> for
0 < t < 2π.

53. Stoke's Theorem
For the line integral:
N=<2x, 2y, 1>
R
σ
C
z = 4 – x2 – y2
∫ ∫curl(F)•<-zx, -zy, 1> dA =
R
∫ ∫
R
(10x + 4y + 3)dA
Put it polar form where R is the circle of radius 2;
∫ ∫
r=0
(10rcos + 4rsin + 3) r drd = 12π
2
=0
2π
C is x2 + y2 = 4, it may be
paramaterized (in 3D) as
<2cos(t), 2sin(t), 0> for
0 < t < 2π.
F = <2z, 3x, 5y>
= <0, 6cos(t), 10sin(t)>

54. Stoke's Theorem
N=<2x, 2y, 1>
R
σ
C
z = 4 – x2 – y2
F•C' = <0, 6cos(t), 10sin(t)>•<-2sin(t), 2cos(t), 0>

55. Stoke's Theorem
N=<2x, 2y, 1>
R
σ
C
z = 4 – x2 – y2
F•C' = <0, 6cos(t), 10sin(t)>•<-2sin(t), 2cos(t), 0>
= 12 cos2(t)
∫F•<x'(t), y'(t), z'(t)>dt
t

56. Stoke's Theorem
N=<2x, 2y, 1>
R
σ
C
z = 4 – x2 – y2
F•C' = <0, 6cos(t), 10sin(t)>•<-2sin(t), 2cos(t), 0>
= 12 cos2(t)
∫F•<x'(t), y'(t), z'(t)>dt
=
t
∫12 cos2(t) dt
t=0
2π

57. Stoke's Theorem
N=<2x, 2y, 1>
R
σ
C
z = 4 – x2 – y2
F•C' = <0, 6cos(t), 10sin(t)>•<-2sin(t), 2cos(t), 0>
= 12 cos2(t)
∫F•<x'(t), y'(t), z'(t)>dt
=
t
∫12 cos2(t) dt
t=0
2π
= 6sin(t)cos(t) + 6t |
t=0
2π

58. Stoke's Theorem
N=<2x, 2y, 1>
R
σ
C
z = 4 – x2 – y2
F•C' = <0, 6cos(t), 10sin(t)>•<-2sin(t), 2cos(t), 0>
= 12 cos2(t)
∫F•<x'(t), y'(t), z'(t)>dt
=
t
∫12 cos2(t) dt
t=0
2π
= 6sin(t)cos(t) + 6t |
t=0
2π
= 12π