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Vector Calculus Gateway Exam
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every 45 minutes.
Num Topic
1 Line Integrals
2 Green’s Theorem
3 Green’s Theorem
4 Surface Integrals
5 Divergence Theorem
6 Stoke’s Theorem
Exam Randomization.
1. There are 6 problems on the exam.
2. One question is randomly selected from
each of the topic listed on the left.
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problems and/or coefficients.
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Webwork Tips. Click HERE for full listing.
- You do not need to simplify.
For: 2 +
1
22
1
26
Enter: 2 + 1/2ˆ2 - 1/2ˆ 6
Not: 2.
Symbol Usage
π pi
e e
Function Usage
p
x sqrt(x) or xˆ(1/2)
ln x ln(x)
e
x
eˆ(x)
tan
− 1
(x) atan(x) or arctan(x)
Contents
Vector Operations 2
Div and Curl 2
Parameterization 3
Line Integral Examples 4
Green’s Theorem 6
Green’s Theorem Examples 6
Surface Integral 7
Divergence Theorem 8
Stoke’s Theorem 9
Green’s Theorem
C
F · d r =
R
∂ F 2
∂ x
∂ F 1
∂ y
d xd y
Divergence Theorem
W
div( F ) dV =
S
F · n d A
Stoke’s Theorem
C
F · d r =
S
curl( F ) · n d A
Idea (Projection). The dot product is the mathematical operation that will allow us to perform a vector
projection. Recall how the dot product is defined
u · v = | u || v | cos( θ ).
Now, fix the tails of u and v at the origin. For a nice diagram, assume v is located on the x-axis and u
is also located in the first quadrant. We may use trigonometry if we view u as the hypotenuse of a right
triangle and θ as the angle between the u and the v. Thus the component of u that lies along v is the scalar
projection of u onto v , scal vu
scal vu = | u | cos( θ ) =
u · v
| v |
The vector projection of u onto v , pr o j vu , is the length (given above) paired with the direction of v :
pr o j vu = scal vu
v
| v |
length ︷︸︸︷ u · v
| v |
v
| v | ︸︷︷︸
direction
u · v
| v |
2
v
Lastly, recall that work only utilizes the component of the force F in the direction of displacement d , so a
projection needs to occur. Hence, the dot product below
W = F · d
For example, if you are pushing a lawn mower, the component of the force pushing the lawnmower down
into the ground is wasted in the sense that it does not contribute to the work done.
Definition (Div). The divergence of F = 〈F 1 , F 2 , F 3 〉 is given as
di v( F ) = ∇ · F =
∂ F 1
∂ x
∂ F 2
∂ y
∂ F 3
∂ z
Interpretation (Line Integral). If F is a force, the line integral over C ,
C
F ( r ) · d r , gives the work done by
the force along C.
Example 1. Calculate the line integral
C
F ( r ) · d r for the given data.
F = 〈y
2 , −x
2
〉, C : line segment from (0, 0) to (1, 4) ♠
Solution. We begin by parameterizing the line segment:
r (t ) = (1 − t )〈0, 0〉 + t 〈1, 4〉 = t 〈1, 4〉, 0 ≤ t ≤ 1
We compose F with r :
( F ◦ r )(t ) = 〈(4t )
2 , −(t )
2 〉
We calculate the derivative along C using the above parameterization:
r
′ (t ) = 〈1, 4〉
The line integral is
C
F ( r ) · d r =
1
0
〈(4t )
2 , −(t )
2 〉 · 〈1, 4〉 d t =
1
0
(4t )
2 (1) − (t )
2 (4) d t
1
0
12 t
2 d t =
t
3
1
0
Solution (variant). — We may also use the following form:
C
F · d r =
C
F 1 d x + F 2 d y
We may parametrize the curve using y = 4 x with d y = 4 d x. Now, we calculate
C
F (r ) · d r =
C
y
2 d x − x
2 d y =
1
0
(4x)
2 d x − x
2 (4d x)
1
0
[16x
2 − 4 x
2 ]d x =
1
0
[12x
2 ]d x =
12 x
3
1
0
Example 2. Calculate the line integral
C
F ( r ) · d r for the given data.
F = 〈e
−x , e
−y , e
−z 〉, C : r (t ) = 〈t , t
2 , t 〉, 0 ≤ t ≤ 2 ♠
Solution. Compose F with r :
( F ◦ r )(t ) = 〈e
−t , e
−t
2 , e
−t 〉
We calculate the derivative along C using the above parameterization:
r
′ (t ) = 〈1, 2t , 1〉
The line integral is
C
F ( r ) · d r =
2
0
〈e
−t , e
−t 2 , e
−t 〉 · 〈1, 2t , 1〉 d t =
2
0
e
−t
−t 2 (2t ) + e
−t d t
2
0
e
−t d t +
4
0
e
−u du = 2[−e
−t
]
2
0
−u
]
4
0
= 2(1 − e
− 2 ) + (1 − e
− 4 ) ■
Definition (Path Independence). A line integral
C
F · d r is path independent in a domain D if for every
pair of endpoints A, B ∈ D the line integral has the same value for all paths in D.
