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Final Exam Review Pack
1
Lecture 23: Vector Fields
Summary of Lecture
1. A vector field is a vector function of position. That is, a vector field is a function from R2 (2 dimensional)
or R3 (3 dimensional) that assigns each point a vector hf (x, y), g(x, y)i (in 3 dimensions it takes the form
hf (x, y, z), g(x, y, z), h(x, y, z)i). We call the functions f, g, h the component functions of the vector field.
2. The domain of the vector field is the intersection of the domains of component functions. A vector field is
continuous/differentiable if and only if all the components are continuous/differentiable. Vector fields are
used in many applications in physics (gravitational, electricity, velocity fields) and other fields. In 2 and 3
dimensional setting, vector fields can be visualized as a set of arrows in 2 or 3 space.
3. A radial vector field has the form F = C
and r = hx, y, zi in 3 dimensions.
r
where p and r are constants. Here r = hx, yi in 2 dimensions
krkp
4. A potential function for a vector field F is a scalar function φ such that F = ∇φ.
5. A vector field F is said to be a conservative vector field if F has a potential function φ
6. A vector field F = hF1 , F2 , F3 i satisfies the cross partial condition (equivalently, irrotational ) if
∂F2
∂F1
=
∂x
∂y
∂F3
∂F2
=
∂y
∂z
∂F1
∂F3
=
∂z
∂x
7. All conservative vector fields satisfy the cross partial condition. But the converse is not true. That is, there
are non conservative vector fields that satisfy the cross partial condition.
8. If the domain of the vector field F is simply connected, then F is conservative if and only if it satisfies the
cross partial condition.
1
Exercises
1. Is the vector field F = hyz, xz, yi conservative?
2. Find a potential function to the vector field F = hx, yi
3. Find a potential function to the vector field F = hyexy, xexy i
4. Find a potential function to the vector field F = hyz 2 , xz 2 , 2xyzi
2
2
5. Find a potential function to the vector field F = h2xzex , 0, ex i
6. Find a potential function φ(x, y) for F =
r
r
and a potential function ψ(x, y, z) G =
in R3 .
krk3
krk4
7. Which of (A) or (B) in the following figure is the contour plot of a potential function for the vector field F.
Recall that the gradient vectors are orthogonal to the level curves.
Answers
1. Not conservative.
2. φ(x, y) =
1 2 1 2
x + y
2
2
3. φ(x, y) = exy
4. φ(x, y, z) = xyz 2
5. φ(x, y, z) = zex
6. φ = −
2
1
1
and ψ = −
2
2krk
3krk3
7. (B)
2
2
Lecture 24: Line integrals of scalar functions and vector fields
Summary of Lecture
1. The scalar line integral of a scalar function f (x, y, z) over a curve C is defined by
Z
f (x, y, z) ds = lim
∆si →0
C
N
X
f (xi , yi , zi )∆si
i=1
where s is the arc length.
2. the computational formula for the scalar line integral of a function f (x, y, z) on a smooth curve C which has
the parametric equation r(t) = hx(t), y(t), z(t)i, t ∈ [α, β] is given by
Z
Z
f (x, y, z) ds =
C
β
f (x(t), y(t), z(t))kr0 (t)k dt
α
R
3. The vector line integral of a vector field F on a smooth curve C is defined by C F · T ds. This sums the
component of F with respect to the unit tangent vector T of the curve C. This is also called the circulation
of F around C.
4. The computational formula for the vector line integral of a vector field F = hF1 , F2 , F3 i on a smooth curve C
which has the parametric equation r(t) = hx(t), y(t), z(t)i, t ∈ [α, β] is given by
Z
C
F · T ds =
Z
β
α
F(x(t), y(t), z(t)) · r0 (t) dt
5. There are various notations for the vector line integral of a vector field F = hf, g, hi over a smooth curve C.
Z
C
F · T ds =
Z
β
α
0
F(r(t)) · r (t) dt =
Z
β
0
0
0
(f x (t) + gy (t) + hz (t)) dt =
α
Z
C
f dx + g dy + h dz =
Z
C
F · dr
6. Properties of vector line integrals: Let C be a smooth oriented curve and F and G are vector fields, then
R
R
R
(a) C (F+G) dr = C (F) dr + C (G) dr (linearity)
R
R
(b) If −C denotes C with reverse orientation, then −C (F) dr = − C (F) dr.
R
R
R
(c) If C is a union of piecewise smooth curves C1 and C2 , then C (F) dr = C1 (F) dr + C2 (F) dr. This can
be extended to any number of pieces.
7. If F represents a force field, the work done moving an object from a point a to b along a path r(t) is given by
Z b
F · T ds.
a
3
Exercises
1. Let f (x, y, z) = x + yz and let C be the line segment from P = (0, 0, 0) to (6, 2, 2).
(a) Calculate f (c(t)) and ds = kc0 (t)kdt for the parametrization c(t) = h6t, 2t, 2ti for 0 ≤ t ≤ 1.
Z
(b) Evaluate
f (x, y, z) ds.
C
2. Let F = hy 2 , x2 i and let C be the curve y = x−1 for 1 ≤ x ≤ 2 oriented from left to right.
(a) Calculate F(c(t)) and ds = c0 (t) dt for the parametrization of C given by c(t) = ht, t−1 i.
Z
0
(b) Calculate the dot product F(c(t)) · c (t) and evaluate
F · dS.
C
2
3. Compute the integral of the scalar function f (x, y, z) = x + y 2 + z 2 on the curve r(t) = hcos(t), sin(t), ti for
t ∈ [0, π]
4. Compute the line integral of the vector field F = hx, y, zi on the curve r(t) = hcos(t), sin(t), ti for t ∈ [0, π].
√
5. Compute the integral of the scalar function f (x, y) = 1 + 9xy on the curve y = x2 for 0 ≤ x ≤ 1.
Z
y3
1
6. Compute the line integral
f ds of the function f (x, y) = 7 on the curve y = s4 for 1 ≤ x ≤ 2.
x
4
C
Z
2
7. Compute the line integral
f ds of the function f (x, y, z) = xez on the piecewise linear path from (0, 0, 1)
C
to (0, 2, 0) to (1, 1, 1).
8. Compute the line integral
0≤t≤1
9. Calculate
Z
C
Z
C
√
f ds of the function f (x, y, z) = x2 z on the curve given by r(t) = het , 2t, e−t i,
√
1ds, where the curve C is parametrized by r(t) = het , 2t, e−t i for t ∈ [0, 2].
10. Compute the vector line integral
(2, 2).
11. Compute the vector line integral
Z
C
F · ds of the vector field F = hx2 , xyi on the line segment from (0, 0) to
Z
C
F · ds of the vector field F = h4, yi on the quarter circle x2 + y 2 = 1 with
x ≤ 0 and y ≤ 0 oriented counterclockwise.
Z
12. Compute the vector line integral
F · ds of the vector field F = hx2 , xyi on the part of the circle x2 + y 2 = 9
C
with x ≤ 0, y ≥ 0 oriented clockwise.
Z
13. Evaluate the integral
y dx + z dy + x dz on the curve parametrized by c(t) = h2 + t−1 , t3 , t2 ) for o ≤ t ≤ 1.
C
14. Calculate the line integral of F = hez , ex−y , ey iover the blue path from P to Q on the figure below.
15. Calculate the work done by the force field F = hx, y, zi along the path r(t) = hcos(t), sin(t), ti from 0 ≤ t ≤ 3π.
4
Answers
√
1. (a) ds = 2 11dt
√
26 11
(b)
3
2. (a) ds = h1, −t−2 idt
1
(b) −
2
√
3. 2(π + π 3 /3)
4.
π4
4
5.
14
5
1
(653/2 − 23/2 )
576
√
3
7.
(e − 1)
2
6.
8.
1 2
(e + 1)
2
9. e2 − e−2
10.
16
3
11. 4.5
12. 0
13.
41
10
14. 2 − e−1 − e
15.
9π 2
2
5
3
Lecture 25: The fundamental theorem of vector line integrals
Summary of Lecture
1. A curve parametrized by r(t), t ∈ [a, b] is called closed if r(a) = r(b). (intersects at endpoints). A curve is
called simple if it intersects itself only at the endpoints. The following figure illustrates a simple closed curve,
a simple non-closed curve, a non-simple, non-closed curve and a non-simple closed curve, respectively
2. a region is open if it consists only of interior points (that is, it does not contain its boundary points.)
3. a region (in R2 or R3 ) is path connected if any two points can be connected by a continuous curve lying in R.
4. A path connected region R (in R2 or R3 ) is simply connected if every closed curve in R encloses only points
lying in R. (in other words, every closed curve in R can be “shrunk to a point” within R).
R, R2 , R3 , open/closed balls and R3 \ {0, 0, 0} are all simply connected. R2 \ {0, 0} and annular regions in 2
dimensions are not simply connected because they have “2 dimensional holes”.
