. . . . . .
.. Chapter 15 Vector Calculus.
......
15.1 Vector Fields
15.2 Line Integrals
15.3 Fundamental Theorem and Independence of Path
15.3 Conservative Fields and Potential Functions
15.4 Green’s Theorem
15.5 Surface Integrals
15.5 Flux of Vector Field across Surface
15.5 Divergence Theorem
15.6 Stokes’ Theorem
15.6 Simply Connected, Orientation
15.6 Irrotational and Conservative Vector Fields
. . . . . .
.. Surface Integrals.
......
If the surface S is given by the graph z = z(x, y), where (x, y) lies inthe domain of D ⊂ R2, i.e. S = { (x, y, z(x, y)) | (x, y) ∈ D }, then the
surface area element dS =√
1 + z2x + z2
y dxdy.
.
......
If the surface S is given by parametric formr(u, v) = x(u, v)i + y(u, v)j + z(u, v)k, where (u, v) lies in the domainof D of uv-plane. Surface area element dS = ∥N∥dudv
=
∣∣∣∣∣∣∣∣ ∂r∂u
× ∂r∂v
∣∣∣∣∣∣∣∣ dudv =
∣∣∣∣∣∣∣∣ ∂(y, z)∂(u, v)
i +∂(z, x)∂(u, v)
j +∂(x, y)∂(u, v)
k∣∣∣∣∣∣∣∣ dudv
=√(yuzv − zuyv)2 + (xuzv − zuxv)2 + (xuyv − yuxv)2 du dv.
.
......
Definition. Let g = g(x, y, z) be a continuous function defined ondomain containing S, the surface integral of the function g on S is∫∫
Sg(x, y, z) dS =
∫∫D
g(x, y, z(x, y))√
1 + z2x + z2
y dxdy, or
=∫∫
Dg(x(u, v), y(u, v), z(u, v))
∣∣∣∣∣∣∣∣ ∂r∂u
× ∂r∂u
∣∣∣∣∣∣∣∣ du dv
. . . . . .
.
......
Definition. Let S be a parameterized surface or a graph, then thesurface area of S is given by∫∫
S1 dS =
∫∫D
√1 + z2
x + z2y dxdy, or
=∫∫
D∥ru × rv∥ dudv.
.
......
Example.Find the area of the part of the surface 2z = x2 that liesdirectly above the triangle in the xy-plane with vertices atA(0, 0), B(1, 0) and C(1, 1).
Solution. The surface is a graph z(x, y) = x2/2 defined on the regionD = { (x, y) | 0 ≤ x ≤ 1, 0 ≤ y ≤ x }, then zx = x and zy = 0. Thesurface area is∫∫∆ABC
√1 + z2
x + x2y dxdy
=∫ 1
0
∫ x
0
√1 + x2 dydx =
∫ 1
0x√
1 + x2dx =
12
∫ 1
0
√1 + x2d(1 + x2) =
12
[(1 + x2)1/2+1
1 + 1/2
]1
0
=13(2√
2 − 1).
. . . . . .
.
......
Definition. Let S be a parameterized surface or a graph, then thesurface area of S is given by∫∫
S1 dS =
∫∫D
√1 + z2
x + z2y dxdy, or
=∫∫
D∥ru × rv∥ dudv.
.
......
Example.Find the area of the part of the surface 2z = x2 that liesdirectly above the triangle in the xy-plane with vertices atA(0, 0), B(1, 0) and C(1, 1).
Solution. The surface is a graph z(x, y) = x2/2 defined on the regionD = { (x, y) | 0 ≤ x ≤ 1, 0 ≤ y ≤ x }, then zx = x and zy = 0. Thesurface area is∫∫∆ABC
√1 + z2
x + x2y dxdy =
∫ 1
0
∫ x
0
√1 + x2 dydx =
∫ 1
0x√
1 + x2dx =
12
∫ 1
0
√1 + x2d(1 + x2) =
12
[(1 + x2)1/2+1
1 + 1/2
]1
0
=13(2√
2 − 1).
. . . . . .
.
......
Example. Find the area of the part of the paraboloidS : z = 9 − x2 − y2 that lies above the plane z = 5.