Theorem (Path Independence Equivalence). Suppose the line integral
C
F ·d r is computed for C in a sim-
ply connected domain D. Then, the following are equivalent:
(a) The line integral
C
F · d r is path independent;
(b) F = ∇ f , for some scalar-valued function f ;
(c) Integration around closed curves C in D is 0;
(d) cur l ( F ) = 0 in D.
Remark. Statement (b) of the above theorem, F = ∇ f , asserts that we have an exact differential form. We
have previously studied the solution of exact differential equations in DEs class with two variables x and
y. In the example below we obtain the exact differential form of three variables x, y and z. ♠
Remark. It is easy to check if the line integral is path independent by verifying cur l ( F ) = 0. ♠
Example 3. Show that the form is exact, and then use the exactness to evaluate the integral
I =
(1,1,1)
(0,2,3)
y z sinh(xz) d x + cosh(xz) d y + x y sinh(xz) d z. ♠
Solution. First, we verify that the form is exact. Below, we show that the curl is the zero vector,
∇ × F =
i
j
k
∂
∂ x
∂
∂ y
∂
∂ z
y z sinh(xz) cosh(xz) x y sinh(xz)
= a(x, y, z) i + b(x, y, z) j + c(x, y, z) k = 0 ,
because
a(x, y, z) = x sinh(xz) − x sinh(xz) = 0
b(x, y, z) = −y sinh(xz) + x y z cosh(xz) − y sinh(xz) − x y z cosh(xz) = 0
c(x, y, z) = z sinh(xz) − z sinh(xz) = 0
Since form is exact, it must be the case that F = ∇ f , i.e.
F 1 =
∂ f
∂ x
F 2 =
∂ f
∂ y
F 3 =
∂ f
∂ z
We integrate to find the scalar-valued function f :
F 1 d x =
y z sinh(xz) d x = y cosh(xz) +G 1 (y, z)
F 2 d y =
cosh(xz) d y = y cosh(xz) +G 2 (x, z)
F 3 d z =
x y sinh(xz) d z = y cosh(xz) +G 3 (x, y)
Where Gi are constants of integration. It must be the case that G 1 (y, z) = G 2 (x, z) = G 3 (x, y) = 0 , so
f (x, y, z) = y cosh(xz).
Since the differential form is exact, we may utilize the following result,
∫ B
A
d f = f (B) − f (A)
to evaluate the line integral. With the following identifications: A = (0, 2, 3) and B = (1, 1, 1), we have
I = f (B) − f (A) = (1) cosh(1) − (2) cosh(0) = cosh(1) − 2 ≈ −0.457 ■
The circle C suggests a change to polar coordinates below. The integrand on the RHS is
∂ F 2
∂ x
∂ F 1
∂ y
= 3 x
2 − (− 3 y
2 ) = 3(x
2
2 ) = 3 r
2 .
The region R is bounded by the downward facing parabola
R = {(r, θ ) | 0 ≤ r ≤ 5, 0 ≤ θ ≤ 2 π }
Hence,
C
F · d r =
2 π
0
5
0
3 r
2 · r dr d θ =
2 π
0
d θ
5
0
3 r
3 dr
= 2 π
[
r
4
] 5
0
3 π
4
1875 π
Example 5. Evaluate
C
F ( r ) · d r counterclockwise around the boundary of C of the region R by Green’s
Theorem, where
F (x, y) = 〈x
2
2 , x
2 − y
2 〉
R : bounded region given by 1 ≤ y ≤ 2 − x
2
. ♠
Solution. We will utilize Green’s Theorem. The region R is bounded by the downward facing parabola
R = {(x, y) | 1 ≤ y ≤ 2 − x
2 , − 1 ≤ x ≤ 1}
The integrand in in the double integral is
∂ F 2
∂ x
∂ F 1
∂ y
= 2 x − 2 y.
Hence,
C
F · d r =
1
− 1
2 −x
2
1
2 x − 2 y d yd x =
1
− 1
[2x y − y
2 ]
2 −x
2
1 d x
1
− 1
−x
4 − 2 x
3
2
1
− 1
−x
4
2 − 3 d x +
1
− 1
− 2 x
3
[
x
5
x
3 − 3 x
] 1
0
[
]
where we have broken the integral into a sum of even and odd parts, respectively, and used
∫ (^) a
−a
g (x)d x = 0
for odd function g. ■
Idea (Surface Integral). We will consider two cases.
Case 1: Scalar-valued function f. Assume continuous f and smooth surface S. Let S by parameterized
by r (u, v) = 〈x(u, v), y(u, v), z(u, v)〉. Assume partial derivatives r u and r v are continuous; then the surface
integral is given as
œ
S
f (x, y, z) dS =
œ
R
f (x(u, v), y(u, v), z(u, v)) | r u × r v | dud v
The parameterization transforms the integral over S to integration in the uv-plane. An infinitesimal patch
on the surface has area approximated by the parallelogram formed by r u , r v. The area of the parallelogram
may be computed with the cross product:
dS = | r u × r v |dud v
Note: the single bars above denote the magnitude of the vector.