I
5. The circulation of a vector field F along a smooth, closed curve C is denoted by the integral
F · T ds
C
6. Fundamental Theorem of vector line integrals If a vector field F is conservative and φ is the potential
function (so that F = ∇φ), then the line integral of F along any path C from a point a to b is given by
Z
I
F · dr = φ(b) − φ(a) . In particular, if C is a closed curve, then
F · dr = 0.
C
C
7. If the domain of a vector field F is simply connected, then F is conservative if and only if F is irrotational.
8. To determine the potential function for a conservative vector field F = hF1 , F2 , F3 i partially integrate F1 with
respect to x, F2 with respect to y, F3 with respect to z and compare.
6
Exercises
1. Evaluate the line integral of F = h3, 6yi over the path given by r(t) = ht, 2t−1 i for 1 ≤ t ≤ 4.
2. Evaluate the line integral of F = hyez , xez , xyez i over the curve c(t) = ht2 , t3 , t − 1i for 1 ≤ t ≤ 2.
3. Find a potential function for F = hz, 1, xi.
4. Find a potential function for F = hx, yi or determine it’s not conservative.
5. Find a potential function for F = y 2 i + (2xy + ez )j + yez k or determine it is not conservative
6. Find a potential function for F = hcos(xz), sin(yz), xy sin(z)i or determine it is not conservative.
7. Find a potential function for F = hyzexy , xzexy − z, exy − yi.
R
2
8. Evaluate C 2xyz dx + x2 z dy + x2 y dz over the path c(t) = ht2 , sin(πt/4), et −2t i for 0 ≤ t ≤ 2.
H
9. Evaluate C sin(x) dx + z cos(y) dy + sin(y) dz Where C is the ellipse 4x2 + 9y 2 = 36, oriented clockwise.
1
1
10. Let F =
,−
. Calculate the work against F required to move an object from (1, 1) to (3, 4) along any
x y
path in the first quadrant.
x
y
11. The vector field F =
,
is defined on the domain D = {(x, y)| x, y 6= 0}.
x2 + y 2 x2 + y 2
(a) Is D simply connected?
(b) Show that F satisfies the cross partial condition. Does this guarantee that F is conservative?
(c) Show that F is conservative by finding a potential function.
(d) Does these results contradict the highlighted statement 7 in Lecture 23 summary?
x
−y
,
is defined on the domain D = {(x, y)| x, y 6= 0}.
12. The vector field F =
x2 + y 2 x2 + y 2
(a) Show that F satisfies the cross partial condition. Does this guarantee that F is conservative?
(b) Is F conservative?
(c) Does these results contradict the highlighted statement 7 in Lecture 23 summary?
7
Answers
1. −
9
4
2. 32e − 1
3. V (x, y, z) = xz + y
4. Not Conservative
5. V (x, y, z) = y 2 z + ez y
6. Not Conservative
7. V (x, y, z) = zexy − yz
8. 16
9. 0
10. ln 4 − ln 3
11. (a) no
(b) can easily verify
1
(c) V (x, y) = ln(x2 + y 2 )
2
(d) No, conservative vector fields can be defined in non simply connected regions.
12. (a) Can easily verify
(b) No
(c) No, because the domain is not simply connected.
8
4
Lecture 26: Surface parametrization and scalar surface integrals
Summary of Lecture
1. A surface can be parametrized by using 2 parameters (variables). A general equation of a surface is r(u, v) = hf (u, v), g(u, v),
where f, g and h are functions of u and v.
2. If a surface is parametrized by r(u, v), the vector ru × rv is a normal vector to the surface at any given point.
Also the vector −ru × rv is another normal. An orientation of the surface is a choice of one normal vector
consistent with the given problem.
3. Some important examples for surfaces are
Surface
Sphere (of radius R)
Cylinder (height H and radius R
Right circular cone (height H)
Helicoid (height H)
Any graph of explicitly defined function z =
k(x, y)
Parametrization
r(φ, θ) = hR cos(θ) sin(φ), R sin(θ) sin(φ), R cos(φ)i
where θ ∈ [0, 2π] and φ ∈ [0, π]
r(θ, z) = hR cos(θ), R sin(θ), zi where θ ∈ [0, 2π],
and z ∈ [0, H]
r(u, v) = hu cos(v), u sin(v), ui where u ∈ [0, 1] and
v ∈ [0, 2π]
r(u, v) = hu cos(v), u sin(v), viwhere u ∈ [−1, 1]
and v ∈ [0, 2π]
r(x, y) = hx, y, k(x, y)i
From the last example, the parametric equation of a paraboloid is r(x, y) = hx, y, x2 + y 2 i, and that of a
parabolic hyperboloid is r(x, y) = hx, y, x2 − y 2 i
4. The surface integral of a scalar function f (x, y, z) on a surface S is defined by the limit
x
f (x, y, z) ds =
S
lim
∆u,∆v→0
n X
m
X
i=1 j=1
f (r(Pij )kru × rv k∆u∆v
5. The computational formula for evaluating the surface integral of a scalar function f (x, y, z) over a surface
r(u, v) on a region R is given by
x
x
f (x, y, z) ds =
f (r(u, v))kru × rv k du dv
S
R
Note that the right hand side is a double integral over a region in the uv-plane.
x
6. The surface area of the surface r(u, v) on a region R is given by
1kru × rv k du dv
R
7. If the surface is explicitly defined in the form of z = k(x, y), then kru × rv k =
q
1 + kx2 + ky2 and the scalar
q
x
x
surface integral can be computed easily by the formula
f (x, y, z) ds =
f (k(x, y)) 1 + kx2 + ky2 dx dy
S
9
R
Exercises
1. Let G(x, y) = hx, y, xyi
(a) Calculate Gx , Gy and n(x, y)
(b) Let S be the part of the surface with parameter domain D = {(x, y) : x2 + y 2 ≤ 1, x ≥ 0, y ≥ 0}, verify
the following formula and evaluate using polar coordinates.
x
xp
1dS =
1 + x2 + y 2 dx dy
S
D
(c) Verify the following formula and evaluate
x
S
z ds =
Z
0
π/2
Z
1
(sin(θ) cos(θ))r3
0
p
1 + r2 dr dθ
2. Find the normal vector to the surface G(u, v) = h2u + v, u − 4v, 3ui at the point u = 1, v = 4. Also find the
equation of the tangent plane to G at that point.
3. Find the normal vector to the surface G(r, θ) = hr cos(θ), r sin(θ), 1 − r2 i at the point r = 1/2, θ = π/4. Also
find the equation of the tangent plane to G at that point.
s
4. Calculate the scalar surface integral S f (x, y, z) dS for the function f (x, y, z) = x(x2 + y 2 ) on the surface
G(u, v) = hu cos(v), u sin(v), ui where 0 ≤ u ≤ 1, 0 ≤ v ≤ 1
p
s
5. Calculate the scalar surface integral S f (x, y, z) dS for the function f (x, y, z) = x2 + y 2 on the surface
G(r, θ) = hr cos(θ), r sin(θ), θi where 0 ≤ r ≤ 1, 0 ≤ θ ≤ 2π
s
6. Calculate the scalar surface integral S f (x, y, z) dS for the function f (x, y, z) = 1 on the surface y = 9 − z 2
where 0 ≤ x ≤ z ≤ 3
s
7. Calculate the scalar surface integral S f (x, y, z) dS for the function f (x, y, z) = z on the part of the plane
x + y + z = 1, where x, y, z ≥ 0
s
8. Calculate the scalar surface integral S f (x, y, z) dS for the function f (x, y, z) = z 2 on part of the plane
x + y + z = 0 contained in the cylinder x2 + y 2 = 1
s
9. Calculate the scalar surface integral S f (x, y, z) dS for the function f (x, y, z) = z on the part of the surface
z = x3 where 0 ≤ x ≤ 1, 0 ≤ y ≤ 1
s
10. Calculate the scalar surface integral S f (x, y, z) dS for the function f (x, y, z) = xy +ez on the triangle shown
below.
11. Find the surface area of the portion S if the cone z 2 = x2 + y 2 where z ≥ 0, contained within the cylinder
y 2 + z 2 ≤ 1.
10
Answers
1. (a) Gx = h1, 0, yi, Gy = h0, 1, xi n(x, y) = h−y, −x, 1i
√
(2 2 − 1)π
(b)
6
√
2+1
(c)
15
2. h12, 3, −9i Plane 4x + y − 3z = 0
√
√
3. h2r2 cos θ, 2r2 sin θ, ri Plane 2 2x + 2 2y + 4z = 5
√
2
4.
5
√
2π(2 2 − 1
5.
3
√
37 37 − 1
6.
12
√
3
7.
6
√
π 3
8.
2
√
1
9.
(10 10 − 1)
54
10. 3e3 − 6e2 + 3e + 1
11. π
11
5
Lecture 27: Surface integrals of vector fields
Summary of Lecture
1. The “amount” of the vector field passing through the surface is called the flux of the vector field across the
surface. For a vector field F defined on a surfec S having the normal n, the flux is calculated by summing up
(integrating) the component of F parallel to n at each point of the surface S.