Solution. For any point P(x, y, 9 − x2 − y2) of the graph lying abovethe plane z = 5, we have 9 − x2 − y2 ≥ 5, so its projection pointQ(x, y) on xy-plane satisfies x2 + y2 ≤ 22. Since z = 9 − x2 − y2, sozx = −2x, and zy = −2y, hence z2
x + z2y = 4(x2 + y2). It follows the
surface area of S is∫∫S
1 · dS =∫∫
x2+y2≤22
√1 + z2
x + z2y dxdy =
∫∫x2+y2≤22
√1 + 4(x2 + y2) dxdy =
∫ 2π
0
∫ 2
0r√
1 + 4r2 drdθ =
18
∫ 2
0
√1 + 4r2 d(1 + 4r2) =
18
[(1 + 4r2)1/2+1
1 + 1/2
]2
0
=
112
(93/2 − 13/2) =2612
=136
.
. . . . . .
.
......
Definition. Let ρ(x, y, z) be the density (mass per unit area) of the a
surface S in space, and the mass of S is given by∫∫
Sg(x, y, z) dS.
.
......
Example. Let ρ(x, y, z) = z2 be the density function of the upperhemisphere S : x2 + y2 + z2 = a2, z ≥ 0. Find its mass.
Solution. Write S as in the graph of the functionz(x, y) =
√a2 − x2 − y2, defined on the region
D = { (x, y) | x2 + y2 ≤ a2 }. For any point (x, y, z) ∈ S, we havez =
√a2 − x2 − y2, zx = − x√
a2−x2−y2, and zy = − y√
a2−x2−y2. Then
1 + z2x + z2
y = 1 + x2+y2
a2−x2−y2 = a2
a2−x2−y2 . Moreover on S, we have
z2 = (√
a2 − x2 − y2)2 = a2 − x2 − y2. Then the mass of S is given by∫∫S
ρ(x, y, z) dS =∫∫
D(a2 − x2 − y2) · a√
a2 − x2 − y2dx dy
= a∫ 2π
0
∫ a
0
√a2 − r2 r dr dθ = −2πa
2
∫ a
0
√a2 − r2 d(a2 − r2)
= πa[(a2 − r2)3/2
]a
0= πa4.
. . . . . .
.
......
Example. Let S be a parametric surface given byr(u, v) = x(u, v)i + y(u, v)j + z(u, v)k, where (u, v) lies in the domainof D of uv-plane. On S, define a vector field
N(u, v) = ru × rv =i j k
∂x∂u
∂y∂u
∂z∂u
∂x∂v
∂y∂v
∂z∂v
=∂(y, z)∂(u, v)
i +∂(z, x)∂(u, v)
j +∂(x, y)∂(u, v)
k.
.
......
Proposition. The vector field N is continuous on S, and is normal tothe tangent plane of the surface S everywhere.
.
......
Definition. A parametric surface S given by r = r(u, v) is calledorientable if there is a continuous unit normal vector field n on S. Achoice of n is called an orientation of S, in this case, S is called anoriented surface.
For an oriented surface S with an orientation n, one defines the unitnormal vector field n(u, v) on S, by
n(u, v) =N
∥N∥ =1
∥N∥
(∂(y, z)∂(u, v)
i +∂(z, x)∂(u, v)
j +∂(x, y)∂(u, v)
k)
.
. . . . . .
..
Surface Integral with respect to coordinate areaelements
.
......
Definition. Let g = g(x, y, z) be a scalar function defined on a domaincontaining S, define the surface integral of S with respect to thecoordinate x-axis, y-axis and z-axis respectively as∫∫
Sg(x, y, z)n · i dS =
∫∫D
g(r(u, v))∂(y, z)∂(u, v)
du dv,∫∫S
g(x, y, z)n · j dS =∫∫
Dg(r(u, v))
∂(z, x)∂(u, v)
du dv, and∫∫S
g(x, y, z)n · k dS =∫∫
Dg(r(u, v))
∂(x, y)∂(u, v)
du dv,
n · i dS =1
∥N∥
(∂(y, z)∂(u, v)
, · · · , · · ·)· (1, 0, 0) ∥N∥ du dv =
∂(y, z)∂(u, v)
du dv = dx dy.
Remark. The jacobian factors are used to make the integralindependent of parametrization of the surface S.
. . . . . .
.. Surface Integral of Vector Field.
......
Let F = (P, Q, R) be a continuous vector field defined in a domaincontaining a smooth oriented surface S, with unit normal vector field.Define the flux of F across S, or the surface integral of F over S as∫∫
SF · n dS =
∫∫D(P, Q, R) ·
(∂(y, z)∂(u, v)
,∂(z, x)∂(u, v)
,∂(x, y)∂(u, v)
)du dv
=∫∫
SP dydz + Q dzdx + R dxdy.
Remark. If we change the orientation of S, the surface integral willchange by a minus sign.
. . . . . .
.
......
