.
Let R be the region in xy-plane bounded by the circles C1 : x2 + y2 = 2x and
2
2
C
. 2 : x + y = 2y. Then R =
. { (x, y) | x2 + y2
2.
{ (x, y) | x2 + y2
3.
{ (x, y) | x2 + y2
4.
{ (x, y) | x2 + y2
1
≤ 2x, x2 + y2 ≤ 2y }
≥ 2x, x2 + y2 ≥ 2y }
≤ 2x, x2 + y2 ≥ 2y }
≥ 2x, x2 + y2 ≤ 2y }
. None of the above
5
.
Review of Matb 201 in 2012
.
.
.
.
.
.
Let R be the region in xy-plane bounded by the circles C1 : x2 + y2 = 2x and
2
2
C
. 2 : x + y = 2y. Then R =
. { (x, y) | x2 + y2
2.
{ (x, y) | x2 + y2
3.
{ (x, y) | x2 + y2
4.
{ (x, y) | x2 + y2
1
≤ 2x, x2 + y2 ≤ 2y }
≥ 2x, x2 + y2 ≥ 2y }
≤ 2x, x2 + y2 ≥ 2y }
≥ 2x, x2 + y2 ≤ 2y }
. None of the above
5
Solution. Rewrite the equation x2 + y2 = 2ax as (x − a)2 + y2 = a2 , which is
equation of a circle C. i.e. the level curve x2 + y2 − 2ax = 0 is a circle on
xy-plane. Point P(x, y) lies on or inside the circle C if and only if
(x − a)2 + y2 ≤ a2 ; and lies on or outside of the circle C if and only if
(x − a)2 + y2 ≥ a2 . In general, let Q(x, y) be a point in the region R, then Q lies
inside both circles C1 and C2 , so we have x2 + y2 ≤ 2x and x2 + y2 ≤ 2y.
.
Review of Matb 201 in 2012
.
.
.
.
.
.
If S is the part of the paraboloid z = x2 + y2 with z ≤ 2, then the surface area of
.S is
. π
. 2π
11
3.
3 π
13
4.
3 π
5. None of the above
1
2
.. Answer
Solution is on the next page!
.
Review of Matb 201 in 2012
.
.
.
.
.
.
Find
the surface area of paraboloid S : z = x2 + y2 with z ≤ 2.
.
Solution. First one notes that surface S is realized as graph of z(x, y) = x2 + y2 ,
then project the curved surface S onto the xy-plane and its shadow
R = {√
(x, y) | x2 + y2 ≤ 2 } in xy-plane. Then surface area element
√
dS = 1 + z2x + z2y dx dy = 1 + 4(x2 + y2 ) dx dy.
So the surface area of S
∫∫
∫∫ √
∫ 2π ∫ √2 √
1 dS =
=
1 + 4x2 + 4y2 dxdy =
1 + 4r2 r drdθ
S
R
0
0
[
] √2
∫ √2 √
2 )1+1/2
2π
π
(
1
+
4r
π 2
=
1 + 4r2 d(1 + 4r2 ) =
= · · (27 − 1)
8 0
4
1 + 1/2
4 3
0
13
= π.
3
.
Review of Matb 201 in 2012
.
.
.
.
.
.
Evaluate the iterated integral
.
.
2.
3.
4.
5.
1
∫ 3∫ x
1
1
0
x
dydx.
− 98
2
ln 3
0
ln 2.
.. Answer
Solution. The iterated integral
∫ 3∫ x
∫ 3 ∫ x
∫ 3
∫ 3
1
1
1
dy dx =
1dydx =
× x dx =
1 dx = 3 − 1 = 2.
1
0 x
1 x 0
1 x
1
.
Review of Matb 201 in 2012
.
.
.
.
.
.
∫∫
Consider the double integral,
R
f (x, y)dA, where R is the portion of the disk
≤ 1, in the upper half-plane, and y ≥ 0. Express the double integral as
an
. iterated integral.
x2
+ y2
.
∫ 1 ∫ √ 1 − x2
√
1
.
−1 − 1−x
∫ 0 ∫ √ 1 − x2
2
2
.
−1 0
∫ 1 ∫ √ 1 − x2
3
.
−1 0
∫ 1 ∫ √ 1 − x2
√
4
.
0
− 1−x
∫ 1 ∫ √ 1 − x2
5
0
0
2
f (x, y)dydx
f (x, y)dydx
f (x, y)dydx
f (x, y)dydx
f (x, y)dydx
.. Answer
.
Review of Matb 201 in 2012
.
.
.
.
.
.
Find a and b for the correct interchange of order of integration:
∫ 2 ∫ 2x
.0
.
.
3.
4.
5.
1
2
x2
∫ 4∫ b
f (x, y)dydx =
0
a
f (x, y)dxdy.
a = y2 , b = 2y
√
a = y/2, b = y
a = y/2, b = y
√
a = y, b = y/2
None of the above
.. Answer
.
Review of Matb 201 in 2012
.
.
.
.
.
.
∫∫
Evaluate the double integral
R
ydA, where R is the region of the xy-plane
.inside the triangle with vertices (0, 0), (2, 0) and (2, 1).
