-
First Year Vector Calculus Mid-course Quiz
26 January 2018
,[2pt],
1
26 January 2018
In all questions, choose the best answer.You will need some
paper to work out things.
Good Luck
We probably won’t get through all these, but they should make
usefulrevision material.
,[2pt],
1
26 January 2018
Mark Gillmarksphysicshelp
Mark GillMPH
-
1. Given that the differentialdf = yz dx + xz dy + (xy + a) dz
is exact, its parentfunction is
(A) f = 3xyz + a
(B) f = yz + xz + xy + a
(C) f = xyz + az
(D) None of the above
,[2pt],
2
26 January 2018
1. Given that the differentialdf = yz dx + xz dy + (xy + a) dz
is exact, its parentfunction is
(A) f = 3xyz + a
(B) f = yz + xz + xy + a
(C) f = xyz + az
(D) None of the above
The correct answer is (D). f = xyz + az + C where C is a
constant.
,[2pt],
2
26 January 2018
Mark GillMPH
-
2. A function f (x , y) is transformed to new
variables (u, v) with x = uv , y = u/v .∂f
∂u=
(A) v∂f
∂x+
1
v
∂f
∂y
(B) v∂f
∂x−
u
v2∂f
∂y
(C)1
v
∂f
∂x+ v
∂f
∂y
(D) −u
v2∂f
∂x+ v
∂f
∂y(E) None of the above
,[2pt],
3
26 January 2018
2. A function f (x , y) is transformed to new
variables (u, v) with x = uv , y = u/v .∂f
∂u=
(A) v∂f
∂x+
1
v
∂f
∂y
(B) v∂f
∂x−
u
v2∂f
∂y
(C)1
v
∂f
∂x+ v
∂f
∂y
(D) −u
v2∂f
∂x+ v
∂f
∂y(E) None of the above
The correct answer is (A).∂f
∂u=
∂f
∂x
∂x
∂u+
∂f
∂y
∂y
∂u
,[2pt],
3
26 January 2018
Mark GillMPH
-
3. A function f (x , y) is transformed to variables(u, v) with x
= uv , y = u/v as in the preceding
question, so∂f
∂u= v
∂f
∂x+
1
v
∂f
∂y. Then
∂2f
∂u2=
(A)∂
∂u
(
v∂f
∂x+
1
v
∂f
∂y
)
(B) v∂
∂u
(
∂f
∂x
)
+1
v
∂
∂u
(
∂f
∂y
)
(C) v
[
v∂2f
∂x2+
1
v
∂2f
∂y ∂x
]
+1
v
[
v∂2f
∂x ∂y+
1
v
∂2f
∂y2
]
(D) All of the above(E) None of the above
,[2pt],
4
26 January 2018
3. A function f (x , y) is transformed to variables(u, v) with x
= uv , y = u/v as in the preceding
question, so∂f
∂u= v
∂f
∂x+
1
v
∂f
∂y. Then
∂2f
∂u2=
(A)∂
∂u
(
v∂f
∂x+
1
v
∂f
∂y
)
(B) v∂
∂u
(
∂f
∂x
)
+1
v
∂
∂u
(
∂f
∂y
)
(C) v
[
v∂2f
∂x2+
1
v
∂2f
∂y ∂x
]
+1
v
[
v∂2f
∂x ∂y+
1
v
∂2f
∂y2
]
(D) All of the above(E) None of the above
The correct answer is (D). These are the steps needed to
calculate
this. Note that by definition ∂2f
∂u2= ∂∂u
∂f∂u so just apply the rule for
changing variables to find ∂/∂u in terms of ∂/∂x and ∂/∂y .
Study thisuntil it is clear!
