1. No category

LAST (family) NAME:
FIRST (given) NAME:
ID # :
MATHEMATICS 2Q04
McMaster University Final Examination
Day Class
Duration of Examination: 3 hours
THIS EXAMINATION PAPER INCLUDES 24 PAGES AND 15 QUESTIONS.
YOU ARE RESPONSIBLE FOR ENSURING THAT YOUR COPY OF THE
PAPER IS COMPLETE. BRING ANY DISCREPANCY TO THE ATTENTION OF YOUR INVIGILATOR.
SAMPLE FINAL EXAM C: SOLUTIONS
Continued. . .
Final Exam / Math 2Q04
-2-
NAME:
ID #:
Part I: Provide all details and fully justify your answer in order to receive credit.
1. Let D be the region in the first quadrant of the x, y plane (i.e. the region where x, y ≥ 0)
where 1 ≤ y ex ≤ 2 and 21 ≤ y e−x ≤ 1. Consider the change of variables
½ x
ye = u
y e−x = v
(a) (4 pts.) Find the region D∗ in the u, v plane that corresponds to D when the change
of variable above is used and find expressions for the variables x and y in terms of u and v
for a point (x, y) in D.
u
y ex
=
= e2x and thus 2 x = ln(u/v), so x =
−x
v
y e√
= y 2 so y = u v.
Solution. Clearly, we have
Also, u v = (y ex ) y e−x

x=
ln(u/v)
2
=
ln u−ln v
.
2
ln u−ln v
,
2

y = u1/2 v 1/2 .
Also,
D∗ = {(u, v), 1 ≤ u ≤ 2,
(b) (2 pts.) Compute the Jacobian
1
≤ v ≤ 1}.
2
∂(x, y)
corresponding to the change of variables above.
∂(u, v)
Solution. We have
¯ ∂x
∂(x, y) ¯¯ ∂u
= ∂y
∂(u, v) ¯ ∂u
=
Continued. . .
¯
∂x ¯
∂v ¯
∂y ¯
∂v
¯
1
¯
2u
= ¯¯ u−1/2
v 1/2
1 −1/2 −1/2
u
v
.
2
2
¯
− 21v ¯¯
u1/2 v −1/2 ¯
2
Final Exam / Math 2Q04
-3-
NAME:
ID #:
(c) (6 pts.) Use the change of variables above to compute the integral
ZZ
y 3 dx dy.
D
Solution. We have
ZZ
ZZ
3
y dx dy =
D
=
1
2
=
1
2
=
Continued. . .
¯
¯
¶
µ
Z 1Z 2
¯ ∂(x, y) ¯
1
−1/2
−1/2
3/2
3/2
¯
¯
u
v
du dv
u v
¯ ∂(u, v) ¯ du dv = 1
2
1
D∗
2
Z 1Z 2
Z
Z 2
1 1
u v du dv =
v dv
u du
1
2 12
1
1
2
· 2 ¸v=1 · 2 ¸u=2 µ ¶ µ ¶ µ ¶
v
u
1
3
3
=
2 v= 1 2 u=1
2
8
2
9
.
32
¡ 1/2 1/2 ¢3
u v
2
Final Exam / Math 2Q04
-4ID #:
NAME:
2. (a) (6 pts.) Let S be the part of the hemisphere (x + 1)2 + y 2 + z 2 = 4, z ≥ 0 which
is inside the cylinder x2 + y 2 = 1. Let C be the curve of intersection of the hemisphere and
the cylinder, oriented counterclockwise as viewed from above. Consider the vector field
F(x, y, z) = (y) i + (x) j + (y z) k.
After finding a suitable parametrization for C, compute the line integral
I
F · dr
C
using the definition of the integral of a vector-field on an oriented curve.
Solution. The two surfaces intersect when x2 + y 2 = 1 and
(x + 1)2 + y 2 + z 2 = 4 or x2 + 2 x + 1 + y 2 + z 2 = 4
This last equation can thus be simplified to z 2 = 2 − 2 x on C or to z =
z ≥ 0 on C. The curve C can be parametrized by
√ √
r(t) = hcos t, sin t, 2 1 − cos ti, 0 ≤ t ≤ 2 π.
We have
1
sin t
r0 (t) = h− sin t, cos t, √ √
i,
2 1 − cos t
Thus,
I
Z
2π
F · dr =
C
Z
√ √
2 1 − x, since
0 ≤ t ≤ 2 π.
F(r(t)) · r0 (t) dt
0
√ √
1
sin t
hsin t, cos t, sin t 2 1 − cos ti · h− sin t, cos t, √ √
i dt
2 1 − cos t
0
Z 2π
Z 2π
2
2
2
=
− sin t + cos t + sin t dt =
cos2 t dt
0
0
·
¸2 π
Z 2π
1 + cos(2 t)
t sin(2 t)
=
dt =
+
2
2
4
0
0
= π.
