LAST (family) NAME:
Test # 2
FIRST (given) NAME:
Math 2MM3
ID # :
TUTORIAL #:
Instructions: You must use permanent ink. Tests submitted in pencil will not be considered
later for remarking. This exam consists of 6 problems on 10 pages (make sure you have all
10 pages). The last page is for scratch or overflow work. A few formulas that you may find
useful are on the last page. The total number of points is 50. Do not add or remove pages
from your test. No books, notes, or “cheat sheets” allowed. The only calculator permitted
is the McMaster Standard Calculator, the Casio fx 991.
GOOD LUCK!
MATH 2MM3: SAMPLE TEST 2 B: SOLUTIONS
#
Mark
1.
2.
3.
4.
5.
6.
TOTAL
Continued. . .
Test # 2 / Math 2MM3
-2-
NAME:
ID #:
PART I: Multiple choice. Indicate your choice very clearly. There is only one correct
answer in each multiple-choice problem. Circle the letter (a,b,c,d or e) corresponding to your
choice. Ambiguous answers will be marked as wrong.
1. (4 pts.) Compute the work done by the conservative force vector field
¡
¢
F(x, y) = (2 x sin(y)) i + x2 cos(y) − 3 y 2 j
on any path starting at (0, 0) and ending at (1, 1). The answer is
(a) 0
(b) 3 + 2 sin(1)
(c) cos(1) + 2
→ (d) sin(1) − 1
(e) 4 cos(1) − 12
Solution. Since
ª
∂
∂ © 2
{2 x sin(y)} = 2 x cos(y) =
x cos(y) − 3 y 2 ,
∂y
∂x
there exists a function f (x, y) such that ∇f = F or
∂f
= 2 x sin(y),
∂x
∂f
= x2 cos(y) − 3 y 2 .
∂y
Solving these equations by integration, we find a solution of the form
f (x, y) = x2 sin(y) − y 3
Hence, the work done by F on any path C starting at (0, 0) and ending at (1, 1) is
Z
F · dr = f (1, 1) − f (0, 0) = sin(1) − 1.
C
Continued. . .
Test # 2 / Math 2MM3
-3-
NAME:
ID #:
2. (4 pts.)
By changing the order of integration, the integral
Z 1Z e
sin(x y 2 ) dy dx
ex
0
can also be expressed as
Z e Z ey
(a)
sin(x y 2 ) dx dy
0
1
Z eZ
ln(y)
→ (b)
1
Z
2Z
0
ln(2)
(c)
1
ln(2)
(d)
0
sin(x y 2 ) dx dy
1
ln(y)
Z
1
(e)
1
sin(x y 2 ) dx dy
ln(y)
Z eZ
Z
sin(x y 2 ) dx dy
sin(x y 2 ) dx dy,
0
Solution. The domain of integration is
R = {(x, y), 0 ≤ x ≤ 1, ex ≤ y ≤ e}
= {(x, y), 1 ≤ y ≤ e, 0 ≤ x ≤ ln(y)}.
3
2.5
2
y 1.5
1
0.5
0
0
0.2
0.6
0.4
x
Continued. . .
0.8
1
Test # 2 / Math 2MM3
-4-
NAME:
ID #:
3. (4 pts.) Suppose that f (x, y) is a continuous function on the x, y-plane. The integral
Z
2
1
√
Z
4−x2
√
− 4−x2
f (x, y) dy dx
can be expressed, using polar coordinates, as
Z
Z
π/3
2
→ (a)
f (r cos θ, r sin θ) r dr dθ
−π/3
Z
Z
π/3
sec θ
2
(b)
f (r cos θ, r sin θ) r dr dθ
−π/3
Z
π
1
Z
2
(c)
f (r cos θ, r sin θ) r dr dθ
−π
Z
0
π/3
Z
2 cos θ
f (r cos θ sin θ) dr dθ
(d)
−π/3
Z
2
1
Z
π/3
(e)
f (r cos θ, r sin θ) r dθ dr
2 sec θ
−π/3
Solution.
1.5
1
0.5
0
0
0.5
1
1.5
2
-0.5
-1
-1.5
The domain of integration is the region inside the disk centered at (0, 0) with radius 2 and
the half-plane x ≥ 1 which can be expressed in polar coordinates as
{(r, θ), −π/3 ≤ θ ≤ π/3, sec(θ) ≤ r ≤ 2}.
Continued. . .
Test # 2 / Math 2MM3
-5-
NAME:
ID #:
Part II: Provide all details and fully justify your answer in order to receive credit.
4. (12 pts.) Use polar coordinates to calculate the integral
ZZ
x
dA,
2
2
R 1+x +y
where R is the circular sector bounded by the lines y = x, y = −x and the curve x =
p
1 − y2.
Solution. The domain of integration can be expressed in polar coordinates as the set
{(r, θ), −π/4 ≤ θ ≤ π/4, 0 ≤ r ≤ 1}.
Letting x = r cos(θ) and y = r sin(θ), we have dA = r drdθ. Hence,
ZZ
Z π/4 Z 1
x
r cos(θ)
dA =
r dr dθ
2
2
1 + r2
R 1+x +y
−π/4 0
Z π/4
Z 1
r2
=
cos(θ) dθ
dr.
2
−π/4
0 1+r
We have
Z
√
π/4
cos(θ) dθ =
−π/4
and
Z
1
0
r2
dr =
1 + r2
Therefore,
Z
1
1−
0
ZZ
0.6
0.4
0.2
0
0
-0.4
-0.6
Continued. . .