Example. Suppose that S is given as a graph of the functionz = z(x, y) where (x, y) is in a domain D in xy-plane. Prove that the
flux of F = (P, Q, R) is given by∫∫
SF · n dS =∫∫
D
(−P(x, y, z(x, y))
∂z∂x
− Q(x, y, z(x, y))∂z∂y
+ R(x, y, z(x, y)))
dxdy.
Proof. Parameterize S by r(x, y) = (x, y, z(x, y)), then∂(y, z)∂(x, y)
=yx yyzx zy
=0 1zx zy
= −zx,∂(z, x)∂(x, y)
=zx zyxx xy
= −zy,
and∂(x, y)∂(x, y)
= 1. Hence,∫∫S
F · n dS =∫∫
D(P, Q, R) ·
(∂(y, z)∂(x, y)
,∂(z, x)∂(x, y)
,∂(x, y)∂(x, y)
)dx dy =∫∫
D
(−P(x, y, z(x, y))
∂z∂x
− Q(x, y, z(x, y))∂z∂y
+ R(x, y, z(x, y)))
dxdy.
Remark. The formula above may be used to calculate the flux of Facross the graph of a function z = z(x, y).
. . . . . .
.
......
Example. Calculate the flux∫∫
SF · n dS, where F = k, and S is the
upper hemi-sphere z =√
a2 − x2 − y2 with D : x2 + y2 ≤ a2, with theouter normal vector field on S.
Solution. As the flux of F = (P, Q, R) = (0, 0, 1) across S is given by∫∫S
F · n dS =∫∫D
(−P(x, y, z(x, y))
∂z∂x
− Q(x, y, z(x, y))∂z∂y
+ R(x, y, z(x, y)))
dxdy
=∫∫
D1 dxdy = Area of D = πa2.
Remark. In the following, we directly compute F · n and dS in terms ofx and y.
. . . . . .
.
......
Example. Calculate the flux∫∫
SF · n dS, where F = k, and S is the
upper hemi-sphere x2 + y2 + z2 = a2, z ≥ 0 of radius a centered at(0, 0, 0), with the outer normal vector field n on S.
Solution. It follows from the gradient thatn(x, y, z) = ∇(x2+y2+z2)
∥∇(x2+y2+z2)∥ = (x,y,z)a , and F · n = z
a .
So one can rewrite S as the graph of the function z =√
a2 − x2 − y2
defined on D = { (x, y) | x2 + y2 ≤ a2}. Then it follows from
dS =√
1 + z2x + z2
y dxdy that∫∫S
F · n dS =∫∫
S
za
dS =∫∫
D
√a2 − x2 − y2
a
√1 + z2
x + z2y dxdy
=∫∫
D
√a2 − x2 − y2
a· a√
a − x2 − y2dxdy =
∫∫D
dxdy = πa2.
. . . . . .
.
......
Example. Find the flux of F(x, y, z) = (x, y, 3) out of the region Dbounded by the paraboloid z = x2 + y2 and the plane z = 4.
Solution. Let S1 be the circular top { (x, y, 4) | x2 + y2 ≤ 22}, and S2be the parabolic part parameterized by z(x, y) = x2 + y2 withx2 + y2 ≤ 4 as shown in the diagram. In this case, the outward unitnormal vector field is given by on S1 : n1(x, y, z) = k, and on S2 by
n2(x, y, z) =∇(−z + x2 + y2)
∥∇(−z + x2 + y2)∥ =(2x, 2y,−1)√4x2 + 4y2 + 1
. In fact, one can
check n2(0, 0, 0) = −1, so F1 · n1 = (x, y, 3) · (0, 0, 1) = 3, and
F2 · n2 = (x, y, 3) · (2x,2y,−1)√4x2+4y2+1
= 2x2+2y2−3√4x2+4y2+1
. On S2, the surface area
element dS =√
1 + z2x + z2
y dxdy =√
4x2 + 4y2 + 1 dxdy for
D = { (x, y) | x2 + y2 ≤ 4}. Then∫∫
SF · n dS =∫∫
S1
F · n1 dS +∫∫
S2
F · n2 dS =∫∫
D3 dxdy +
∫∫D(2x2 + 2y2 − 3) dxdy
=∫∫
D(2x2 + 2y2) dxdy =
∫ 2π
0
∫ 2
02r2 · r dr dθ = 2π
[2r4
4
]2
0= 16π.
. . . . . .
.
......
Example. Suppose that a body has temperature u(x, y, z) at its point(x, y, z). The flow of heat in the body is described by the heat-flowvector q(x, y, z) = −k∇u(x, y, z) for some constant k.