Case 2: Vector-valued function F. (Flux Integral) Assume F is continuous and S is a smooth surface.
Let S by parameterized by r (u, v) = 〈x(u, v), y(u, v), z(u, v)〉. Assume partial derivatives r u and r v are contin-
uous; then the surface integral is given as
œ
S
F · n dS =
œ
R
F · n | r u × r v | dud v (Flux Integral)
Remark. The area of a surface may be computed with 1 as the integrand:
A(S) =
œ
S
1 dS =
œ
R
| r u × r v | dud v ♠
Remark. The unit normal vector is n (u, v) =
r u × r v
| r u × r v |
and the normal vector N (u, v) = r u × r v. We may rewrite
the flux integral as
œ
S
F · n dS =
œ
R
F · n | r u × r v | dud v
œ
R
F ·
r u × r v
| r u × r v |
| r u × r v | dud v
œ
R
F · ( r u × r v ) dud v =
œ
R
F · N (u, v) dud v ♠
Theorem (Divergence Theorem). Let W be a closed, simply connected region with smooth boundary S.
Further, let F has continuous partial derivatives, then
W
div( F ) dV =
œ
S
F · n d A. (1)
Remark. The unit normal vector n is chosen so that it points away from the solid not into the solid. ♠
Example 6. Evaluate the surface integral
∂ W
F · n d A, where
F = 〈 13 y
2
p
7 z
3 , x
3
3 ), xz〉,
W : x
2
2 ≤ z ≤ 2, x ≥ 0. ♠
Solution. We will utilize the Divergence theorem. The divergence of F is
div(F ) = 0 + 0 + x = r cos( θ ).
The solid given by the paraboloid W is best rewritten in cylindrical coordinates. It may be written in each
of the following ways
W^ ˜ = {(r, θ , z) |r 2 ≤ z ≤ 2, 0 ≤ r ≤
p
2, − π /2 ≤ θ ≤ π /2}
Theorem (Stoke’s Theorem). Let S be a piecewise smooth orientable surface with smooth boundary C.
Then,
C
F · d r =
œ
S
curl( F ) · n d A (2)
Example 8. Calculate the line integral
C
F ( r ) · d r by Stoke’s Theorem for the given F and C. Assume the
z-component of the surface normal to be positive, where
F = 〈y
2 , x
2 , z + x〉,
C : triangle with vertices (0, 0, 0), (1, 0, 0), (1, 1, 0) ♠
Solution. We utilize Stokes’s theorem. The curl is computed as
curl( F ) = 〈0, −1, 2x − 2 y〉.
The surface normal is perpendicular to the x y-plane:
n = 〈1, 0, 0〉 × 〈1, 0, 0〉 = 〈0, 0, 1〉.
We may characterize S as
S = {(x, y, z) | 0 ≤ y ≤ x, 0 ≤ x ≤ 1, z = 0}.
We may now compute the integral
C
F · d r =
œ
S
curl( F ) · n d A =
œ
S
〈0, −1, 2x − 2 y〉 · 〈0, 0, 1〉 d A
œ
S
2 x − 2 y d A =
1
0
x
0
2 x − 2 y d y d x
1
0
x y −
y
2
x
0
d x = 2
1
0
x
2 −
x
2
d x
0
x
2 d x =
x
3
1
0
Example 9. Evaluate the line integral
C F ( r )^ ·^ d r^ by using Stoke’s Theorem for the given^ F^ and^ C^.
F = 〈x, y, y
2 〉,
C : boundary curve of paraboloid
z = 9 − x
2 − y
2 , 0 ≤ z ≤ 9
Assume C is oriented counterclockwise when viewed from above. ♠
Solution. Parameterize the paraboloid
r (u, v) = 〈u, v, 9 − u
2 − v
2 〉, (u, v) ∈ R
R = {(u, v) | u
2
2 ≤ 9}
The surface normal is
N = r u × r v =
i j k
1 0 − 2 u
0 1 − 2 v
= 〈 2 u, 2v, 1〉,
This vector is oriented upward and consistent with orientation of C. The curl is computed as
curl( F ) =
i j k
∂ ∂ x
∂ ∂ y
∂ ∂ z
x y y
2
= 〈 2 y, 0, 0〉 = 〈 2 v, 0, 0〉
We may now compute the integral by ultimately changing to polar coordinates
C
F · d r =
œ
R
curl( F ) · N du d v =
œ
R
〈 2 v, 0, 0〉 · 〈 2 u, 2v, 1〉 dud v
œ
R
4 uv dud v = 4
2 π
0
3
0
r
2 cos( θ ) sin( θ )r dr d θ
2 π
0
cos( θ ) sin( θ )d θ
3
0
r
3 dr = 4
[
sin
2 ( θ )
] 2 π
0
[
r
4
] 3
0
4