. 2
8
2.
3
2
3.
3
1
4.
3
5.
1.
1
.. Answer
.
Review of Matb 201 in 2012
.
.
.
.
.
.
∫∫
Evaluate the double integral R ydA, where R is the region of the xy-plane
.inside the triangle with vertices O(0, 0), A(2, 0) and B(2, 1).
Solution. By marking the points on xy-plane, we know that the region R is a
right-angle triangle as shown in (a), One can project the region R onto y-axis
as shown in (b), or onto the x-axis as shown in (c). Then R can be described
as follows:
R
= { (x, y) | 0 ≤ x ≤ 2, 0 ≤ y ≤ x/2 }
= { (x, y) | 0 ≤ y ≤ 1, 0 ≤ x ≤ 2y }.
∫∫
The area of R is given by
∫ 1 ∫ 2y
R
1 dA =
0
0
∫ 1
y dx dy =
.
Review of Matb 201 in 2012
0
.
2y2 dy =
.
.
2
2
× 13 = .
3
3
.
.
.
The volume of the solid region in the first octant bounded above by the
parabolic sheet z = 1 − x2 , below by the xy-plane, and on the sides by the
planes
y = 0 and y = x is given by the double integral
.
.
∫ 1∫ x
1
0
.
0
(1 − x2 )dydx
∫ 1 ∫ 1 − x2
2
.
∫01 ∫0 x
3
.
x
∫−11∫ −
0
4
.
0
x
0
x
(1 − x2 )dydx
(1 − x2 )dydx
∫ 1 ∫ 1 − x2
5
xdydx
dydx
.. Answer
.
Review of Matb 201 in 2012
.
.
.
.
.
.
Determine the volume of the solid region D in the first octant bounded above
by the parabolic sheet z = 1 − x2 , below by the xy-plane, and on the sides by
.the planes y = 0 and y = x.
Solution. First sketch the 3 planes: xyplane (i.e. z = 0), y = 0 and y = x (in
green). Next draw curve C : z = 1 − x2 on
the xz-plane, which is independent of variable y, and then move the curve C along
the direction given by y-axis, and the locus
of the curve traces a cylinder (pink top). Third, we project the solid D onto
xy-plane, and its shadow is the triangle R bounded by the line y = x, x-axis and
the intersection line of the surfaces z = 0 and z = 1 − x2 (i.e. x = 1). More
precisely, R = { (x, y) | 0 ≤ x ≤ 1, 0 ≤ y ≤ x }. Four, the solid region
D = { (x, y, z) | 0 ≤ x ≤ 1, 0 ≤ y ≤ x, 0 ≤ z ≤ 1 − x2 }, and the volume of D
∫∫∫
is
D
1 dV =
∫ 1 ∫ x ∫ 1−x2
0
0
0
∫ 1∫ x
1 dz dy dx =
0
0
1 − x2 dy dx.
.
Review of Matb 201 in 2012
.
.
.
.
.
.
The area of one leaf R of the three-leave rose bounded by the graph of
.r = 5 sin 3θ is
. 5π
6
.2 25π
12
25π
3.
3
5π
4.
3
25π
5.
3
1
.. Answer
.
Review of Matb 201 in 2012
.
.
.
.
.
.
Find the area of one leaf R of the three-leave rose bounded by the graph of
r. = 5 sin 3θ.
∫ π/3 ∫ 5 sin 3θ
1
Solution. Area of R =
r drdθ =
2
0
0
∫
25 π
25
25 π/3 1 − cos 6θ
=
dθ =
· = π.
2 0
2
2 3
12
∫ π/3
0
25 sin2 3θ dθ
.
Review of Matb 201 in 2012
.
.
.
.
.
.
Find the area of the portion S of the plane x + 3y + 2z = 6 that lies in the first
octant.
.
√
1. 3
√11
2. 6
√7
3. 6
√14
4. 3
√14
5. 6
11
.. Answer
.
Review of Matb 201 in 2012
.
.
.
.
.
.
Find the area of the portion S of the plane x + 3y + 2z = 6 that lies in the first
octant.
.
Solution. First draw the plane: the plane x + 3y + 2z = 6
meets x-axis at (6, 0, 0), y-axis at (0, 2, 0), and z-axis at
(0, 0, 3) respectively as shown in the figure on the right.
Then these 3 intersection point determine a triangular region S in the plane x + 3y + 2z = 6. As we are interested
in the point (x, y, z) in the first octane, i.e. x, y, z ≥ 0, so
6 = x + 3y + 2z ≥ 3y + 2z. This determines shadow of S as
R = { (y, z) | y, z ≥ 0 and 3y + 2z ≤ 6 }
= { (x, y) | 0 ≤ y ≤ 2, 0 ≤ x ≤ 6−23y }.
Note that the surface S is the graph of x(y, z) = 6 − 3y − 2z, where (y, z) is in
R. Then xy = −3 and xz = −2, and the surface area element
√
√
dS = 1 + (xy )2 + (xz )2 dy dz = 14 dy dz. The area of the triangle S is
∫∫ √
∫ 2 ∫ 6−3y √
√
√
2
1
1 + x2y + x2z dydz =
14 dzdy = 14 · · 2 × 3 = 3 14.