.
......
Definition. Let S be a closed surface within the body bounding thesolid region D and let n denote the outer unit normal vector for S.Then the net heat flow across the sphere S out of the region D across
its boundary D is defined by∫∫
Sq · n dS = −
∫∫S
k∇u · n dS.
. . . . . .
.
......
Example. Suppose that a uniform solid ball B of radius R is centeredat the origin, and at its point (x, y, z) the temperatureu(x, y, z) = c(R2 − x2 − y2). Find the rate of the flow of heat acrossthe sphere S of radius a < R centered at O(0, 0, 0).
Solution. The heat flow across the sphereS = { (x, y, z) | x2 + y2 + z2 = a2 } is given by∫∫
Sq · n dS = −
∫∫S
k∇u · n dS = −k∫∫
S∇(R2 − x2 − y2 − z2) · n dS =
−k∫∫
S(−2x,−2y,−2z) · (x, y, z)
adS = k
∫∫S
2(x2 + y2 + z2)
adS =
2ak∫∫
SdS = 2ak(4πa2) = 8kπa3.
. . . . . .
.. Divergence Theorem.
......
Definitions. A surface S is called piecewise smooth, if it consists of afinite number of smooth parametric surfaces.A surface S is called closed if it is the boundary of a bounded solidregion in space.
.
......
Divergence Theorem. Suppose that S is a closed piecewise smoothsurface that bounds a space region D. Let n be the outer unit normalvector field, which is continuous on each smooth piece of S. IfF = (P, Q, R) is continuously differentiable on T, then∫∫
SF · n dS =
∫∫∫D∇F dV, i.e.∫∫
SP dydz + Q dzdx + R dxdy =
∫∫∫D
(∂P∂x
+∂Q∂y
+∂R∂z
)dV.
Remark. The condition that the surface S is closed plays a crucialroles in divergence theorem.
. . . . . .
.
......
Example. Let S be the surface (with outer unit normal vector n) of theregion D bounded by the planes z = 0, y = 0, y = 2 and the paraboliccylinder z = 1 − x2. Apply the divergence theorem to compute∫∫
SF · n dS, where F(x, y, z) = (x + cos y)i + (y + sin z)j + (z + ex)k.
Solution. For any point (x, y, z) in D, we have 0 ≤ z ≤ 1 − x2, hence1 − x2 ≥ 0, so x2 ≤ 1, i.e. −1 ≤ x ≤ 1, hence we haveD = { (x, y, z) | − 1 ≤ x ≤ 1, 0 ≤ y ≤ 2, 0 ≤ z ≤ 1 − x2 }.∇ · F(x, y, z) = (x + cos y)x + (y + sin z)y + (z + ex)z = 1 + 1 + 1 = 3.Instead of evaluating the surface integral directly, we can apply the
divergence theorem that∫∫
SF · n dS =
∫∫∫D∇ · F dV =
∫∫∫D
3 dV =
3∫ 1
−1
∫ 2
0
∫ 1−x2
0dz dy dx = 3 × 2
∫ 1
−1(1 − x2) dx = 6(2 − 2
3) = 8.
Remark. Though we had not determined the outer normal vector fieldn, but it is necessary to know that n is pointing outward on theboundary S of the solid region D, before we apply divergencetheorem.
. . . . . .
.
......
Example. Let S be the surface of the solid cylinder D bounded by theplanes z = 0, z = 3 and the cylinder x2 + y2 = 4. Calculate the
outward flux∫∫
SF · n dS, where
F(x, y, z) = (x2 + y2 + z2)(xi + yj + zk).
Solution. Note that the divergence ∇ · F(x, y, z)= (x3 + xy2 + xz2)x + (x2y + y3 + yz2)y + (x2z + y2z + z3)z= 5(x2 + y2 + z2). It follows from the divergence theorem and thenuse cylindrical coordinates that∫∫
SF · n dS =
∫∫∫D∇ · F dV =
∫∫∫D
5(x2 + y2 + z2) dV
=∫ 2π
0
∫ 2
0
∫ 3
05(r2 + z2) r dz dr dθ
= 10π∫ 2
0
[r3z +
13
rz3]3
0dr = 10π
∫ 2
0(3r3 + 9r) dr
= 10π
[34
r4 +92
r2]2
0= 300π.
. . . . . .
.
......
Example. Let S be the sphere x2 + y2 + z2 = 4, which bounds a solid
ball D. Calculate the outward flux∫∫
SF · n dS, where
F(x, y, z) = (x2 + y2 + z2)(xi + yj + zk).