2. When choosing the orientation for the surface S having the parametric equation r(u, v) (you have two choices
to start with, ru × rv or −ru × rv ), follow these steps;
(a) See whether the orientation is specified in the problem, if so, pick that orientation.
(b) If the surface is a closed surface, pick the orientation so that the normal vector points outward from the
surface.
(c) If the surface is not closed, pick the orientation so that the 3rd component of the normal vector is positive.
In all the formulas given below, a star
∗
indicates “assuming correct orientation”.
3. The flux of a vector field F on a surface S is denoted by the integral
x
x
F · n ds =
F · dS
S
4. For a surface S having the parametric equation r(u, v) where u ∈ [a, b] and v ∈ [c, d], the computational
formula for the flux over S is given by
x
S
F · n ds =
Z
d
v=c
Z
b
u=a
F(r(u, v)) · (ru × rv )∗ du dv
5. For an explicitly defined surface z = k(x, y) x ∈ [a, b], y ∈ [c, d], the normal vector is h−kx , −ky , 1i (for the
orientation with z component positive) and the flux can be computed by the easier formula
x
S
F · n ds =
Z
d
y=c
Z
b
x=a
F(x, y, k(x, y) · h−kx , −ky , 1i dx dy
6. Some applications of the flux are
(a) Outward flux of heat
(b) Flow rate of a fluid across a surface.
12
Exercises
1. Compute the flux
x
S
F · dS for the function F = hy, z, xi on the plane 3x − 4y + z = 1 with upward pointing
normal , where x, y ∈ [0, 1]
x
2. Compute the flux
F · dS for the function F = hez , z, xi on the surface parametrized by G(r, s) = hrs, r+s, ri
S
oriented by Gr × Gs , where r, s ∈ [0, 1]
x
3. Compute the flux
F · dS for the function F = h0, 3, xi on the part of the sphere of radius 3 on the first
S
octant with outward pointing normal.
x
4. Compute the flux
F · dS for the function F = hx, y, zi on the part of the unit sphere with outward pointing
S
√
normal, where 1/2 ≤ z ≤ 3/2
x
5. Compute the flux
F · dS for the function F = hz, z, xi on the surface z = 9 − x2 − y 2 with upward pointing
S
normal, where x, y, z ≥ 0
x
F · dS for the function F = hsin(y), sin(z), yzi on the rectangle 0 ≤ y ≤ 2, 0 ≤ z ≤ 3 in
6. Compute the flux
S
the yz plane, normal pointing in the negative x direction.
x
7. Compute the flux
F · dS for the function F = hy 2 , 2, −xi on the portion of the plane x + y + z = 1 with
S
upward pointing normal in the first octant.
x
8. Compute the flux
F · dS for the function F = hx, y, ez i on the cylinder x2 + y 2 = 4 with outward pointing
S
normal , where 1 ≤ z ≤ 5
x
9. Compute the flux
F · dS for the function F = hxz, yz, z −1 i on the disk of radius 3 at height 4 parallel to
S
the xy plane, with upward pointing normal.
x
10. Compute the flux
F · dS for the function F = hxy, y, 0i on the cone z 2 = x2 + y 2 , where x2 + y 2 ≤ 4, z ≥ 0
S
with downward pointing normal
x
11. Compute the flux
F · dS for the function F = h0, 0, ey+z i on the boundary of the unit cube 0 ≤ x ≤ 1, 0 ≤
y ≤ 1, 0 ≤ z ≤ 1.
12. Compute the flux
S
x
S
F · dS for the function F = h0, 0, z 2 i on the helicoid parametrized by G(u, v) =
hu cos(v), u sin(v), vi , where 0 ≤ u ≤ 1 and 0 ≤ v ≤ 2π with upward pointing normal.
13
Answers
1. −4
2. 4/3 − e
27
(3π + 4)
12
√
4. π(1 − 3)
3.
5. 312.25
6. 3(cos 2 − 1)
7.
11
12
8. 32π
9.
9π
4
10.
8π
3
11. (e − 1)2
12.
4π 3
3
14
6
Lecture 28: Green’s Theorem
Summary of Lecture
1. The Green’s Theorem is used in two situations,
(a) To evaluate a vector line integral (circulation) enclosing a planar region of the xy plane, by converting
it to a double integral.
(b) To evaluate the area of an irregular region on the xy-plane by using a vector line integral.
2. Green’s Theorem: If ∂D is a smooth closed curve oriented counterclockwise that encloses a connected, simply
connected region D on the plane, and F = hf, gi is a vector field where g and h are continuously differentiable
on D, then
I
x
F · dr =
(gx − fy ) dA
∂D
D
Note: Green’s theorem is valid even if ∂D is a piecewise smooth curve. (eg: a triangle). 3 dimensional version
(generalization) of the Green’s Theorem is called Stoke’s Theorem.
3. A 2 dimensional vector field F = hf, gi is called irrotational if gx − fy = 0
4. If F = hf, gi is any vector field such that gx − fy = 1, then the area of the region D with a boundary ∂D is
I
F · dr .
given by A =
∂D
There are many such vector fields F, (eg: F = h0, xi, F = h−y, 0i, F = h−y/2, x/2i)
Exercises
For the following exercises (1) - (8), use Green’s theorem to evaluate the line integral. Orient the curve counterclockwise unless otherwise indicated.
H
1. C y 2 dx + x2 dy where C is the boundary of the unit square x, y ∈ [0, 1]
H
2. C e2x+y dx + e−y dy where C is the triangle with vertices (0, 0), (1, 0) and (1, 1).
H
3. C x2 ydx where C is the unit circle centered at the origin.
H
√
4. C F · dS where F = hx + y, x2 − yi where C is the boundary of the region enclosed by y = x2 and y = x for
0≤x≤1
H
5. C F · dS where F = hx2 , x2 i where C consists of the arcs y = x2 and y = x for 0 ≤ x ≤ 1.
H
6. C (ln(x) + y) dx − x2 dy where C is the rectangle with vertices (1, 1), (3, 1) (1, 4) and (3, 4)
H
7. C F · dS for F = hex+y , ex−y i where C is the curve consisting of line segments joining the points (2, 2), (4, 2),
(2, 0) and back to (0, 0), oriented clockwise
H
8. C xy dx( x2 + x) dy where C is the triangle with vertices (−1, 0), (1, 0), (0, 1) oriented clockwise.
9. Use Green’s Theorem to calculate the area of a circle of radius 3 centered at the origin.
10. Use Green’s Theorem to calculate the area of the triangle with vertices (0, 0), (1, 0) and (1, 1)
11. Use Green’s Theorem to calculate the area of the region between the x-axis and the cycloid parametrized by
c(t) = ht − sin(t), 1 − cos(t)i (cycloid) for 0 ≤ t ≤ 2π
12. Use Green’s Theorem to calculate the area of the region between the graph of y = x2 and the x-axis for
0 ≤ x ≤ 2.
15
13. Let x3 + y 3 = 3xy be the follum of Descartes
(a) Show that the follum has a parametrization in terms of t = y/x given by x =
t ∈ (−∞, ∞), t 6== −1
(b) Show that x dy − y dx =
9t2
dt
(1 + t3 )2
(c) Find the area of the loop of the follum
Answers
1. 0
2. e2 /2 − e3 /3 − 1/6
3. −π/4
4. −1/30
5. 1/6
6. −30
7.
(e2 − 1)(5 − e4 )
2
8. 1
9. 9π
10. 1/2
11. 3π
12. 8/3
13. (c) : 3/2
16
3t2
3t
, y =
3
1+t
1 + t3
7
Lecture 29: Stoke’s Theorem
Summary of Lecture
1. The Curl of a vector field F = hF1 , F2 , F3 i is given by the symbolic cross product:
i
∂
CurlF = ∇ × F = ∂x
F
1
The symbol ∇ =
∂ ∂ ∂
,
,
∂x ∂y ∂z
j
∂
∂y
F2
k
∂ ∂z F3 is called the del operator. A vector field F is called irrotational if ∇ × F = 0
2. Conservative vector fields are irrotational, but not all irrotational vector fields are conservative. But an
irrotational vector field defined on a simply connected domain is always conservative.
3. The Stoke’s Theorem is used mainly in 2 cases
(a) To evaluate the flux of the curl of a certain vector field by converting it to a vector line integral.
(b) To evaluate a vector line integral (usually a union of lines having different parametrizations, like a triangle) as a single vector surface integral.
Also, if we can find a vector potential G to a vector field F (so that F = ∇ × G), then we can evaluate
the flux of F across any surface as a line integral using the Stoke’s theorem.
(c) Stoke’s Thoerem Let S be a smooth oriented surface in R3 , with a (piecewise) smooth closed boundary
curve ∂S whose orientation is consistent with that of S. Assume that F = hF1 , F2 , F3 i is a vector fields
whose componets are continuously differentiable on S, then
I
x
(∇ × F) · n ds
Fdr =
∂S
In particular, if S is a closed surface, then
S
s
S
(∇ × F) · n ds = 0
Exercises
1. Find ∇ × F for F = z − y 2 , x + z 3 , y + x2
Dy y z E
2. Find ∇ × F for F =
, ,
x z x
3. Find ∇ × F for F = hey , sin(x), cos(x)i
x
y
4. Find ∇ × F for F =
,
,0
x2 + y 2 x2 + y 2
2
2
5. Compute the flux of curl(F) of the vector field F = hez − y, ez + x, cos(xz)i on the upper hemesphere using
Stoke’s Theorem.