2
0
0
R
.
Review of Matb 201 in 2012
.
.
.
.
.
.
A solid region D in the first octant is bounded by the surfaces
2
.z = y , y = x, y = 0, z = 0 and x = 4. The volume of the region is
. 64
64
2.
3
32
3.
3
4. 32
16
5.
3
1
.. Answer
.
Review of Matb 201 in 2012
.
.
.
.
.
.
Find the volume of the solid region D in the first octant is bounded by the
surfaces
z = y2 , y = x, y = 0, z = 0 and x = 4.
.
Solution. The planes z = 0 is the base of the solid D, and the 3 planes
y = 0, x = 4 and y = x bounds a triangular prism which gives the ‘3’ side faces
of D. And finally the surface z = y2 is the top of the solid D. Hence, we have
D = { (x, y, z) | 0 ≤ x ≤ 4, 0 ≤ y ≤ x, 0 ≤ z ≤ y2 }. Then the volume of D
[ 4 ]4
∫ 4∫ x
∫ 4 3
∫ 4 ∫ x ∫ y2
x
x
64
2
y dydx =
1 dzdydx =
dx =
=
= .
3
12
3
0
0
0
0
0
0
0
.
Review of Matb 201 in 2012
.
.
.
.
.
.
Let D be a region√bounded above by the sphere x2 + y2 + z2 = 32 and below
by the cone z = x2 + y2 . The mass density at any point in D is its distance
.from the xy-plane. The total mass m of D is
∫ 4 ∫ √16−x2 ∫ √32−x2 −y2
1.
z dz dy dx
√
√
−4 − 16−x2
.
2
.
3
.
0
√
z dz dy dx
z dz dy dx
z dz dy dx
x2 +y2
∫ 4 ∫ √16−x2 ∫ √32−x2 −y2
√2 2
√
−4 − 16−x2
x +y
0
.
x2 + y2
√2 2
√
−4 − 16−x2
x +y
√
∫ 2 ∫ 4−x2 ∫ √32−x2 −y2
√
√
−2 − 4−x2 − x2 +y2
√
√
∫ 4 ∫ 16−x2 ∫ 32−x2 −y2
4
5
−
∫ 4 ∫ √16−x2 ∫ √32−x2 −y2
z dz dy dx
.. Answer
.
Review of Matb 201 in 2012
.
.
.
.
.
.
Let D be a region√bounded above by the sphere x2 + y2 + z2 = 32 and below
by the cone z = x2 + y2 . The mass density at any point in D is its distance
.from the xy-plane. Find the total mass m of D.
Solution. Let P(x, y, z) be an arbitrary point in the
intersection of the cone and the sphere, then both
x2 + y2 + z2 = 32 and z2 = x2 + y2 hold, it follows
that z2 = x2 + y2 = 16, so the projection of P onto
2
2 = 42 . For any
xy-plane will be on the
√ circle x + y √
point in D, we have x2 + y2 ≤ z ≤ 32 − x2 + y2
and hence √
x2 + y2 ≤ 32 − x2√+ y2 , i.e. x2 + y2 ≤ 42 .
2
2
So D = { (x, y, z) | 0 ≤ x + y ≤ 16,√ x2 + y2 ≤ z ≤ 32 − x2 + y2 }. Then
∫∫∫
mass of D =
=
∫∫
D
∫
√
z dV =
x2 +y2 ≤4
32−x2 −y2
x2 +y2
∫ 4 ∫ √16−x2 ∫ √32−x2 −y2
√
−4 − 16−x2
√
x2 + y2
z dz dA
z dz dy dx.
.
Review of Matb 201 in 2012
.
.
.
.
.
.
Let D be a region√bounded above by the sphere x2 + y2 + z2 = 32 and below
by the cone z = x2 + y2 . The mass density at any point in D is its distance
.from the xy-plane. The total mass m of D, in terms of spherical coordinates is
.
∫ 2π ∫ π/4 ∫ √32
1
0
.
0
∫ 2π ∫ π/4 ∫
0√
32
2
.
0
0
0
∫ 2π ∫ π/4 ∫ √32
3
0
.
0
∫ 2π ∫ π/2 ∫
0√
32
4
0
.
0
∫ 2π ∫ π/4 ∫
0√
5
0
0
0
32
ρ3 cos φ sin φ dρdφdθ
ρ cos φ sin φ dρdφdθ
ρ3 sin2 φ dρdφdθ
ρ3 cos φ sin φ dρdφdθ
ρ cos φ sin φ dρdφdθ
.. Answer
.
Review of Matb 201 in 2012
.
.
.
.
.
.
Let D be a region√bounded above by the sphere x2 + y2 + z2 = 32 and below
by the cone z = x2 + y2 . The mass density at any point in D is its distance
from the xy-plane. Express the total mass m of D, as iterated integral in terms
.of spherical coordinates.
Solution. By using
√ (x, y, z) = (ρ sin ϕ cos θ, ρ sin ϕ sin θ, ρ cos ϕ), one can rewrite
the cone z√= x2 + y2 as
ρ cos ϕ = (ρ sin ϕ cos θ )2 + (ρ sin ϕ sin θ )2 = |ρ sin ϕ| = ρ sin ϕ (as
0 ≤ ϕ ≤ π ) in spherical coordinates, it follows
√that tan ϕ = 1, i.e. ϕ = π/4.