2π
=
2.0
1.5
1.0
0.5
0.0
−1.0
1.0
0.5
−0.5
t
0.0
0.0
−0.5
0.5
1.0
Continued. . .
−1.0
z
Final Exam / Math 2Q04
NAME:
-5ID #:
(b) (6 pts.) Compute the line integral in part (a) using Stokes’ theorem applied to the
surface S (with an orientation compatible with the orientation of the boundary curve C in
part (a) ).
Solution. The surface S is parametrized by the vector function
p
r(x, y) = hx, y, 4 − (x + 1)2 − y 2 i, where x2 + y 2 ≤ 1.
We have
x+1
rx = h1, 0, − p
4 − (x + 1)2 − y 2
i and ry = h0, 1, − p
y
4 − (x + 1)2 − y 2
i.
Thus
¯
¯
¯i j
¯
k
¯
¯
x+1
x+1
y
¯1 0 − √
¯
2
2
rx × ry = ¯
, p
, 1i.
4−(x+1) −y ¯ = h p
2
2
¯
¯
y
4 − (x + 1) − y
4 − (x + 1)2 − y 2
¯0 1 − √
¯
2
2
4−(x+1) −y
Note that this normal on S points towards the z-axis and corresponds to the orientation of
the curve C in part (a). Furthermore,
¯
¯
¯i
¯
¯ ∂ ∂j k
¯
∂ ¯
¯
∇ × F = ¯ ∂x ∂y ∂z ¯ = (z) i + (0) j + (0) k = (z) i
¯ y x y z¯
Thus, using Stokes’ theorem, we have, letting D = {(x, y), x2 + y 2 ≤ 1},
I
ZZ
ZZ
F · dr =
(∇ × F) · n dS =
(∇ × F)(r(x, y)) · rx × ry dx dy
CZZ
S
D
p
x+1
y
, p
, 1i dx dy
=
h 4 − (x + 1)2 − y 2 , 0, 0i · h p
4 − (x + 1)2 − y 2
4 − (x + 1)2 − y 2
D
ZZ
Z 2π Z 1
Z 2π
Z 1
Z 1
2
=
x + 1 dx dy =
(r cos θ + 1) r dr dθ =
cos θ dθ
r dr + 2 π
r dr
D
0
0
0
0
0
· 2 ¸r=1
r
= 2π
= π.
2 r=0
Continued. . .
Final Exam / Math 2Q04
-6-
NAME:
ID #:
3. Consider the wave equation
∂ 2u
∂2u
=
4
,
∂t2
∂x2
0 < x < π, t > 0,
(1)
together with the boundary conditions
u(0, t) = u(π, t) = 0,
t > 0,
(2)
0 < x < π.
(3)
and the initial conditions
u(x, 0) = 0, ut (x, 0) = 1,
(a) (3 pts.) Using the method of separation of variables, find particular non-trivial solutions
of the equations (1) and (2) of the form u(x, t) = X(x) T (t). Show that this leads to the
eigenvalue problem X 00 (x)+λ X(x) = 0 with boundary condition X(0) = X(π) = 0 for X(x)
together with the differential equation T 00 (y) + 4 λ T (t) = 0 for T (t).
Solution. If u(x, t) = X(x) T (t), the equation (1) becomes X(x) T 00 (t) = 4 X 00 (x) T (t) = 0
or, dividing both sides by 4 X(x) T (t), X 00 (x)/X(x) = T 00 (t)/(4 T (t)). Since the left-hand side
of the last equation is a function of x only while the right-hand side is a function of y only,
we deduce the existence of a constant λ such that X 00 (x)/X(x) = T 00 (t)/(4 T (t)) = −λ which
yields the required differential equations that X(x) and Y (y) must satisfy. The condition
(2) yields X(0) T (t) = X(π) T (t) = 0 which, if we want non-trivial solutions, is equivalent
to X(0) = X(π) = 0.
(b) (3 pts.) Find all eigenvalues and eigenfunctions for the eigenvalue problem associated
with X(x) in the case where λ > 0. (They are no eigenvalue when λ ≤ 0, but you dont need
to show that.)
Solution. If λ > 0, we can let λ = ω 2 , where ω > 0 and the general solution of the DE
X 00 (x) + ω 2 X(x) = 0 is X(x) = C1 cos(ω x) + C2 sin(ω x). The condition X(0) = 0 yields
C1 = 0 and the condition X(π) = 0 then implies that C2 sin(ω π) = 0. For a non-trivial
solution to exists, we need sin(ω π) = 0 which is possible only when ω = k, a positive integer.
This yields the eigenfunctions Xk (x) = sin(k x) corresponding to the eigenvalue λ = k 2 ,
for each integer k ≥ 1.
Continued. . .
Final Exam / Math 2Q04
-7-
NAME:
ID #:
(c) (3 pts.) For each eigenvalue λ found in part (b), find the non-trivial solutions of the
form u(x, t) = X(x) T (t) of (1) and (2) associated with λ and use the superposition principle
to find the general solution of (1) and (2).