Solution. Note that the divergence ∇ · F(x, y, z)= (x3 + xy2 + xz2)x + (x2y + y3 + yz2)y + (x2z + y2z + z3)z= 5(x2 + y2 + z2). It follows from the divergence theorem and thenuse spherical coordinates that∫∫
SF · n dS =
∫∫∫D∇ · F dV =
∫∫∫D
5(x2 + y2 + z2) dV
=∫ 2π
0
∫ π
0
∫ a
05ρ2 · ρ2 sin ϕ dρ dϕ dθ
=∫ a
05ρ4 dρ ×
∫ π
0sin ϕ dϕ ×
∫ 2π
0dθ
= a5 × [1 − (−1)]× 2π = 4πa5.
. . . . . .
.
......
Example. Let D be a solid region bounded by a smooth parametricclosed surface S, with the outer unit normal vector field n to S. Prove
that the volume of D is13
∫∫S
x dydz + y dzdx + z dxdy.
Solution. Let F(x, y, z) = (x, y, z), and ∇ · F(x, y, z) = ∇ · (x, y, z)= (x)x + (y)y + (z)z = 3. Then it follows from the definition of flux
integral and divergence theorem that13
∫∫S
x dydz + y dzdx + z dxdy =
13
∫∫S
F · n dS =13
∫∫∫D∇ · F dV =
13
∫∫∫D
3 dV = volume of D.
.
......
Example. Let V and A be the volume and the surface area of thesphere of radius a centered at (0, 0, 0). Prove that 3V = aA.
Solution. Let S be the sphere given by { (x, y, z) | x2 + y2 + z2 = a2 },
hence the outer normal vector n(x, y, z) =(x, y, z)
a. Then 3V
=13
∫∫S
x dydz + y dzdx + z dxdy =∫∫
S(x, y, z) · n dS =∫∫
S(x, y, z) · (x, y, z)
adS =
1a
∫∫S(x2 + y2 + z2) dS = a
∫∫S
dS = aA.
. . . . . .
.
......
Example. Show that the divergence of the continuously differentiable
vector field F at the point P is given by ∇ · F(P) = limr→0
1|Br|
∫∫Sr
F · n dS,
where Sr is the sphere of radius r centered at P and |Br| = 43 πr3 is the
volume of the ball Br that the sphere bounds.
Solution. As F is continuously differentiable at P(a, b, c), so ∇ · F iscontinuous at P. Then for any ε > 0, there exists δ > 0 such that|∇ · F(x, y, z)−∇ · F(a, b, c)| < ε for all point (x, y, z) in the open ballcentered at P(a, b, c) with radius δ > 0. Then∣∣∣∣∇ · F(P)− 1
|Br|
∫∫Sr
F · n dS∣∣∣∣ = 1
|Br|
∣∣∣∣ |Br|∇ · F(P)−∫∫
SrF · n dS
∣∣∣∣=
1|Br|
∣∣∣∣∫∫∫Br∇ · F(P)dV −
∫∫∫Br∇ · F dS
∣∣∣∣=
1|Br|
∫∫∫Br|∇ · F(P)−∇ · F| dS <
1|Br|
∫∫∫Br
ε dV =|Br||Br|
ε = ε, and
hence the result follows..
......
Remark. Let F be a vector field defined on a region containing a pointP, then P is called a source if ∇ · F(P) > 0, and sink if ∇ · F(P) < 0.
. . . . . .
.
......
Example. The outward flux of vector field F(x, y, z) = xi + yj + zkacross the sphere x2 + y2 + z2 − 2x − 2y − 2z = 3 is .
Solution. Let D be the solid bounded by the sphere, then fromdivergence theorem we have the outward flux of F is∫∫
SF · n dS =
∫∫∫D∇ · F dV =
∫∫∫D(
∂x∂x
+∂x∂x
+∂x∂x
) dV =∫∫∫
D3 dV
= 3 volume of D = 3 × 4(√
6)3π/3 = 24π√
6.
. . . . . .
.
......
Example. Let F(x, y, z) = (x, y, z), and T be the surface defined by{ (x, y, z) ∈ R3 | |x|+ |y|+ |z| = 1 }, with outward pointing unit
normal n on T. Evaluate the surface integral∫∫
TF · ndS.
Solution. Let D be the solid bounded by the surface T, i.e.D = { (x, y, z) ∈ R3 | |x|+ |y|+ |z| ≤ 1 }. One can check that thecondition of the divergence theorem holds, and hence∫∫
TF · ndS =
∫∫∫D
divFdV =∫∫∫
D(
∂x∂x
+∂x∂x
+∂x∂x
) dV
= 3 ×∫∫∫
DdV = 3Vol(D) = 3 × 8 × 1
3!× 1 × 1 × 1 = 4.