Hence, the unbounded cone { (x, y, z) | z = x2 + y2 } can be described as
{ (ρ, ϕ, θ ) | ϕ =√
π/4 }. By the same reason, unbounded solid cone
{ (x, y, z) | z ≥ x2 + y2 } can be described as { (ρ, ϕ, θ ) | ϕ = π/4 } in
spherical coordinates.
√ Eventually,
√ the bounded cone
√
2
2
D = { (x, y, z) | z = x + y , x2√+ y2 + z2 ≤ ( 32)2 } can be described as
{ (ρ, ϕ, θ ) | 0 ≤ ϕ ≤ π/4, 0 ≤ ρ ≤ 32, 0 ≤√θ ≤ 2π } in spherical coordinates.
∫∫∫
The mass of D =
=
z dV =
1
2
∫ 2π ∫ π/4 ∫
D
0
0
∫ 2π ∫ π/4 ∫ √32
3
0
0
0
32
0
(ρ cos ϕ) · ρ2 sin ϕ dρ dϕ dθ
ρ sin(2ϕ) dρ dϕ dθ.
.
Review of Matb 201 in 2012
.
.
.
.
.
.
The double integral
∫ 1 ∫ √1−x2
0
coordinates
becomes
.
.
∫ π∫ 1
1
.
∫0
2
.
.
∫0
π∫ 1
∫0
0
π/2 ∫ 1
0
.
π/2
∫0
5
0
r9 sin2 θdrdθ
r8 sin θdrdθ
4
∫0
y2 (x2 + y2 )3 dydx when converted to polar
r9 sin2 θdrdθ
0
π/2 ∫ 1
3
0
0
1
r8 sin θdrdθ
r8 sin2 θdrdθ
.. Answer
.
Review of Matb 201 in 2012
.
.
.
.
.
.
Rewrite the iterated integral
terms
of polar coordinates.
.
∫ 1 ∫ √ 1 − x2
0
0
y2 (x2 + y2 )3 dydx as iterated integral in
Solution. Recall that (x, y) = (r cos√θ, r sin θ ). First the domain of integration is
R = { (x, y) | 0 ≤ x ≤ 1, 0 ≤ y ≤ 1 − x2 }. It follows that R is bounded
√ by the
x-axis (x = 0), y-axis (y = ymin (x) = 0), and the curve y = ymax (x) = 1 − x2
i.e. x2 + y2 = 1 (unit circle). Hence R can be written as
{ (r, θ ) | 0√≤ r ≤ 1, 0 ≤ θ ≤ π/2 } in polar coordinates. Then
∫ 1∫
=
1 − x2
0
0∫
∫ π/2
1
∫0
=
0
π/2
∫0
0
1
y2 (x2 + y2 )3 dydx
r2 sin2 θ (r2 )3 r dr dθ
r9 sin2 θ dr dθ.
.
Review of Matb 201 in 2012
.
.
.
.
.
.
Line integrals, Surface integrals,
Green’s Theorem, Divergence Theorem
.
and Stokes’ Theorem
.
.
.
.
.
.
. Surface Integrals
.
If the surface S is given by the graph z = z(x, y), where (x, y) lies in the domain
2
of D ⊂
√R , i.e. S = { (x, y, z(x, y)) | (x, y) ∈ D }, then the surface area element
dS =
.
1 + z2x + z2y dxdy.
.
If the surface S is given by parametric form
r(u, v) = x(u, v)i + y(u, v)j + z(u, v)k, where (u, v) lies in the domain
of D
of uv-plane.
Surface
area element dS = ∥N∥dudv
∂r
∂(y, z)
∂r ∂(z, x)
∂(x, y) = × dudv = i+
j+
k dudv
∂(u, v)
∂(u, v)
∂(u, v) √∂u ∂v
2
2
2
.= (yu zv − zu yv ) + (xu zv − zu xv ) + (xu yv − yu xv ) du dv.
.
Definition. Let g = g(x, y, z) be a continuous function defined on domain
containing
S, the surface
integral of the
∫∫
∫∫
√ function g on S is
g(x, y, z(x, y)) 1 + z2x + z2y dxdy, or
∂r
∂r × du dv
g(x(u, v), y(u, v), z(u, v)) ∂u ∂u
D
g(x, y, z) dS =
S
∫∫
=
.
D
.
.
.
.
.
.
.
Definition. Let S be a parametric surface given by
r(u, v) = x(u, v)i + y(u, v)j + z(u, v)k, where (u, v) lies in the domain of D of
uv-plane. On S, define a vector field
i
j
k
∂(y, z)
∂(z, x)
∂(x, y)
∂y
∂x
∂z
N(u, v) = ru × rv = ∂u
=
i+
j+
k.
∂u
∂u
∂y
∂x
∂z
∂
(
u,
v
)
∂
(
u,
v
)
∂
(u, v)
∂v
∂v
∂v
.
.