17
p
6. Compute the flux of curl(F) of the vector field F = hx + y, z 2 − 4, x y 2 + 1i over the surface of the wedge
shaped box in the following figure with outward pointing normal.
2
7. Let I be the flux of F = hey , 2xex , z 2 i through the upper hemesphere S of the unit sphere.
2
(a) Let G = hey , 2xex , 0i. Find a vector field A such that ∇ × A = G
(b) Use Stoke’s Theorem to show that the flux of G through S is zero. (Hint: Calculate the circulation of A
around ∂S)
(c) Calculate I. (Hint: Use (b) to show thatI is equal to the flux of (0, 0, z 2 ) through S.
8. Let F = h−xR2 y, x, 0i. Referring to the following figure, let C be the closed curve ABCD. Use Stoke’s Theorem
to evaluate C F · dS in two ways. First, regard C as the boundary of the rectangle with vertices A, B, C, and
D. Then treat C as the boundary of the wedge shaped box with open top.
9. Let F = h−z 2 , 2zx, 4y − x2 i and let C be a simple
closed curve in the plane x + y + z = 4 that encloses a
I
region of area 16 (see figure below). Calculate
F · ds where C is oriented in the counterclockwise direction
C
when viewed from above the plane.
10. Let F = h0, −z, 1i. Let S be the spherical cap x2 + y 2 + z 2 ≤ 1 where z ≥ 1/2. Evaluate
x
S
F · dS directly as
a surface integral. Then verify that F = ∇ × A, where A = h0, x, xzi and evaluate the surface integral again
using the Stoke’s Theorem.
18
Answers
1. h1 − 3z 2 , 1 − 2x, 1 + 2yi
2. hy/z 2 , z/x2 , −1/xi
3. h0, sin(x), cos(x) − ey i
4. h0, 0, 0i
5. 2π
6. 1/2
7. (a) A = h0, 0, ey − ex i (b) 0 (c)π/2
8. 234
9. 32
10. 3π/4
19
8
Lecture 30: The Divergence Theorem
Summary of Lecture
1. The divergence of a vector field F = hF1 , F2 , F3 i is given by the symbolic dot product
div(F) = ∇ · F =
∂F1
∂F2
∂F3
+
+
∂x
∂y
∂z
This results in a scalar valued function. If for a vector field F, div(F) = 0, we say F is incompressible.
2. For any vector field F with continuous second order partial derivatives, ∇ · (∇ × F) = 0.
However, if for a given vector field G such that div(G) = 0, G = curl(F) for some vector field F only if G is
defined on a simply connected region.
3. The divergence Theorem is used to evaluate a vector surface integral (flux) over a closed, bounded surface S,
by converting it to a triple integral.
4. The divergence Theorem: Let F be a vector field defined on a connected, simply connected region D in R3
enclosed by a smooth oriented surface S. Also assume that the components of F are continuously differentiable.
Then
x
y
F · n ds =
(∇ · F)dV
S
D
Where n is the outward unit normal vector to S.
Exercises
Use the divergence Theorem to evaluate the following surface integrals for the function F given on the surface
S.
1. F = h0, 0, z 3 /3i where S is the sphere x2 + y 2 + z 2 = 1
2. F = hy, z, xi where S is the sphere x2 + y 2 + z 2 = 1
3. F = hx3 , 0, z 3 i S is the first octant of the sphere x2 + y 2 + z 2 = 4
4. F = hex+y , ex+z , ex+y i where S is the boundary of the unit cube x, y, z ∈ [0, 1]
5. F = hx, y 2 , z + yi where S is the boundary of the region contained in the cylinder x2 + y 2 = 4 between the
planes z = x and z = 8
2
6. F = hx2 − z 2 , ez − cos(x), y 3 i where S is the boundary of the region bounded by x + 2y + 4z = 12 and the
coordinate planes in the first octant.
7. F = hx + y, z, z − xi where S is the boundary of the region between the paraboloid z = 9 − x2 − y 2 and the
xy-plane.
p
2
8. F = hez , 2y + sin(x2 z), 4z + x2 + 9y 2 i where S is the region x2 + y 2 ≤ z ≤ 8 − x2 − y 2
9. Calculate the flux of the vector field F = 2xyi − y 2 j + k through the surface S in the following figure.
20
Answers
1.
4π
15
2. 0
3.
128π
5
4. (e − 1)2
5. 64π
6. 216
7. 81π
8. 96π
9. π
21
MAC2313 Final A
(5 pts) 1. Let f (x, y, z) be a function continuous in R3 and let S be a surface parameterized
by r(u, v) with the domain of the parameterization given by R; how many of the following are true?
(Assume that the order of the product du dv is consistent with the limits in the double integral.)
a. The vector ru × rv lies in the tangent plane of S at a given point.
b.
∫∫
S
f(x, y, z) dS =
∫∫
R
f(r(u, v))|ru × rv | du dv.
c. The vectors ru and rv are orthogonal to the tangent plane of S at a given point.
d. The surface area of S is given by the integral
A. 0
B. 1
C. 2
∫∫
R
|ru × rv | du dv.
D. 3
E. 4
(5 pts) 2. A surface S is given by the graph of z = 3x2 + 3y 2 − 5, {(x, y) | x2 + y 2 ≤ 9} and
oriented so that the z component of the unit normal vector is negative; a unit normal vector to the
surface is given by:
A. (36x2 + 36y 2 + 1)−1/2 ⟨6x, 6y, −1⟩
B. (36x2 + 36y 2 + 1)−1/2 ⟨−6x, −6y, −1⟩
C. (36x2 + 36y 2 + 1)−1/2 ⟨−6x, −6y, 1⟩
D. (36x2 + 36y 2 + 1)−1/2 ⟨6x, 6y, 1⟩
E. none of the above
(5 pts) 3. Let S be the part of the surface x + y + z = 1 which lies above the triangular region
of the x, y - plane with vertices (0, 0), (1, 0), and (0, 1) and oriented so that the normal vector has
a positive z component. If F = ⟨2, −3, 1⟩ , the flux of F through S is given by:
A. 0
B. 1/2
C. −1/2
D. −2
E. 2
(5 pts) 4. Let C be the square with vertices (0, 0), (1, 0), (1, 1) and (0, 1) and having a counterclockwise orientation and let F = ⟨y + x2 , ey − x⟩. Which of the following could be used to compute
the circulation of F around C?
A. Bubba’s Theorem
B. Clairaut’s Theorem
C. Green’s Theorem
D. the Divergence Theorem
E. the Fundamental Theorem of Line Integrals
(5 pts) 5. The circulation of F around C in problem 4 has the value:
B. −2
A. 0
to:
D. −6
E. −8
(5 pts) 6. If F = ⟨x2 e3y , e3y cos(πz), z ln x⟩, then the divergence of F at the point (1, 0, 1) is equal
A. 0
to:
C. −4
B. 1
C. 5
D. e3
E. none of the above
(5 pts) 7. If F = ⟨2xy3 , xy + zy2 , z3 (x − 2y)⟩, then the curl of F at the point (1, 2, −1) is equal
A. ⟨−2, 1, −22⟩
B. ⟨2, 1, −22⟩
C. ⟨−2, 1, −22⟩
D. ⟨2, −1, 22⟩
E. ⟨−2, −1, 22⟩
(5 pts) 8. If a curve C is given ∫parametrically by r(t) = ⟨3, 2t, t2 ⟩ , t : 0 →
f (x, y, z) = 6xy, then the value of C f (x, y, z) ds is:
A. 144
B. 146
C. 168
D. 172
√
3 and
E. none of the above
(5 pts) 9. Let F be a conservative vector field in R3 and let C1 and ∫C2 be two simple,
smooth
∫
curves which share the same initial point P and terminal point Q; then C1 F · dr = C2 F · dr.
A. true
B. false
(5 pts) 10. Let F = ⟨x − y, x + y⟩; the circulation of F around the unit circle with counterclockwise orientation is equal to:
A. 0
C. −2π
B. 2π
D. 2π + 1
E. −2π − 1
(5 pts) 11. Let f (x, y, z) be a function continuous in R3 and let S be a surface given as the
graph of the differentiable function z = k(x, y) with domain R; how many of the following are true?
(Assume that the order of the product dy dx is consistent with the limits in the double integral.)
a.
b.
∫∫
S
∫∫
S
f(x, y, z) dS =
∫∫
[f(x, y, z)]2 dS =
R
∫∫
√
f(x, y, k(x, y)) k2x + k2y + 1 dy dx.
R
√
[f(x, y, k(x, y))]2 k2x + k2y + 1 dy dx.
c. If r(x, y) = ⟨x, y, k(x, y)⟩, then rx × ry is normal to the surface.
d. The vector ⟨kx , ky , 1⟩ is orthogonal to the tangent plane of S at a given point.