. . . . . .
.
......
Example. Let T be the solid cylinder defined by the inequalitiesx2 + y2 ≤ 1 and 0 ≤ z ≤ 4. Let S be the entire boundary of T. LetF(x, y, z) = (5z + 3)k. Find the outward flux of F through S.
Solution I. Let S1 = { (x, y, 4) | x2 + y2 ≤ 4 } be the top of T, andS2 = { (x, y, 0) | x2 + y2 ≤ 4 } be the bottom of T. The outer unitnormal vector field on S1 and S2 are n1 = k, and n2 = −krespectively. The outward flux of F throughS1 =
∫∫S1
F · n dS =∫∫
S1(5 × 4 + 3)k · k dS = 23×Area of S1 = 23π.
The outward flux of F throughS2 =
∫∫S2
F · n dS =∫∫
S2(5 × 0 + 3)k · (−k) dS = 3×Area of S2 = 3π.
The flux through the curved part of the boundary is 0 since n ishorizontal there, so F · n = 0, So the total flux of F across S is23π − 3π = 20π.Solution II. We have div F = 0 + 0 + 5 = 5, so by the divergence
theorem, the flux equals∫∫
SF · n dS =
∫∫∫T∇ · F dV = 5× Volume of
T= 5 × π × 1 × 4 = 20π.
. . . . . .
.. Stokes’ Theorem.
......
Definition. A piecewise smooth surface S in space is called anoriented surface, if one chooses a continuous unit normal vector fieldon each smooth piece S.A positive orientation of the boundary C of an oriented surface is aunit tangent vector T such that n × T always points into S.
Remark. Think of a man walks along C as T, and heads up in n, thenthe region is always on the left hand side of the man..
......
Stokes’ Theorem. Let S be an oriented, bounded, and piecewisesmooth surface in space with positive oriented boundary C withrespect to the unit normal vector field n on S. Suppose that T is apositively oriented unit vector field tangent to C in the right orientationof C. If F is a continuously differentiable vector field defined in a
region containing S, then ∫
CF · T ds =
∫∫S(curl F) · n dS.
Remark. The condition that the positive orientation of C agrees withthe choice of unit normal vector field n is essential.
. . . . . .
.
......
Example. Let F(x, y, z) = (2y,−2x, sin z). Let S be the upper half ofthe sphere x2 + y2 + z2 = 9. Evaluate the outward (i.e., upward) fluxof curl F through S.
Solution. The surface S is given by{ (x, y, z) | x2 + y2 + z2 = 9, z ≥ 0 }, then its boundary C is the circlex2 + y2 = 9 in the xy-plane oriented counterclockwise. So C isparameterized by r(t) = (3 cos t, 3 sin t, 0) for t ∈ [0, 2π]. The flux is∫∫
ScurlF · n dS =
∮C
F · T ds
=∫ 2π
0(2 · 3 sin t,−2 · 3 cos t, 0) · (−3 sin t, 3 cos t, 0) dt
=∫ 2π
0−18(cos2 t + sin2 t) dt = −36π.
Remark. As n is already fixed in the question, you can use the sameidea of a man walking along the boundary C so that the surfaceappears in his left hand side to determine the positive orientation of Cwith respect to n of S.
. . . . . .
.
......
Example. Let F(x, y, z) = 3zi + 5xj − 2yk. Evaluate the line integral∮C
F · T ds where C is the ellipse in which the plane z = y + 3
intersects the cylinder x2 + y2 = 1. Orient C counterclockwise as viewfrom above.
Solution. Let S be the region on the plane z = y + 3 bounded by thecylinder x2 + y2 = 1. Then S = { (x, y, z) | z − y = 3, x2 + y2 ≤ 1 } ispart of a level surface z − y = 3, with a unit normal vector field
n(x, y, z) =∇(z − y)
∥∇(z − y)∥ =(0,−1, 1)√
2on S. Next the
curl F(x, y, z) =i j k∂
∂x∂∂y
∂∂z
3z 5x −2y= −2i + 3j + 5k, hence
curl F · n = (−2, 3, 5) · (0,−1, 1)/√
2 = (−3 + 5)/√
2 =√
2.
∫
CF · T ds =
∫∫S(curl F) · n dS =
∫∫S
√2 dS =
√2Area of (S) =
√2 × 1 ×
√2π = 2π.