Proposition. The vector field N is continuous on S, and is normal to the tangent
.plane of the surface S everywhere.
.
Definition. A parametric surface S given by r = r(u, v) is called orientable if
there is a continuous unit normal vector field n on S. A choice of n is called an
orientation
of S, in this case, S is called an oriented surface.
.
For an oriented surface S with an orientation n, one defines the unit normal
vector field n(u, v) on S,(by
)
N
1
∂(z, x)
∂(x, y)
∂(y, z)
n(u, v) =
=
i+
j+
k .
∥N∥
∥N∥ ∂(u, v)
∂(u, v)
∂(u, v)
.
.
.
.
.
.
. Surface Integral of Vector Field
.
Let F = (P, Q, R) be a continuous vector field defined in a domain containing 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
∫∫
∫∫
∂(y, z) ∂(z, x) ∂(x, y)
F · n dS =
(P, Q, R) ·
,
,
du dv
∂(u, v) ∂(u, v) ∂(u, v)
S
D
∫∫
=
.
S
P dydz + Q dzdx + R dxdy.
Remark. If we change the orientation of S, the surface integral will change by a
minus sign.
.
.
.
.
.
.
. Gradient, Curl and Divergence
Let f (x, y, z) and F(x, y, z) = P(x, y, z)i + Q(x, y, z)j + R(x, y, z)k be
differentiable function, and differentiable vector filed defined on a domain D in
Rn .
.
Definition. Gradient vector field of f , denoted by ∇f is given by
.∇f (x, y, z) = ( fx (x, y, z), fy (x, y, z), fz (x, y, z) ), which is a vector field on D.
.
Definition. The curl of F, denoted curl F is defined by curl F(x, y, z)
= ∇×F =
i
∂
∂x
P
j
∂
∂y
Q
k
∂
∂z
R
= (Ry − Qz , Pz − Rx , Qx − Py ), which is a vector field
on
. D.
.
Definition. The divergence of F is defined to be
∂Q
∂R
divF(x, y, z) = ∇ · F(x, y, z) = ∂P
∂x + ∂y + ∂z , which is a scalar function on D.
.
.
.
.
.
.
.
. Summary of Operations on vector fields and functions
.
One can put these three operations
grad(·)
{ Scalar differentiable functions } −→ { Vector Fields }
curl(·)
{ Vector Fields } −→ { Vector fields }, and
div(·)
.
{ Vector Fields } −→ { Scalar differentiable functions },
into a chain
.
grad(·)
.
curl(·)
div(·)
{ Diff. Fun. } −→ { V.Fields } −→ { V.Fields } −→ { Diff. Fun. }.
such that
.
curl ◦ grad(f ) = ∇ × (∇f ) = 0 for any smooth function f ;
div ◦ curl(F) = ∇ · (∇ × F) = 0 for any smooth vector field F,
where smooth means that all the scalar components are at least twice
.continuously differentiable functions.
.
.
.
.
.
.
. Line Integrals of Vector Fields
.
Let F = (P, Q, R) be a continuous vector field defined on a region D, and C be
a piecewise continuously differentiable curve in a domain D, parameterized by
r′ (t)
r(t) = x(t)i + y(t)j + z(t)k for a ≤ t ≤ b, and T(t) = ′
be the unit
∥r (t)∥
.tangent vector of C at r(t).
.
r′ (t )
∥r′ (t)∥ dt
∥r′ (t)∥
dx
dy
dz
= ( P(r(t)) + Q(r(t)) + R(r(t)) ) dt
dt
dt
dt
= Pdx + Qdy + Rdz
Definition. F · T ds = F · dr = (P, Q, R) ·
.
.
Definition. The line integral of F along the curve C is defined to be
∫
.
C
F · T ds =
∫ b
a
( P(r(t))x′ (t) + Q(r(t))y′ (t) + R(r(t))z′ (t) ) dt.
Remark. The line integral is sometimes called the work done of F along the
path C.
.
.
.
.
.
.
.
Fundamental Theorem for Line Integrals
.
Suppose that F = ∇f ,∫for some scalar
function
∫
∫ f , then the line integral (or work
done of F along C) is
C
F · Tds =
C
F · dr =
.where C is parameterized by r(t) (a ≤ t ≤ b).
C
∇f · dr = f (r(b)) − f (r(a)),
Proof. The proof is just a matter of notation and application of fundamental
theorem
of calculus.
)
∫ b
∫
∫ b ( Recall that
∂f dy ∂f dz
d
∂f dx
∇f · Tds =
·
+
·
+
·
dt =
(f (r(t))) dt
∂x dt
∂y dt
∂z dt
a dt
C
a
[
]b
= f (r(t)) = f (r(b)) − f (r(a)).
a
∫
Remark. One should remember that
C
∇f · Tds = f (B) − f (A) where A and B
are the starting point and terminal point of the curve C, as it is independent of
the parametrization of C.
.
.
.
.
.
.
.
Definition. Vector field F defined on a region D is conservative, if there exists a
scalar function f defined on D such that F = ∇f at each point of D. In this
.case, f is called a potential function of F.