,[2pt],
4
26 January 2018
Mark GillMPH
-
4. A surface is given by z(x , y) = xe−y over theregion bounded
by x = 0, y = 0, and x + 2y = 1.The volume occupied above the x − y
plane isfound by evaluating
(A)
ˆ 1/2
y=0
ˆ 1
x=0xe−y dx dy
(B)
ˆ (1−x)/2
y=0
ˆ 1
x=0xe−y dx dy
(C)
ˆ 1/2
y=0
ˆ 1−2y
x=0xe−y dx dy
(D) More than one of the above
(E) None of the above
,[2pt],
5
26 January 2018
4. A surface is given by z(x , y) = xe−y over theregion bounded
by x = 0, y = 0, and x + 2y = 1.The volume occupied above the x − y
plane isfound by evaluating
(A)
ˆ 1/2
y=0
ˆ 1
x=0xe−y dx dy
(B)
ˆ (1−x)/2
y=0
ˆ 1
x=0xe−y dx dy
(C)
ˆ 1/2
y=0
ˆ 1−2y
x=0xe−y dx dy
(D) More than one of the above
(E) None of the above
The correct answer is (C). If the y integration were to be done
first,
then the limits in (B) would be ok, i.e.
ˆ 1
x=0
ˆ (1−x)/2
y=0xe−y dy dx .
Multiple integration is always done inside → out, or right →
left. Youshould draw a picture of the region of integration.
,[2pt],
5
26 January 2018
Mark GillMPH
-
x
y
x+2y=1dydx
x=0 x=1-2y 1
y=1/2x integration first
x
y
x+2y=1dy
dxy=0
y=(1-x)/2
1
1/2y integration first
Mark GillMPH
-
5. A double integral is given in plane polar coordinates by
I =
ˆ π/2
φ=0
ˆ 1/(cos φ+sinφ)
ρ=0ρ5 cos2 φ sin3 φdρdφ. Expressed in cartesian
coordinates this can be written
(A)
ˆ 1
y=0
ˆ 1−y
x=0x2y3 dx dy
(B)
ˆ 1
y=0
ˆ
√1−y2
x=0x2y3 dx dy
(C)
ˆ 1
y=0
ˆ 1−y
x=0x2y2 dx dy
(D)
ˆ 1
y=0
ˆ 1−y
x=0xy2 dx dy
(E) None of the above
,[2pt],
6
26 January 2018
5. A double integral is given in plane polar coordinates by
I =
ˆ π/2
φ=0
ˆ 1/(cos φ+sinφ)
ρ=0ρ5 cos2 φ sin3 φdρdφ. Expressed in cartesian
coordinates this can be written
(A)
ˆ 1
y=0
ˆ 1−y
x=0x2y3 dx dy
(B)
ˆ 1
y=0
ˆ
√1−y2
x=0x2y3 dx dy
(C)
ˆ 1
y=0
ˆ 1−y
x=0x2y2 dx dy
(D)
ˆ 1
y=0
ˆ 1−y
x=0xy2 dx dy
(E) None of the above
The correct answer is (E).The ρ limit is ρ(cosφ+ sinφ) = 1 = x +
y whichis a straight line, so R is the triangle bounded by this
line in the 1st quadrant.
However, the Jacobian for the transformation gives dρ dφ =
(1/ρ)dx dy so
the correct answer should be
ˆ 1
y=0
ˆ 1−y
x=0
x2y3√
x2 + y2dx dy
,[2pt],
6
26 January 2018
Mark GillMPH
-
6. A cylindrically symmetric volume is bounded bythe plane z =
a/2 and the surface
z = a3/(x2 + y2 + a2). Given∂(x , y , z)
∂(ρ,φ, z)= ρ in
cylindrical polar coordinates, the volume enclosedis
(A)
ˆ a
z=a/2πa2
(a
z− 1
)
dz
(B)
ˆ a
ρ=0
ˆ a3/(ρ2+a2)
z=a/2πρ2 dz dρ
(C) 2π
ˆ a
ρ=0
ˆ a3/(ρ2+a2)
z=a/2ρdz dρ
(D) More than one of the above
(E) None of the above
,[2pt],
7
26 January 2018
6. A cylindrically symmetric volume is bounded bythe plane z =
a/2 and the surface
z = a3/(x2 + y2 + a2). Given∂(x , y , z)
∂(ρ,φ, z)= ρ in
cylindrical polar coordinates, the volume enclosedis
(A)
ˆ a
z=a/2πa2
(a
z− 1
)
dz
(B)
ˆ a
ρ=0
ˆ a3/(ρ2+a2)
z=a/2πρ2 dz dρ
(C) 2π
ˆ a
ρ=0
ˆ a3/(ρ2+a2)
z=a/2ρdz dρ
(D) More than one of the above
(E) None of the above
The correct answer is (D). (A) follows if you do the ρ (and φ)
integrationfirst, noting on the surface ρ2 = a2[(a/z) − 1], so this
just sums upvolumes dV = πρ2dz. (C) is the result of doing the z
(and φ) integralfirst. In both cases the φ integral just gives
2π.