Remark. S can be parameterized by r(x, y) = (x, y, y + 3) defined onD = { (x, y) | x2 + y2 ≤ 1 }.
. . . . . .
.
......
Example. Let F(x, y, z) = (y, z,−2x). Let C be a simple closed curve
contained in the plane x + y + z = 1. Show that∫
CF · T ds = 0.
Solution. Let By Stokes’ theorem, ∫
CF · T ds =
∫∫S(curl F) · n dS. Here
n is perpendicular to S everywhere, and hence perpendicular to the
plane with the unit normal vector n =∇(x + y + z)
∥∇(x + y + z)∥ =(1, 1, 1)√
3to
the plane. On the other hand,curl F(x, y, z) = curl (y, z,−2x) = (−1, 2, 1), so curl F(x, y, z) · n = 0,
thus∫
CF · T ds =
∫C
0 ds = 0.
. . . . . .
.
......
Example. Evaluate the surface integral∫∫
S(∇× F) · n dS, where
F(x, y, z) = 3zi + 5xj − 2yk, and S is the parabolic surface z = x2 + y2
that lie below the plane z = 4 and whose orientation is given by theupper unit normal vector.
Solution. The boundary of S is the circle C parameterized byr(t) = (2 cos t, 2 sin t, 4), where 0 ≤ t ≤ 2π. It follows from the Stokes’
theorem that∫∫
S(∇× F) · n dS =
∮C
F · T ds =∮
C3z dx + 5x dy − 2y dz
=∫ 2π
03 · 4(−2 sin t dt) + 5 · (2 cos t) · (2 cos t dt) + 2 · (2 sin t) · (0 dt)
=∫ 2π
0(−24 sin t + 20 cos2 t) dt =
∫ 2π
010(1 + cos 2t) dt = 20π.
. . . . . .
.. Conservative and Irrotational Fields.
......Definition. A differentiable vector field F is called irrotational if∇× F = 0.
.
......
Example. The vector field F(x, y) =−yi + xjx2 + y2 is irrotational on
D = R2 \ {(0, 0)}. , but we had proved that F is not conservative on D.
.
......
Definition. Let D be a region in space, D is called simply connected, ifevery simple closed curve in D can be continuously shrunk to a pointwhile staying inside D.
.
......
Examples. (a) The entire plane, entire space are simply connected;(b) Rectangle, the sphere and ball are simply connected;(c) The interior of a torus, and the punctured plane are not a simplyconnected.
. . . . . .
.
......
Proposition. Let F be a continuous vector field defined on a region D,prove that the line integral of F is independent of path if and only if∫
CF · T ds = 0 for any piecewise smooth closed curve C in D.
Proof. Suppose that line integral of F is independent of path, then letC be any closed curve with the same starting and terminal point A,then the constant path C′ with A for all t is also a curve with the samestarting and terminal point A. It follows from the path independence of
the line integral of F that∫
CF · T ds =
∫C′
F · T ds =∫
C′F · 0 ds = 0.
Conversely, suppose C1 and C2 are two paths, both of them startsfrom the same point A, and terminates at point B. Let C = C1 ∪ (−C2)
be a closed path from A to B via C1, and back from B to A via −C2 (inreverse direction of C2). Then C is a piecewise smooth closed curvein D, hence one has0 =
∫C
F · T ds =∫
C1
F · T ds +∫−C2
F · T ds =∫
C1
F · T ds −∫
C2
F · T ds.
Hence∫
C1
F · T ds =∫
C2
F · T ds. So the line integral of F is
independence of path.
. . . . . .
.
......
Theorem. Let F be a continuously differentiable vector field in asimply connected region D in space. Then the vector field F isirrotational if and only if it is conservative; that is ∇× F = 0 in D if andonly if F = ∇f for some scalar function f defined on D.
Solution. If F = ∇f , then ∇× F = ∇×∇f = 0 on D. Conversely,suppose that F is irrotational on D, one wants to define a functionf (x, y, z) such that ∇f = F on D. Following the same idea in theplanar case, Let A(a, b, c) and B(x, y, z) be two points in D, and C be apath from A to B. If one can show that the line integral of F is path
independent, then define f (x, y, z) =∫ (x,y,z)
(a,b,c)∇F · T ds, which depends
only on the end points of C. Then it follows from continuity of thevector field F that ∇f = F on D. It remains to show that the lineintegral of F is independent of path, which is equivalent to the fact that∮
CF · T ds = 0 for any closed path C in D. As D is simply connected, it
follows one can shrink the curve C continuously to a path in such waythat C bounds a surface S via the shrinking, then∮
CF · T ds =
∫∫S(∇× F) · n dS =
∫∫S
0 · n dS = 0.