Let F = (P, Q, R) = ∇f for some differentiable function f on R3 , where P, Q and
R are functions defined on R3 . Assume that O(0, 0, 0) ∈ D. For any point
E(a, b, c) in D, one can travel from O(0, 0, 0) to A(x, 0, 0) along the x-axis, then
from A(a, 0, 0) to B(a, b, 0), and finally from B(a, b, 0) to E(a, b, c). In this way, we
call the path ∫C. It follows from
the fundamental theorem of line integral and
∫
F = ∇f that
C
F · T ds =
C
∇ · T ds = f (E) − f (O) = f (a, b, c) − f (0, 0, 0).
On the other hand, as the line integral of F is independent of path, it follows
from
that f (a, b,∫c) − f (0, 0, 0)
∫ the definition
∫ of line integral
∫
F · T ds =
=
∫Ca
=
|0
−→ F · T ds + −
→ F · T ds + −
→ F · T ds
OA ∫
AB
BE
∫
c
b
R(a, b, z) dz.
Q(a, y, 0) dy +
P(x, 0, 0)dx +
{z
}
{z
}
|0
{z
}
|0
depends on aonly
depends on b and aonly
(♡).
depends on a,b,c
.
.
.
.
.
.
. Summary of Line Integral of Vector Field
Let F be a vector field defined on a region D. Then we can summary our
important result as follows:
.
F is conservative on D
⇕
∮
F · T ds = 0
.
C
for any closed path in D
⇐⇒
F = ∇f for some function f
⇓(−)xy = (−)yx
D simply connected
∇ × F = 0 on D
⇐=
Stokes Thm
Remark. We have proved all the blue arrows, except the red arrow. One can
−yi+xj
easily checked that F(x, y) = x2 +y2 on D = R2 \ {(0, 0)} satisfies curl F = 0
on D, but there is no function f (x, y) defined on D such that F = ∇f on D.
.
.
.
.
.
.
. Three Important Theorems
.
∫∫ (
)
∂Q ∂P
Green’s Theorem.
P dx + Q dy =
−
dA, where R is the
∂x
∂y
C
R
region bounded by the simple closed curve C, and P, Q are differentiable
functions
on R.
.
∮
.
∫∫
∫∫∫
Divergence Theorem.
F · n dS =
div F dV, where D is the (solid)
S
D
region bounded by the closed surface S with outward unit normal vector field n,
and
F are continuously differentiable vector field on D.
.
.
∫
Stokes’ Theorem.
C
F · T ds =
∫∫
S
curlF · n dS, where the curve C is the
boundary of the surface S in positive orientation, and F are continuously
differentiable
vector field on a domain containing S.
.
.
.
.
.
.
.
.
If S is the part of the paraboloid z = x2 + y2 with z ≤ 2, then the surface area of
.S is
B. 13
C. 13
D. 13
E. 13
A. 13
6 π
3 π
5 π
7 π
9 π
Solution. First note that realizes the surface S can be realized as the graph of
z(x, y) = x2 + y2 , so one can project the curved surface S onto the xy-plane
2
2
and obtain a shadow
√R = { (x, y) | x + y√≤ 2 } in xy-plane. Then the surface
area element dS = 1 + z2x + z2y dx dy = 1 + 4(x2 + y2 ) dx dy.
∫∫
∫∫ √
Surface area of S =
1 dS =
1 + 4x2 + 4y2 dxdy
∫ 2π ∫ √2 √
S
R
∫ √
2√
2π
1 + 4r2 d(1 + 4r2 )
=
r 1 + 4r2 drdθ =
8 0
0
0
√
[
] 2
π (1 + 4r2 )1+1/2
π
2
13
=
= × × (27 − 1) = π.
4
1 + 1/2
4
3
3
0
.
.
.
.
.
.
.
Find
the area of a triangle with vertices P(1, 1, 1), Q(1, 2, 3) and R(2, 3, 1).
.
√
√
√
√
√
A. 17 B. 17/2 C. 21 D. 21/2 E. 11.
Solution. The area of the triangle is given by
√
→ −
→
21
1 −
1
1
2 ∥PQ × PR∥ = 2 ∥(0, 1, 2) × (1, 2, 0)∥ = 2 ∥(−4, 2, −1)∥ = 2 .
.
.
.
.
.
.
.
2 2
2
Let
. F = (yz , x z, xy ), then curl F at (x, y, z) = (1, 2, 3) is
A. (3, −8, 3) B. (3, 8, 3) C. (3, 8, −3)
D. (−3, 8, 3) E. (3, −8, −3).
Solution. Recall that
curl(P, Q, R) = ∇ × (P, Q, R) = (Ry − Qz , Pz − Rx , Qx − Py ), hence
curl F(x, y, z)
= ((xy2 )y − (x2 z)z , (yz2 )z − (xy2 )x , (x2 z)x − (yz2 )y )
= (2xy − x2 , 2yz − y2 , 2xz − z2 ).
Then curl F(1, 2, 3) = (4 − 12 , 12 − 22 , 6 − 32 ) = (3, 8, −3).
.
.
.
.
.
.
.
.Which of the following make sense for a scalar function f ?
1. div(∇f )
2. ∇(div(f ))
3. curl (∇f )
4. ∇(curl(f ))
5. div(curl(f ))
6. curl (div(f )).