,[2pt],
7
26 January 2018
Mark GillMPH
-
xy
z
a
dz
ρ = a [ (a/z) - 1 ]1/2
z=a/2
ρ integration first
ρ=0
xy
z
a
adρ
z=a3/(ρ2+a2)z=a/2
z integration first
Mark GillMPH
-
7. A surface S is defined by z(x , y) = y2ex . An
element−→dS on the surface is
(A) k̂ dx dy
(B) y2ex (̂i dy dz + ĵ dx dz + k̂ dx dy)
(C) (−y2ex î − 2yex ĵ)dx dy
(D) (−y2ex î − 2yex ĵ + k̂)dx dy
(E) None of the above
,[2pt],
8
26 January 2018
7. A surface S is defined by z(x , y) = y2ex . An
element−→dS on the surface is
(A) k̂ dx dy
(B) y2ex (̂i dy dz + ĵ dx dz + k̂ dx dy)
(C) (−y2ex î − 2yex ĵ)dx dy
(D) (−y2ex î − 2yex ĵ + k̂)dx dy
(E) None of the above
The correct answer is (D). Write r = x î + y ĵ + z(x , y) k̂
to give a pointon the surface parameterised by (x , y). Then
dS =
(
∂r
∂x
∂r
∂y
)
dx dy =
∣
∣
∣
∣
∣
∣
î ĵ k̂1 0 y2ex
0 1 2yex
∣
∣
∣
∣
∣
∣
dx dy . Or the negative of this.
,[2pt],
8
26 January 2018
Mark GillMPH
-
8. For the surface z(x , y) = y2ex with−→dS = (−y2ex î − 2yex
ĵ + k̂) dx dy from the precedingquestion, the flux FB of a
constant vector field−→B = Bo k̂ through the portion of S bounded
byx = 0, x = 1, y = 0, y = 1 is
(A) Bo
(B) Bo´ 1
y=0
´ 1x=0(−y
2ex − 2yex + 1)dx dy
(C) Bo´ 1
y=0
´ 1x=0 2y
3e2x dx dy
(D) None of the above
,[2pt],
9
26 January 2018
8. For the surface z(x , y) = y2ex with−→dS = (−y2ex î − 2yex
ĵ + k̂) dx dy from the precedingquestion, the flux FB of a
constant vector field−→B = Bo k̂ through the portion of S bounded
byx = 0, x = 1, y = 0, y = 1 is
(A) Bo
(B) Bo´ 1
y=0
´ 1x=0(−y
2ex − 2yex + 1)dx dy
(C) Bo´ 1
y=0
´ 1x=0 2y
3e2x dx dy
(D) None of the above
The correct answer is (A).FB ≡
˜
R B · dS =˜
R Bo k̂ · (−y2ex î − 2yex ĵ + k̂)dx dy =
˜
R Bo dx dy = Bo˜
R dx dy = Bo since R has area 1 (m2).
,[2pt],
9
26 January 2018
Mark GillMPH
-
9. The line integral of the vector field
B = y î − 2x ĵ + z k̂ along the straight line from theorigin
to the point (1, 1, 0) is
(A) -1
(B) 1
(C) 1/2
(D) -1/2
(E) None of the above
,[2pt],
10
26 January 2018
9. The line integral of the vector field
B = y î − 2x ĵ + z k̂ along the straight line from theorigin
to the point (1, 1, 0) is
(A) -1
(B) 1
(C) 1/2
(D) -1/2
(E) None of the above
The correct answer is (D). The path is the line y = x . Along
this path,dy = dx and dz = 0 soB · dr = Bx dx + By dy + Bz dz = x
dx − 2x dx + 0 dz = −x dx . Hence´
C B · dr =´ 1
x=0 −x dx = −1/2.
,[2pt],
10
26 January 2018
Mark GillMPH