. . . . . .
.. Summary of Line Integral of Vector Field
Let F be a vector field defined on a region D. Then we can summaryour important result as follows:.
......
F is conservative on D ⇐⇒ F = ∇f for some function f⇕ ⇓ (−)xy = (−)yx∮
CF · T ds = 0
for any closed path in D
Simply connected⇐=Stokes Thm
∇× F = 0 on D
Remark. Conservative vector field is irrotational; but the converse isnot true. In fact, it depends on the domain of the vector field (compare the lower horizontal arrow). An example is F(x, y) = (−y, x)
x2+y2
on R2 \ {(0, 0)}.
. . . . . .
.
......
Example. (a) Show that the vector fieldF(x, y, z) = 3x2i + 5z2j + 10yzk is irrotational on R3.(b) Find a potential function f (x, y, z) such that ∇f = F.
Solution. (a) One needs to show that∇× F(x, y, z) = ∇× (3x3, 5z2, 10yz) = (10z − 10z)i = 0.(b) One can use the line integral of F along the segment from (0, 0, 0)to (x, y, z) by r(t) = (tx, ty, tz) for 0 ≤ t ≤ 1 as follows:
f (x, y, z) =∫ (x,y,z)
(0,0,0)F · T ds
=∫ 1
0( 3(tx)2, 5(tz)2, 10(ty)(tz) ) · (x, y, z) dt
=∫ 1
0(3x3 + 15yz2)t2 dt = x3 + 5yz2 + C.
. . . . . .
.
......
Example. Let T be the solid bounded by the paraboloidsS1 : z = x2 + 2y2 and S2 : z = 12 − 2x2 − y2. Evaluate the outward fluxof F(x, y, z) = xi + yj + zk across the boundary of T.
Solution. Project the solid onto the xy-plane with its shadow R. LetQ(x, y, z) be the intersection of S1 and S2, thenz = x2 + 2y2 = z = 12 − 2x2 − y2, so 3(x2 + y2) = 12, thenx2 + y2 = 4. Then the image Q′(x, y) of Q in R satisfies the equationx2 + y2 = 22, i.e. Q′ lies on a circle. Inside the circular disc R, we havex2 + y2 ≤ 2, so it follows that x2 + 2y2 ≤ 12 − 2x2 − y2, and henceT = { (x, y, z) | 0 ≤ x2 + y2 ≤ 22, x2 + 2y2 ≤ z ≤ 12 − 2x2 − y2 }. Itfollows from divergence theorem that∫∫
SF · n dS =
∫∫∫T∇ · F dV =
∫∫x2+y2≤22
∫ 12−2x2−y2
x2+2y2(1 + 1 + 1) dV
= 3∫ 2π
0
∫ 2
0(12 − 3r2)r drdθ = 6π
[6r2 − 3
4r4]2
0= 6π(24 − 12) = 72π.
. . . . . .
.
......
Example. Let F(x, y, z) = (x + y)i + (y − x)j + zk be vector field. LetS be the hemisphere x2 + y2 + z2 = 4, z ≤ 0, and n be the outward
pointing unit normal vector field of S. Then∫∫
ScurlF · ndσ is .
Solution. The boundary of S is the circle given byC : r(t) = ( x(t), y(t), z(t) ) = (2 cos t, 2 sin t, 0), 0 ≤ t ≤ 2π. From theStokes’ theorem, we have∫∫
ScurlF · ndσ =
∫C
F · Tds
=∫ 2π
t=0[ (x(t) + y(t))x′(t) + (y(t)− x(t))y′(t) + z(t)z′(t) ]dt
=∫ 2π
0−4(cos t + sin t) sin t + 4(sin t − cos t) cos tdt
=∫ 2π
0(−4 sin2 t − 4 cos2 t)dt = −8π.
. . . . . .
.
......
Example. Let F(x, y, z) = f (x, y, z)k, where f (x, y, z) is a differentiable
function defined in R3. Then the outward flux∫∫
ScurlF · n dS of curlF
across the upper hemisphereS = { (x, y, z) | x2 + y2 + z2 = 1, z ≥ 0 } isA. 0 B. f (0, 0, 0) C. grad f (0, 0, 0) D. None of the above.
Solution. Let C be the unit circle x2 + y2 = 1 in the usualcounterclockwise direction, one can apply Stokes’ theorem to obtain
the following∫∫
ScurlF · n dS =
∫C
F · T ds =∫
C0 ds = 0.