A. 1, 3, 5; B. 2, 4, 6; C. 1, 3 ; D. 4, 6 E. 2, 5.
Solution. Recall that ∇f = grad(f ) = fx i + fy j + fz k.
div(P, Q, R) = ∇ · (P, Q, R) = Px + Qy + Rz , and
curl(P, Q, R) = ∇ × (P, Q, R) = (Ry − Qz , Pz − Rx , Qx − Py ).
So curl(f ) and div(f ) don’t make sense, 2, 4, 5 and 6 are not defined.
For (1), div(∇f ) = div(fx , fy , fz ) = fxx + fyy + fyy . For (3), one can easily check
that curl (∇f ) = ((fz )y − (fy )z ), · · · , · · · ) = (0, 0, 0).
.
.
.
.
.
.
.
∫
Evaluate
C
xyds where C = { (x, y) | x2 + y2 = 4, x ≥ 0, y ≥ 0 } oriented
.counterclockwise.
A. 0; B. 1; C. 2; D. 4; E. 8.
Solution. Parameterize the curve C as follows: r(t) = (2 cos t, 2 sin t) where
0 ≤ t ≤ π/2. (In general, check the orientation
of C.) Then
√
√
2
′
′
2
′
2
ds = ∥r (t)∥ dt = x (t) + y (t) dt = 4 sin t + 4 cos2 t = 2dt. and hence
∫
xyds =
∫C π/2
∫ π/2
0
(2 cos t) · (2 sin t) · (2 dt)
sin(2t) d(2t) = 2 [− cos(2t)]0π/2 = 2(1 − (−1)) = 4.
∫
Remark. In fact,
∫ the integral C f (x, y, z)ds does not depend on the orientation
of C, whereas C F · dr will change by a sign if one reverse the orientation of C.
=2
0
.
.
.
.
.
.
.
∫
Evaluate line integral
C
ydx + x2 ydy where C consists of the line segments
.from A(0, 0) to B(1, 0), and from B(1, 0) to C(1, 2).
A. 0; B. 2/3; C. 2; D. 5/6; E. 3.
Solution. Divide the curve C into two segments: AB, and BC; For any point on
AB, its y-coordinates
does not change and y(t) ≡ 0, so dy = y′ (t)dt = 0 · dt, it
∫
2
−
→ (ydx + x ydy) = 0. And similarly, x ≡ 1 on BC,
AB
∫
∫
∫
so −→ (ydx + x2 ydy) = −→ x2 ydy = −→ ydy.
BC ∫
∫BC
∫ BC
ydx + x2 ydy = −→ (ydx + x2 ydy) + −→ (ydx + x2 ydy)
C
AB
BC
[
]
∫
∫ 2
1 2 2
2
= 0 + −→ x ydy =
y dy =
y
= 2.
2
BC
1
0
follows that
.
.
.
so dx = 0, and
.
.
.
.
∫
Evaluate line integral
C
. : r(t) = te
√
C
(y + 1)dx + xdy where
t−1 i + t2 sin( πt/2)j,
where 0 ≤ t ≤ 1.
A. 0; B. 1; C. e; D. 2; E. −e + 1.
Solution. It is not obvious to evaluate the integral by means of parametrization,
so we resort another method. Let f (x, y) = xy + x, then ∇f (x, y) = (y + 1, x). It
follows
theorem of line integral that
∫ from the fundamental
∫
(y + 1)dx + xdy = ∇(xy + y) · Tds
C
]1
= f (r(t))
= f (r(1)) = f (r(0)) = f (1, 1) − f (0, 0) = 1 · 1 + 1 − 0 = 2.
[
C
t=0
.
Remark. The method of rewriting the given vector field F = ∇f does not work
in general, as vector field in general not equal to the gradient vector field. For
example F(x, y) = (y, −x). If there exists (fx , fy ) = ∇f = F = (y, −x), so
fxy = (fx )y = (y)y = 1, and fyx = (fy )x = (−x)x = −1. However, as both mixed
derivatives are continuous, so they should be equal, and this contradicts to the
.result above.
.
.
.
.
.
.
.
2
2
If C goes from (1,
∫ 0) counterclockwise once around the circle x + y = 1, then
the line integral
.
C
(x2 + 2xy)dx + (x2 + 2x + y)dy =
A. 4π; B. −4π; C. −2π; D. 0; E. 2π.
Solution. Let R be the region
bounded by ∫the
∫
∫ closed curve C on xy-plane. One
can use Green theorem
C
Pdx + Qdy =
∫
integral to double integral as follows:
∫∫
C
=
R
R
(Qx − Py )dxdyto turn the line
(x2 + 2xy)dx + (x2 + 2x + y)dy
(x2 + 2x + y)x − (x2 + 2xy)y dxdy =
∫∫
R
(2) dxdy = 2 · π × 12 = 2π.
.
.
.
.
.
.
.
Let S : r(u, v) = u cos vi + u sin vj + vk be parametric surface, where
0. ≤ u ≤ π, 0 ≤ v ≤ π. Find the area of S.