A. 0
B. 1
C. 2
D. 3
E. 4
(5 pts) 12. Let S be the unit sphere centered at the origin and let F = ⟨−2x, 3y, 5z⟩, then the
flux across the surface is given by:
A. 0
B. 6π
C. 8π
D. 12π
E. 16π
(5 pts) 13. Let S be the part of the cylinder of radius 5 centered on the z axis with x ≤ 0 and
|z| ≤ 1; then S is given parametrically by r(θ, z) = ⟨5 cos θ, 5 sin θ, z⟩ with domain
R = { (θ, z) | π ≤ θ ≤ 2π, −1 ≤ z ≤ 1 }.
A. true
B. false
(5 pts) 14. If curl F = 0, then F is called irrotational.
A. true
B. false
(5 pts) 15. Let F be any continuous vector field given by F = ∇ϕ(x, y); let C be any simple,
smooth, open, oriented curve in R2 , then
∫
A. true
C
F · dr = 0.
B. false
⟨
⟩
(5 pts) 16. If F = ⟨y, x + z2 , 2zy⟩ and C is given parametrically by r(t) = 2, cos3 t, sin3 t ,
∫
t : 0 → π/2, then what is the value of C F · T ds?
A. −3
B. −2
C. −1
D. 0
E. 4
∫∫
√
(5 pts) 17. If f (x, y, z) = z(x2 + y 2 ) and S the part of the cone z = 2 x2 + y 2 , 0 ≤ z ≤ 4, then
S f (x, y, z) dS is equal to which of the following:
√ ∫ ∫
A. 2 5 02π 02 r4 dr dθ
√ ∫ ∫
B. 2 5 02π 02 r3 dr dθ
√ ∫ ∫
D. 5 2 02π 04 r4 dr dθ
√ ∫ ∫
C. 2 5 02π 02 r5 dr dθ
E. none of the above
(5 pts) 18. If F(x, y, z) = ⟨x2 , 3y + x, 2y⟩ and S is a surface given by r(u, v) = ⟨3u, 2u sin v, 0⟩ ,
0 ≤ u∫ ≤
1, 0 ≤ v ≤ π/2 and oriented so that the z component of the normal vector is positive;
∫
then S F(x, y, z) · n̂ dS is equal to:
A. 0
B. 2
C. 4
D. 6
E. 8
(5 pts) ∫19.
Let F be a constant vector field and let S be a smooth, closed surface; then the
∫
statement S F · n̂ dS = 0 is true:
A. sometimes
B. always
C. never
(5 pts) 20. Assuming the limits for the integrals are consistent with the given properties, which
of the following are true:
I. Stoke’s Theorem is given by
∫∫
S
(∇ × F) · n̂ dS =
∫
C
F · T ds.
II. If F is constant and S is a smooth, bounded, closed surface, then
III. If F is constant and C is a smooth, simple closed curve, then
IV. The Divergence Theorem is given by
A. only I
B. only III
∫∫
C. only I and II
S
∇ · F dS =
∫∫∫
D
∫
C
∫∫
S
(∇ × F) · n̂ dS = 0.
F · dr = 0.
F · n̂ dV.
D. only I, II, and IV
E. I, II, and III
Bonus (5 pts) 21. Let S be the part of the unit sphere with z ≥ 0 oriented ∫so∫ that the z component of the unit normal vector is positive; if F = ⟨−y/2, x/2, z2 (x + y)⟩, then S (∇ × F) · n̂ dS
is equal to: (Don’t compute this integral directly.)
A. 0
B. −π
C. π
D. −2π
Bonus (5 pts) 22. Let F be a radial vector field given by
A. (1/3)(x2 + y 2 + z 2 )−3/2
r
; a potential function for F is:
|r|5
B. (2/3)(x2 + y 2 + z 2 )−3/2
D. (2/7)(x2 + y 2 + z 2 )−7/2
E. 2π
C. (1/7)(x2 + y 2 + z 2 )−7/2
E. none of the above
MAC2313 Final A
(5 pts) 1. How many of the following are necessarily true?
i. The vector field F = ⟨−2x + 3y, 3x − 5y⟩ is conservative.
ii. The vector field F = 5(x2 + y2 )−3/2 ⟨x, y⟩ is radial.
iii. All constant vector fields in R3 are conservative.
iv. Gravitational fields in R3 are conservative and radial.
A. 0
B. 1
C. 2
D. 3
E. 4
(5 pts) 2. How many of the following are necessarily true?
i. The potential function for a conservative vector field is unique.
ii. The divergence of a constant vector field in R3 is equal to the zero vector.
iii. The divergence of a conservative vector field in R3 is a constant.
iv. The curl of a conservative vector field in R3 is equal to the zero vector.
A. 0
B. 1
C. 2
D. 3
E. 4
(5 pts) 3. Let C be the curve
given parametrically by r(t) = ⟨t + 3, 4 − 2t⟩ , t : 1 → 2; if
∫
f (x, y) = 2x + 4y, the value of C f (x, y) ds is:
A. 0
√
B. 13 5
√
C. 4 5
√
D. 26 5
√
E. 22 5
(5 pts)
4. Let C be the curve given parametrically by r(t) = ⟨3t, 1 − 2t2 , 4⟩ , t : 0 → 2; the
∫
value of C y dx + x2 dy + z dz is:
B. −40
A. 0
C. −86
D. −154
E. −210
(5 pts) 5. ∫Let C be the upper half of the unit circle oriented clockwise and let F = ⟨2y + 1, −2x − 1⟩,
the value of C F · dr is:
A. 0
B. −2 − π
C. 2 + 2π
D. −2 − 2π
E. 2 + π
(5 pts) 6. Let D = { (x, y) | |x| < 1, |y| < 2}; how many of the following are true?
i. D is simply connected.
ii. D is connected.
iii. D is open.
iv. The boundary curve(s) for D is closed but not simple.
A. 0
B. 1
C. 2
D. 3
E. 4
(5 pts) 7. Let C be a simple, smooth curve with initial
point (1, 0, 1) and terminal point
∫
2 2xy
2 2xy
2xy
(0, 0, −1); if F = ⟨6yz e , 6xz e , 6ze + 1⟩, the value of C F · dr is:
A. 0
B. −2
C. −4
D. 3
E. none of the above
(5 pts) 8. How many of the following are necessarily true?
i. The line integral of a conservative vector field is independent of the path connecting an initial
point P to a terminal point Q.
ii. The curl of a constant vector field in R3 is equal to the scalar zero.
iii. A vector field F = ⟨f(x, y), g(x, y)⟩ is conservative if fx = gy .
iv. The circulation of a conservative vector field along a smooth, oriented curve is equal to zero.
A. 0
B. 1
C. 2
D. 3
E. 4
(5 pts) 9. The part of the unit sphere with x ≤ 0 and z ≤ 0 is parameterized by
r(θ, ϕ) = ⟨cosθ sin ϕ, sin θ sin ϕ, cos ϕ⟩; the domain for this parameterization is equal to:
A. D = { (θ, ϕ) | 0 ≤ θ ≤ 2π, 0 ≤ ϕ ≤ π }
B. D = { (θ, ϕ) | π/2 ≤ θ ≤ π, 0 ≤ ϕ ≤ π/2 }
C. D = { (θ, ϕ) | π ≤ θ ≤ 2π, 0 ≤ ϕ ≤ π/2 }
D. D = { (θ, ϕ) | π/2 ≤ θ ≤ 3π/2, π/2 ≤ ϕ ≤ π }
E. D = { (θ, ϕ) | π ≤ θ ≤ 2π, π/2 ≤ ϕ ≤ π }
(5 pts) 10. Let f (x, y, z) = z + y 2 and let S be the surface parameterized
by
∫∫
r(u, v) = ⟨2u, −3v, u + v⟩ with 0 ≤ u ≤ 1, 0 ≤ v ≤ 2. The integral S f (x, y, z) dS is equal to:
A. 0
B. 160
C. 189
D. 84
E. 62
(5 pts) 11. Let S be the surface parameterized by r(u, v) = ⟨2v, u2 + v2 , u⟩ with
0 ≤ u ≤ 2, 0 ≤ v ≤ 2. A vector normal to the tangent plane of the surface at u = 1, v = 1 is:
A. ⟨−2, 2, −4⟩
B. ⟨1, −2, 2⟩
C. ⟨1, 2, −4⟩
D. ⟨−2, 1, −4⟩
E. ⟨2, 2, −4⟩
(5 pts) 12. Let f (x, y, z)
= xyez and let S be the part of the paraboloid z = x2 + y 2 + 8 where
∫∫
2
x + y ≤ 5. The integral S f (x, y, z) dS is equal to:
2
A.
∫ 5 ∫ √5−x2
C.
∫ 5 ∫ √5−x2
√
−5 − 5−x2
√
−5 − 5−x2
xyez dy dx
xyex
2 +y 2 +8
∫ √5 ∫ √5−x2
B.