0.2
0.4
0.6
2
=
−
2
µ
√ ¶
√
2
−
= 2
2
1
π
dr = [r − arctan(r)]10 = 1 − .
2
1+r
4
√
x
π
dA
=
2 (1 − ).
2
2
1+x +y
4
R
-0.2
π/4
[sin(θ)]−π/4
0.8
1
Test # 2 / Math 2MM3
-6-
NAME:
ID #:
y
x
and Q(x, y) = .
2
2
(a) (4 pts.) Show that the path integral
Z
5. Let P (x, y) = −
P dx + Q dy
L
on any line segment L parametrized by x(t) = k t, y(t) = m t, 0 ≤ t ≤ a, where a > 0 and
k, m are constants, is always zero.
Solution. We have x0 (t) = k and y 0 (t) = m. Hence,
Z
Z a
P dx + Q dy =
P (x(t), y(t)) x0 (t) + Q(x(t), y(t)) y 0 (t) dt
L
0
Z a
mt
kt
)k +
m dt
=
(−
2
2
0
Z a
=
0 dt = 0.
0
(b) (2 pts.) Compute
Solution. We have
Continued. . .
∂Q ∂P
−
for the given functions P , Q.
∂x
∂y
µ ¶
1
1
∂Q ∂P
−
= − −
= 1.
∂x
∂y
2
2
Test # 2 / Math 2MM3
-7-
NAME:
ID #:
(c) (6 pts.) For a > 0, consider the region R bounded by the hyperbolic arc C parametrized
by
x = cosh t, y = sinh t,
0 ≤ t ≤ a,
and the two line segments connecting the endpoints of C with (0, 0). Use Green’s theorem
to compute the area of R.
Hint: Use the results in parts (a) and (b).
Solution. The area of R is
ZZ
A(R) =
1 dA
R
and by Green’s theorem, letting P and Q as in part (a) and using (b),
Z
Z
Z
∂Q ∂P
1 dA =
−
dA =
P dx + Q dy
∂x
∂y
R
R
Γ
Z
1
=
(−y) dx + x dy,
2 Γ
where Γ denotes the curve bounding R described in the positive direction and made up of
the curve C and the two line segments connecting the endpoints of C with (0, 0). Since by
part (a), the integral of P dx + Q dy on each of these line segments is 0, we have
Z
Z
1
1
(−y) dx + x dy =
(−y) dx + x dy
2 Γ
2 C
Z
1 a
=
(− sinh t) cosh0 t + cosh t sinh0 t dt
2 0
Z
1 a
=
(− sinh t) sinh t + cosh t cosh t dt
2 0
Z
Z
1 a
1 a
a
2
2
=
cosh t − sinh t dt =
1 dt = .
2 0
2 0
2
1
0.8
0.6
0.4
0.2
0
0
Continued. . .
0.4
0.8
1.2
Test # 2 / Math 2MM3
-8-
NAME:
6.
ID #:
(14 pts.) Let S be the part of the paraboloid z = 1 − x2 − y 2 that lies in the first
octant x, y, z ≥ 0 and choose the orientation of S so that the unit normal on S has a positive
z-component. Compute the surface integral
ZZ
F · n dS
S
where F is the vector field defined by
F(x, y, z) = x i + y j + 2 z k.
1.4
1.2
1
0.8
0.6
0.4
0.2
0
0.2
0.4
0.6
0.8 x
1
1.2
1.4
0
0
0.2 0.4
0.6
0.8
y
1
1.2 1.4
Solution. Note that the paraboloid z = 1 − x2 − y 2 intersect the x, y-plane when z = 0, i.e.
when x2 + y 2 = 1. Hence the surface S is the graph of the function z = 1 − x2 − y 2 defined
in the domain
D = {x, y), x, y ≥ 0, x2 + y 2 ≤ 1}.
The surface S can be parametrized by
r(x, y) =< x, y, 1 − x2 − y 2 >,
(x, y) ∈ D.
We have thus
rx =< 1, 0, −2 x >
ry =< 0, 1, −2 y >
Test # 2 / Math 2MM3
-9-
NAME:
ID #:
and
¯
¯
¯i j
¯
k
¯
¯
¯
rx × ry = ¯1 0 −2 x¯¯ =< 2 x, 2 y, 1 > .
¯0 1 −2 y ¯
Note that rx × ry gives the correct orientation of S since its z-component, 1, is positive. We
have thus
ZZ
ZZ
F · n dS =
F(r(x, y)) · (rx × ry )(x, y) dA
D
ZS Z
=
hx, y, 2 (1 − x2 − y 2 )i · h2 x, 2 y, 1i dA
Z ZD
ZZ
2
2
2
2
=
2 x + 2 y + 2 (1 − x − y ) dA =
2 dA
D
D
Since D can be expressed in polar coordinates as
D∗ = {(r, θ), 0 ≤ θ ≤
π
, 0 ≤ r ≤ 1},
2
this last integral equals
Z
π/2
Z
1
2 r dr dθ =
0
Thus,
0
π £ 2 ¤r=1 π
r r=0 = .
2
2
ZZ
F · n dS =
S
π
.
2
Test # 2 / Math 2MM3
-10-
NAME:
ID #:
Some formulas you may use:
cosh t =
et + e−t
,
2
sinh t =
et − e−t
2
d
d
(cosh t) = sinh t,
(sinh t) = cosh t, cosh2 t − sinh2 t = 1
dt
dt
I
ZZ
∂Q ∂P
P dx + Q dy =
−
dA
∂y
C
R ∂x
SCRATCH
THE END