√
A. 32 (1 + π 2 )3/2 ; B. 13 ((1 + π 2 )3/2 − 1); C. 12 ( 1 + π 2 − 1); D.
1
2 3/2 − 1); E. None of the above.
5 ((1 + π )
Solution. First ru (u, v) = (cos v, sin v, 0) and rv (u, v) = (−u sin v, u cos v, 1), and
ru × rv (u, v) = (cos v, sin v, 0) × (−u sin v, u cos v, 1)√= (sin v, − cos v, u). Then
2
dS = ∥ru × r∫v∫∥ du dv = ∫∥(sin
∫ v, − cos v, u)∥ du dv = 1 + u du dv. The surface
π
π
area of S is
1 dS =
∥ru × rv ∥ du dv
0
0∫
∫ π∫ π√ S
π√
=
1 + u2 du dv = π
1 + u2 du
0[ 0
0
]π
√
u√
1
=π
1 + u2 + ln |u + 1 + u2 |
2
2
0
√
π2 √
π
2
2
=
1 + π + ln(π + 1 + π ).
2
2
.
.
.
.
.
.
.
If F(x, y, z) = P(x, y, z)i + Q(x, y, z)j + R(x, y, z)k is defined everywhere in R3
.and curl F = 0, then which of the following are true?
A. divF = 0; B. ∇(divF) = 0; C. F = ∇f for some f ;
D. both A and C; E. both B and C.
Solution. As F satisfies curlF = (0, 0, 0), and its domain is the entire space R3 ,
which is simply connected, so there is a potential function f defined on R3
such that F = ∇f . Hence (C) holds.
Observe that divF = div(fx , fy , fz ) = fxx + fyy + fzz which is not necessarily 0,
and
∇(divF) = ∇(div(fx , fy , fz )) = ∇(fxx + fyy + fzz ) = (fxxx + fxxy + · · · , · · · , · · · )
which is not necessarily 0. Hence both (A) and (B) don’t hold.
.
.
.
.
.
.
.
∫∫
Evaluate
S
.octant.
z dS, where S is part of the plane 2x + 2y + z = 4 in the first
√
A. 8/3; B. 6; C. 8; D. -8/3; E. 16 2/3.
Solution. Rewrite the given equation of the plane in graph form:
z = z(x, y) = 4 − 2x − 2y, so
S = { (x, y, z) | x, y ≥ 0 and z = 4 − 2x − 2y ≥ 0 }
= { (x, y, 4 − 2x − 2y) | 0 ≤ x ≤ 2, 0 ≤ y ≤ 2 − x }. Then
∫∫
∫ 2 ∫ 2−x √
z dS =
z 1 + z2x + z2y dy dx
0
0
∫ 2 ∫S 2−x
√
=
(4 − 2x − 2y) 1 + (−2)2 + (−2)2 dy dx
∫ 0 2 ∫ 0 2−x √
∫ 2[
] 2−x
=
9(4 − 2x − 2y) dy dx = 3
4y − 2xy − y2
0∫
2
0
0
0
(4(2 − x) − 2x(2 − x) − (2 − x) ) dx
[
]2
∫ 2
x3
= 3 (4 + 2x + x2 ) dx = 3 4x + x2 +
= 36 + 8 = 44.
3 0
0
=3
2
0
.
.
.
.
.
.
.
Let F(x, y, z) = xi ∫+∫ yj + zk, and R be the surface z = 1 − x2 − y2 with z ≥ 0.
Evaluate the flux
.
R
F · ndS, where n points upward.
√
A. π; B. 2π; C. 4π/3; D. 5π/3; 3π/2. 16 2/3.
Solution.
S : z = 1 − x2 − y2 , z ≥ 0 gives x2 + y2 ≤ 1, and
√ The graph surface
√
dS = 1 + z2x + z2y = 1 + (−2x)2 + (−2y)2 dxdy. Rewrite S as level surface
z + x2 + y2 = 1, and hence ∇(z + x2 + y2 ) = (2x, 2y, 1) is the normal vector
∫∫
∫∫
√
(x, y, z) · (2x, 2y, 1)
√
field on S. Then
F · ndS =
1 + 4(x2 + y2 )dxdy
2 + y2 )
S
1
+
4
(
x
2
2
∫∫
2
2
x +y ≤1
∫ 2π ∫ 1
(2x + 2y + z)dxdy =
=
x2 + y2 ≤ 1
0
0
(1 + r2 )r drdθ
[
]1
= 2π r2 /2 + r3 /3 = 5π/3.
0
.
.
.
.
.
.
.
∫∫
Evaluate the flux
sphere
.
x2
+ y2
+ z2
R
F · ndS, where F(x, y, z) = xzi − 2yj + 3zk, and R is the
= 4 with normal pointing outward.
A. 32π/3; B. 44π/3; C. 12π; D. 11π; E. 10π.
Solution.
It follows
from Divergence theorem
that
∫∫∫
∫∫∫
∫∫
R
F · ndS =
∫∫∫
1 dV =
=
x2 +y2 +z2 ≤22
(z − 2 + 3) dV
divF dV =
x2 +y2 +z2 ≤4
x2 +y2 +z2 ≤4
4π × 23
32π
=
.
3
3
.
.
.
.
.
.