√ 2
4x + 4y 2 + 1 dy dx
√
√
− 5 − 5−x2
D.
xyex
∫ √5 ∫ √5−x2
√
√
− 5 − 5−x2
2 +y 2 +8
xyex
dy dx
2 +y 2 +8
√
2x + 2y + 1 dy dx
E. none of the above
(5 pts) 13. Let F = ⟨z, x, y⟩ and let S be the surface parameterized by r(u, v) = ⟨cos u, sin u, 2v + 3⟩
with domain R = { (u, v) | 0 ≤ u ≤ 2π, 0 ≤ v ≤ 2 } ∫and
oriented so that normal vectors to the
∫
surface are pointing away from the z-axis; the integral S F · n̂ dS is given by:
A.
C.
E.
∫ 2π ∫ 2
0
0
∫ 2π ∫ 2
0
0
∫ 2π ∫ 2
0
0
6v cos u + 6 sin u dv du
B.
∫ 2π ∫ 2
0
4v cos u + 6 cos u + 2 sin u cos u dv du
12v cos u sin u dv du
0
6 sin u + 6v + v sin u cos u dv du
D.
∫ 2π ∫ 2
0
0
6 cos u + sin u cos u dv du
(5 pts) 14. Let F = ⟨x, y, z⟩ and let S be the part of the plane x + y + z = 1 defined above the
triangular region in the x, y-plane with vertices (0, 0), (1, 0), and (0, 1) and oriented so that the z
component of the normal vector is positive; the flux of F across S is given by:
A.
D.
∫1∫1
0
0
1 dy dx
∫ 1 ∫ −x+1
0
0
B.
∫ 1 ∫ −x+1
0
0
1 − x − y dy dx
1 dy dx
C.
∫ 1 ∫ −x+1
0
0
x + y dy dx
E. none of the above
(5 pts) 15. Let C be the closed curve in the x, y-plane given by the triangle with vertices (−1, 0),
(0, 2), and (1, 0) and oriented counter-clockwise. If F = ⟨4y − 1, 3 − 2x⟩ , then the circulation of F
around C is equal to:
A. 0
B. −6
C. 12
D. −12
E. 6
(5 pts) 16. Let F = ⟨2xy + 4z2 , x2 + 2z, 2y + 8xz⟩; the divergence of F at the point
(x, y, z) = (1, 1, 1) is equal to:
A. 0
B. 4
C. 10
D. 12
(5 pts) Bonus 17. The vector field in problem 16 is conservative.
A. True
B. False
E. 6
(5 pts) 18. Let F = ⟨xz, xy, yz⟩; the curl of F at the point (x, y, z) = (0, 0, 0) is equal to:
A. ⟨−1, −1, −1⟩
B. ⟨1, 1, 1⟩
C. ⟨1, −1, 1⟩
D. ⟨0, 0, 0⟩
E. ⟨1, 1, 1⟩
(5 pts) Bonus 19. The vector field in problem 18 is conservative.
A. True
B. False
(5 pts) 20. Let S be a closed surface bounded by a cylinder of radius one centered on the z-axis,
z = 0, and z = 2 and let F = ⟨−6y, 2x, z2 ⟩, then the flux across the surface is equal to:
A. 0
B. π
C. 4π
D. 3π
E. 2π
(5 pts) 21. Let S be the part of the paraboloid z = 9 − ⟨x2 − y 2 with z ≥⟩ 0 and oriented so
2
2
that the z component of the normal vector is positive; if F = −y + z, x, ex +y , then the integral
∫∫
S (∇ × F) · n̂ dS is equal to:
A. 0
B. 6π
C. 3π
D. 2π
E. 18π
(5 pts) 22. Which of the following are true:
I. The Divergence Theorem is given by
II. Stoke’s Theorem is given by
∫
C
∫∫∫
D
F · T ds =
F · n̂ dV =
∫∫
S
∫∫
S
∇ · F dS.
(∇ × F) · n̂ dS.
III. If F is conservative and S is a smooth bounded surface, then
∫∫
S
(∇ × F) · n̂ dS = 0.
IV. The flux of a constant vector field out of a solid is always greater than zero. (Hint: consider
the Divergence Theorem)
A. only II and III
E. I, II, III, and IV
B. only I and III
C. only I, II, and IV
D. only I, II, and III
MAC2313 Final A
(5 pts) 1. Let C be the bottom half of the circle of radius
3 centered at the origin in the x, y∫
plane and with a counterclockwise orientation; the integral C (2x2 y − y)2 ds is equal to:
A. 9
C. 3
∫ 2π
π
∫ 2π
π
(6 cos2 t − sin t)2 dt
B. 3
(18 cos2 t + 3 sin t)2 dt
D.
E. none of the above
(5 pts) 2. The function ϕ(x) =
√
k
x2 +y 2 +z 2
∫ 2π
π
∫ 2π
π
(18 cos2 t − sin t)2 dt
(54 cos2 t + 9 sin t)2 dt
is a potential function for which of the following
vector fields:
A. F =
−2kr
|r|3
B. F =
−kr
2|r|3
−kr
C. F = |r|
D. F =
−kr
|r|3
E. F =
−2kr
|r|
(5∫ pts) 3. If F = ⟨2y, −4x, y⟩ and C is given parametrically by r(t) = ⟨et , 3, −t⟩ , t : 0 → 1,
then C F · dr is equal to:
A. 2e − 3
B. 6e − 9
C. 6e − 3
D. 2e + 3
E. 6e + 9
(5 pts) 4. Let F ⟨f, g, h⟩ and G = ⟨l, m, n⟩ be a differentiable vector fields and let C be a simple,
smooth curve given parametrically by r(t) = ⟨x(t), y(t), z(t)⟩ , t : a → b; which of the following are
true:
I.
∫
II.
F · dr =
C
∫
III.
C
∫b
a
∫
C
f dx + g dy + h dz
F · T ds = −
∫
−C
F · T ds
[(f + l)x′ (t) + (g + m)y ′ (t) + (h + n)z ′ (t)] dt =
IV. If F is continuous, then
A. only III
∫
C
B. only I and II
∫
C
F · dr +
∫
C
G · dr
F · T ds = 0
C. only III and IV
D. only I, II, and III
E. I, II, III, and IV
(5 pts) 5. Let D be the annular region in the x, y-plane which lies between the circles x2 + y 2 = 1
and x2 + y 2 = 4; which of the following are true:
I. D is connected.
II. D is simply connected.
III. The curve x2 + y 2 = 4 is closed but not simple.
IV. The curve x2 + y 2 = 1 is closed and simple.
A. only III
B. only IV
C. only III and IV
D. only I and III
E. only I and IV
∫
C
(5 pts) 6. If a curve C is given parametrically by r(t) = ⟨t2 , 3t2 ⟩ , t : 0 → 1 then the integral
y dx + x dy is equal to:
A. −1
∫
C
B. 0
C. 1
D. 2
E. 3
(5 pts) 7. Let F = ∇ϕ and let C be a smooth curve parametrized by r(t), t : a → b, then
F · T ds = ϕ(r(b)) − ϕ(r(a)).
A. true
B. false
(5 pts) 8. The vector field F = ⟨2yex + 2xz2 , ex + z2 cos y, 2z + sin y + 2zx⟩ is conservative.
A. true
B. false
(5 pts) 9. Let F = ⟨18xy2 , 18x2 y + 12y2 ⟩ and let C be the square in the x, y-plane with vertices
(0, 0), (3, 0), (0, 3), and (3, 3) and with a counter-clockwise orientation; then the circulation of F
about C is given by:
A. −2/3
B. −1/3
C. 0
D. 1/3
E. 4/9
(5 pts) 10. Let F ∫= ⟨2x, 6y⟩ and let C be a simple, smooth curve with initial point (1, 3) and
terminal point (2, 5); C F · T ds is equal to:
A. 23
B. 47
C. 51
D. 61
E. 63
(5 pts) 11. Let a surface S be given parametrically by r(u, v) = ⟨2u + v, 3u, 4v⟩ ,
−∞ < u, v < ∞; a normal vector to the surface at (3, 3, 4) is given by:
A. ⟨12, −8, 3⟩
B. ⟨6, −4, 3⟩
C. ⟨−12, 8, 3⟩
D. ⟨−6, 4, −3⟩
E. ⟨−6, −4, 3⟩
(5 pts) 12. Let S be the part of the plane 3x + 3y + z = 6 which lies in the first octant and
oriented
so that the z component of the normal vector is positive; if F = k, what the value of
∫
F
·
n
dS?
S
A. −1
B. 0
C. 1
D. 2
E. 4
(5 pts) 13. Let a surface S be given parametrically
by r(u, v) = ⟨3 cos u, 3 sin u, v⟩ ,
∫
0 ≤ u ≤ 2π, 0 ≤ v ≤ 1; if f (x, y, z) = z, then S f (x, y, z) dS is equal to:
A. 0
B. π
C. 2π
D. 3π
E. 4π
(5 pts) 14. The standard orientation of a closed surface has the normal vectors pointing into
the region bounded by the surface.
A. true
B. false
(5 pts) 15. Let D be a connected and simply connected bounded region in the x, y-plane and
let ∂D be smooth with a counter-clockwise orientation. If F = ⟨f, g⟩ with f and g differentiable,
which of the following are true:
H
I. If
∂D
F · dr =
∫∫
D
gy − fx dA
II. If gy − fx = 1, then Area(D) =
H
∂D
F · dr
III. If F is conservative, Area(D) = 0
IV.
H
∂D
F · dr =
A. only IV
H
∂D
f dx + g dy
B. only I and II
C. only I and III
D. only II and III
E. only I, II, and III
(5 pts) 16. If F = ⟨x, y, 1⟩ and S is a surface given by r(u, v) = ⟨u cos v, u sin v, v⟩∫, 0 ≤ u ≤ 1,
0 ≤ v ≤ 2π oriented so that the z component of the normal vector is positive; then S F · n dS is
equal to which of the following:
A. 0
B. −2π
C. 2π
D. π
E. −π
(5 pts) 17. Let S be a surface given by the graph of z = xey , 0 ≤ x ≤ 1, 0 ≤ y ≤ 1, oriented
so that the z component of the normal vector is positive; if F = ⟨xy, 4x2 , xyey ⟩ then the flux of F
across S is equal to:
A. −1 − e
B. −1 + e
C. 1 − e
D. e + 1
E. 0
(5 pts) 18. Let C be the curve defined as the triangle with Hcounterclockwise orientation with
vertices (−2, 0), (2, 0), and (0, 2); if F = ⟨x2 + 2y, x − y2 ⟩, then C F · dr is equal to: (use Green’s
Theorem)
B. −4
A. 0
C. −6
D. −8
E. −10
(5 pts) 19. Let S be a smooth oriented surface which bounds the solid R and let F be a vector
field whose components have continuous first partial derivatives in R; the Divergence Theorem is
given by:
A.
B.
C.
D.
E.
∫∫
∫∫
F dS =
S
div F dS =
∫∫
R
div F dV
∫∫∫
F · n̂ dS =
∫∫∫
S
∇ · F dS =
∫∫
S
F · n̂ dS =
S
∫∫
∫∫
∫∫∫
S
S
∫∫∫
R
R
F · n̂ dV
div F dV
F · n̂ dS
R
(div F) · n̂ dV
(5 pts) 20. Let D be the box in the first octant bounded by the three coordinate planes and the
planes z = 1, x = 2, and y = 3 and let F = ⟨x, z, y⟩; calculate the flux out of the surface of the box
(Hint: use the divergence theorem.)
A. 6
B. 1
C. 0
D. 12
E. 3
Bonus (5 pts) 21. Let S be a hemisphere given parametrically by r(θ, ϕ) = ⟨cos θ sin ϕ, sin θ sin ϕ, cos ϕ⟩
with a domain given by R = {(θ, ϕ) | 0 ≤ θ ≤ 2π, 0 ≤ ϕ ≤ π/2} and suppose the surface is oriented
so that the
z component of the normal vector is positive; if F = ⟨−x, y, 3⟩, what is value of the
∫∫
integral S (∇ × F) · n̂ dS?
A. 0
B. 1
C. 2π
D. −1
E. −2π
Bonus (5 pts) 22. Let F be a vector field with differentiable component functions; if curl F = 0
on a simply connected domain, then the vector field is conservative on that domain.
A. true
B. false
I made the following practice exams about an year ago for cy summer calc III class. I do not have the worked
out solutions for them. The answers are in the last page.
67
Fall 2014
FINAL: REVIEW 1
1. Evaluate the line integral
Z
C
(2xy + 5)dx + (x2 − 4z)dy − (4y)dz where C is parametrized
by c(t) = (t sin πt, (t2 + 1) cos πt, t2 + t) for 0 ≤ t ≤ 1
(a) −8
(b) 16
(c) 8π
(d) −16
(e) 0
2. If the curve CZ is given by c(t) = (cos t, sin t, t) where 0 ≤ t ≤ π and f (x, y, z) = xy + z,
the value of
f (x, y, z)ds is given by;
C
(a) π/2
(b) −π 2
√
(c) π 2 / 2
(d) π 2 /2
√
(e) 2π 2
R
3. Compute F. dS for the vector function F = h4, yi for the quarter circle x2 + y 2 = 1
with x ≤ 0, y ≤ 0 oriented counterclockwise.
(a) −4
(b) −1
(c) 9
(d) 4.5
(e) 0
t t
4. Let F = h3zy −1 , 4x, −yi
Z and let C be the curve given parametrically by r(t) = (e , e , t)
for t ∈ [−1, 1]. Then
F . T ds is equal to;
C
(a) 0
(b) e4 − 1
(c) e2 − e−2 − e + e−1
(d) e2 − e−2 − 2e + 2e−1
(e) 2e2 − 2e−2 − e + e−1
5. Find a potential function V for F or determine that F is not conservative.
F = hyzexy , xzexy − z, exy − yi
(a) F is not conservative
(b) V (x, y, z) = zey − yz + C
(c) V (x, y, z) = zexy − yz + C
(d) V (x, y, z) = zex + yz + C
(e) V (x, y, z) = zexy + yz + C
6. Let F be a two dimensional vector field and C, D be two simple smooth curves in R2
going from P to Q. Which of the following statements are true.
Z
Z
A. If F is conservative, then
F. dS =
F. dS.
C
D
Z
B. If P = Q, then
F. dS = 0
Z C
F. dS gives the net change in potential.
C. The integral
C
D. If F is conservative and C is closed, the circulation of F around C is zero.
(a) A only
(b) A and D only
(c) A, C and D only
(d) A, B and C only
(e) B and C only
7. If a curve C is given parametrically by r(t) = hcos t, sin ti, where 0 ≤ t ≤ 2π and
F = hx, 0i, then the flux of F across C is given by;
(a) 0
(b) π
(c) −π
(d) 2π
(e) −2π
2
8. Let F be a two dimensional vector field and let C be a simple smooth closed curve in
R2 ; which of the following statements is/are true?
i. If C is oriented clockwise, then the normal vector textbf n points outward
from the curve.
Z
F.nds
ii. The circulation of F around C is given by
C
Z
iii. The flux of F around C is given by
F.Tds
C
Z
iv. If F is conservative, then
F.Tds = 0
C
(a) only iv
(b) only i and iv
(c) only iii and iv
(d) only i, ii and iv
(e) all four statements are true.
x
−y
,
in R2 \ {0}. Find the circulation of F arounf the unit
9. Let F =
x2 + y 2 x2 + y 2
circle oriented counterclockwise.
(a) 0
(b) 1
(c) π
(d) −π
(e) −1
10. G(r, θ) = (r cos θ, r sin θ, 1 − r2 ) parametrizes the paraboloid z = 1 − x2 − y 2 .
(a) TRUE
(b) FALSE
11. Let S be the portion of the sphere x2 + y 2 + z 2 = 9 where 1 ≤ x2 + y 2 ≤ 4 and z ≥ 0.
What is the area of S? (Hint: Find a parametrization of S in polar coordinates)
(a) π ln(0.8)
(b) π ln(1.2)
(c) 6π ln(1.6)
(d) 3π ln(1.6)
(e) 2π ln(1.2)
3
x2
. Find the surface integral
12. Let f (x, y, z) =
4z
by z = 4 − x2 − y 2 where 0 ≤ z ≤ 3.
√
√
(a) π/6(17 17 − 5 5)
√
√
(b) π/12(17 17 − 5 5)
√
√
(c) π/2(17 17 − 5 5)
√
√
(d) −π/12(17 17 − 5 5)
√
(e) π/12(5 5))
3
ZZ
f (x, y, z) dS for the surface given
S
13. Let G(u, v) = (u, v , u + v), where 0 ≤ u ≤ 1, 0 ≤ v ≤ 1. Find
ZZ
y dS
S
√
19 19 − 1
(a)
108
√
6 6−1
(b)
54
√
19 19 + 1
(c)
108
√
19 19 − 1
(d)
72
√
19 − 1
(e)
108
Z
14. Calculate xy dx + (x2 + x) dy where C is the triangular path traversed in the order
C
of the vertices (0, 1), (−1, 0), (1, 0).
(a) 3
(b) 2
(c) −1
(d) 0
(e) 1
15. Calculate the area of the region between the x axis and the cycloid parametrized by
c(t) = ht − sin t, 1 − cos ti for 0 ≤ t ≤ 2π
(a) 2π
(b) 3π
(c) 6π
(d) π
(e) 4π
4
16. Find the surfac integral
ZZ
S
F.dS for F = hez , z, xi where S is the surface parametrized
by G(r, s) = (rs, r + s, r) oriented by Tr × Ts .
(a) 4/3 − e
(b) e − 1/3
(c) 4/3 − 1/e
(d) e2
(e) 0
RR
17. Calculate S zdS where S is the part of the surface z = x3 where x, y ∈ [0, 1].
√
5 10
1
(a)
+
27
54
√
5 10
(b)
27
1
(c)
54
(d) 0
√
5 10
1
(e)
−
27
54
ZZ
18. Find
F. dS for the vector field F = h−y, z, −xi on the surface x2 + y 2 + z 4 = 4,
S
z ≥ 0 with outward pointing normal.
(a) 27
(b) 54
(c) -27
(d) 9
(e) -3
19. Use the Stokes theorem to comupute the flux of curl(F) through the upper hemesphere
2
3
x2 + y 2 + z 2 = 1, z ≥ 0 with outward pointing normal for F = hez − y, ez + x, cos(xz)i
(a) 0
(b) −π
(c) eπ
(d) 2π
(e) π
5
20. Let F = hx, y 2 , z + yi and let S be the boundary of the region contained in the cylinder
x2 + y 2 = 4 between the planes z = x and z = 8. Find the flux of F across the surface
of S
(a) 64π
(b) 0
(c) 32π
(d) −64π
(e) 11π
21. Let F = hz sec2 x, z, y + tan xi, and let S be the surface of the upper ellipsoid 9x2 +
y 2 + 4z 2 = 36 , z ≥ 0. find the circulation of F around the boundary of S
(a) 0
(b) π
(c) pi
√
(d) 8π + 2 2
√
(e) 8π − 2 2
22. Let F = hz, −z, −y 2 i and let C consist of the three line segments that bound the plane
z = 8 − 4x − 2y in the first octant, and
R suppose that C is oriented counterclockwise
when viewed from above. The value of C F. dr is given by:
(a) 0
88
3
71
(c) −
6
23
(d) −
3
71
(e) −
3
(b) −
6
Answer Key
Multiple Choice Questions
1. B
2. C
3. D
4. E
5. C
6. B
7. B
8. D
9. C
10. A
11. D
12. B
13. A
14. E
15. C
16. A
17. E
18. B
19. D
20. A
21. A
22. B
7
Fall 2014
FINAL: REVIEW 2
1. Let C be the left half of the
Z circle of radius 2 centered at (3, 2) with counterclockwise
x2 − 6x + y 2 − 4y + 13 ds is equal to
orientation. The integral
C
(a) 4π
(b) 8π
(c) −2π
(d) 2π
(e) π
2. Calculate
Z
C
F . Tds for the vector field F = hez , ex−y , ey i on the curve given by the
path from P to Q in the figure below.
(a) 2 − e2
(b) 2 − e2 − e−1
(c) 2 − e − e−1
(d) e − e−1
(e) 2 − e−1
3. Find a potential function for the vector field below or show that it is not conservative.
y
−1
F=
, tan x, 2z
1 + x2
(a) V (x, y, z) = y tan−1 (x) − z 2 + C
(b) V (x, y, z) = y 2 tan−1 (x) − z 2 + C
(c) V (x, y, z) = y tan−1 (x) + z + C
(d) V (x, y, z) = y tan−1 (x) + z 2 + C
(e) The vector field F is not conservative
p
3
y 2 + 1 + z 2 , z 3 sin(z 2 )ez i.
4. Calculate Curl(F) for F = hx2 + y 2 , ln
(a) h2z, −2y, 0i
(b) h0, 0, 0i
(c) h2z, 0, −2yi
(d) h2z, 2y, 0i
(e) h2z, 0, 2yi
2
5. Let f (x, y, z) = 6xz − 2y . Calculate
t goes from 0 to 2
Z
√
f (x, y, z) ds along the curve (t, t2 / 2, t3 /3) as
C
(a) 0
(b) π
(c) 864/35
(d) 123/63
(e) 8π
6. For the vector field v = hey , 2x − 1i, calculate the flux across the parabola y = x2 for
0 ≤ x ≤ 1 oriented from left to right.
(a) 0
(b) e − 1
(c) 1
(d) 1 − e
(e) e2
7. Calculate the work done by the force field F = hx, y, zi along the path c(t) = (cos t, sin t, t)
for 0 ≤ t ≤ 6π.
(a) 9π 2
(b) 18π 2
(c) 9π
(d) 36 − 9π 2
(e) 0
2
8. If a curve C is given parametrically by r(t) = hcos t, sin ti, t ∈ [0, 2π] and F = hx, 0i,
then the flux across C is given by
(a) 0
(b) −π
(c) 2π
(d) π
(e) −2π
9. Calculate the line integral
Z
C
y dx − x dy on the parabola y = x2 for 0 ≤ x ≤ 2.
(a) 1
(b) 0
(c) −8/3
(d) −4/3
(e) 6
10. Evaluate
0≤t≤2
Z
2 −2t
2xyz dx + x2 z dy + x2 y dz over the path c(t) = (t2 , sin((πt)/4), et
C
(a) 16
(b) 0
(c) −8
(d) 2π
(e) −9
11. Let f (x, y, z) = 3x2 yez , find Div(∇(f (−1, 1, 3))).
(a) 6e3
(b) 3e3
(c) e3
(d) 9e3
(e) 0
3
) for
12. Find a normal vector to the surface of the paraboloid z = x2 + y 2 at the point (3, 1, 10).
(a) (−6, 2, 1)
(b) (6, 2, −1)
(c) (−6, −2, −1)
(d) (−3, 2, −1)
(e) (6, −2, 1)
RR
x2
, calculate
f (x, y, z)ds where S is the surface given by z = 4 −
4−z
x2 − y 2 for 0 ≤ z ≤ 3.
√
√
(a) π/12(17 17 − 5 5)
√
√
(b) π/12(17 17 + 5 5)
√
√
(c) π/6(17 17 − 5 5)
√
√
(d) π 2 /12(17 17 − 5 5)
13. Let f (x) =
(e) 0
14. Calculate the flux of the vector field F = h0, 3, xi through the surface defined by the
part of the sphere x2 + y 2 + z 2 = 9 where x, y, z ≥ 0 with outward pointing normal.
(a) 27/4(3π − 4)
(b) 27/12(3π − 4)
(c) 27/4(3π + 4)
(d) 27/12(3π − 12)
(e) 9/4(3π + 4)
15. Let S be the portion of the sphere x2 + y 2 + z 2 = 9 where 1 ≤ x2 + y 2 ≤ 4 adn z ≥ 0.
What is the area of S? (Hint: Use a surface integral)
√
√
(a) 6π( 8 + 5)
√
√
(b) 6π( 2 − 5)
√
√
(c) 6π( 8 − 5)
√
√
(d) 3π( 8 − 5)
(e) 3π
4
16. Let S be a surface given by the graph of z = xey , x, y ∈ [0, 1] oriented so that the z
component of the normal vector is positive. If F = hxy, 4x2 , xyey i the flux of F across
S is equal to;
(a) −1 − e
(b) −1 + e
(c) 1 − e
(d) e + 1
(e) 0
17. Let S be the annular region given by 1 ≤ x2 + y 2 ≤ 4 and let F = h−x2 y, xy 2 i be
a vector field. Suppose the boundary of the inner circle is oriented clockwise
and the
H
boundary of the outer circle is oriented counterclockwise. Calculate ∂S F. dr.
(a) 0
(b) π 2
(c) −π
(d) 103
(e) 126.5π
18. Calculate the area of the region between the x axis and the cycloid parametrized by
c(t) = ht − sin t, 1 − cos ti for 0 ≤ t ≤ 2π
(a) 0
(b) 6π
(c) 3π
(d) π
(e) 3π 2
19. Calculate
(a)
(b)
(c)
(d)
(e)
ZZ
√
5 10
27
√
5 10
9
√
5 10
27
√
5 10
27
√
5 15
27
S
−
+
+
+
−
zdS where S is the part of the surface z = x3 where x, y ∈ [0, 1].
1
54
1
54
5
54
1
54
1
54
5
20. Let F = hx2 − y 2 , 2xy, zi and S be the surface that bounds the ellipsoid 9x2 + 16y 2 +
4z 2 = 36. The flux of Curl(F) over S is given by
(a) 0
(b) π
(c) −π
(d) 2π
(e) 9π 2
21. Let F = h−z 2 , 2zx, 4y − xri and Let C be a simple closed curve bounded by the plane
x + y + z = 4 in the first octant, and suppose C is counterclockwise when viewed from
above.
Z
The value of
F.dr is given by;
C
(a) 32
(b) 16
(c) 8
(d) 4
(e) 2
22. Let F be a vector filed with differentiable component functions. If Curl(F)=0 in a
connected domain, then F is conservative
(a) TRUE
(b) FALSE
23. Let F = hz, −z, x2 − y 2 i, let C consist of the three line segments that bound the plane
z = 8 − 4x − 2y in the first octant,
and suppose C is counterclockwise when viewed
Z
from above. Find The value of
F.dr.
C
(a) 0
88
3
71
(c) −
6
23
(d) −
3
71
(e) −
3
(b) −
6
Answer Key
Multiple Choice Questions
1. B
2. C
3. D
4. E
5. C
6. B
7. B
8. D
9. C
10. A
11. D
12. B
13. A
14. E
15. C
16. A
17. E
18. B
19. D
20. A
21. A
22. B
7