Solution If λ = k 2 where k ≥ 1 is an integer, the corresponding DE satisfied by T (t) is
T 00 (t) + 4 k 2 T (t) = 0 and has solution Tk (y) = Ak cos(2 k t) + Bk sin(2 k t). According to the
superposition principle, the general solution of (1) and (2) has the form
u(x, t) =
∞
X
Xk (t) Tk (t) =
k=1
∞
X
(Ak cos(2 k t) + Bk sin(2 k t)) sin(k x)
k=1
(d) (3 pts.) Use the result in part (c) to obtain the unique solution of (1) and (2) that also
satisfies the boundary conditions (3).
P
Solution. The initial condition u(x, 0) = 0 for 0 < x < π yields ∞
k=1 Ak sin(k x) = 0 for
0 < x < π. It follows that the left-hand side of the previous equation has to be the sine
Fourier series of the function f (x) = 0 on the interval (0, π). By the uniqueness of the sine
Fourier series coefficients, we deduce that Ak = 0 for all integers k ≥ 1. Thus the solution
has the form
∞
X
Bk sin(2 k t) sin(k x), 0 < x < π.
u(x, t) =
k=1
P
The initial condition ut (x, 0) = 0 for 0 < x < π then shows that ∞
k=1 2 k Bk sin(k x) = 1
for 0 < x < π. It follows that 2 k Bk is the k-th sine Fourier coefficients of the function 1 on
(0, π), so
·
¸π
Z
2 π
2
2 1 − cos(k π)
cos(k x)
2 k Bk =
sin(k x) dx =
=
−
π 0
π
k
π
k
0
Thus,
1 1 − cos(k π)
Bk =
=
π
k2
(
0,
2
,
π k2
k even
k odd,
and
∞
2 X 1
sin(2 k t) sin(k x)
u(x, t) =
π k=1 k 2
k odd
¾
½
2
1
1
=
sin(2 t) sin(x) + 2 sin(6 t) sin(3 x) + 2 sin(10 t) sin(5 x) + . . . ,
π
3
5
Continued. . .
0<x<π
Final Exam / Math 2Q04
-8-
NAME:
ID #:
4. (12 pts.) Let R be the solid region inside the sphere x2 + y 2 + z 2 = 4 and above the
plane z = −1. Let S be the boundary surface of R. Consider the vector field
F(x, y, z) = x z 2 i + y z 2 j + 2 z (x2 + y 2 ) k.
If S is oriented using the outward pointing normal, compute the surface integral
ZZ
F · n dS
S
using the divergence theorem.
Warning: Do not waste your time calculating the surface integral directly. No credit will
be given for that.
Solution We have
∇·F=
∂(x z 2 ) ∂(y z 2 ) ∂(2 z (x2 + y 2 ))
+
+
= z 2 + z 2 + 2 (x2 + y 2 ) = 2 (x2 + y 2 + z 2 ).
∂x
∂y
∂z
Thus, according to the divergence theorem,
ZZ
ZZZ
ZZZ
F · n dS =
∇ · F dV =
2 (x2 + y 2 + z 2 ) dV
S
R
R
Passing to polar coordinates, note that the plane z = −1 is expressed as ρ cos φ = −1.
The region R is thus expressed by the inequalities ρ ≤ 2 and ρ cos φ ≥ −1 or by ρ ≤ 2 if
0 ≤ φ ≤ 23π and ρ ≤ − cos1 φ if 23π ≤ φ ≤ π. this last integral become
ZZZ
2 (x2 + y 2 + z 2 ) dV
Z
R
2π
Z
2π
3
=
0
0
Z
2
Z
2
Z
2π
2
π
− cos1 φ
2 ρ (ρ sin φ) dρ dφ dθ +
0
2π
3
0
= I1 + I2 .
2
1
−2
−1
0
0
1
−1
−2
Continued. . .
Z
−1
0
1
2
2
0
2 ρ2 (ρ2 sin φ) dρ dφ dθ
Final Exam / Math 2Q04
-9-
NAME:
ID #:
We have
Z
Z
2π
I1 =
Z
2π
3
2
4
1 dθ
sin φ dφ
2 ρ dρ = 2 π
0
0
µ ¶µ ¶
3
64
192 π
= (2 π)
=
2
5
5
0
φ= 2 π
[− cos φ]φ=03
·
2 ρ5
5
¸ρ=2
ρ=0
and
Z
Z
2π
I2 =
Z
π
1 dθ
0
Z
4π
=
5
sin φ
2π
3
µ
π
sin φ
2π
3
Therefore,
Z
− cos1 φ
0
1
− 5
cos φ
¶
π
4
2 ρ dρ dφ = 2 π
π
dφ =
5
·
ZZZ
2 (x2 + y 2 + z 2 ) dV =
R
2π
3
−1
cos4 φ
·
2 ρ5
sin φ
5
¸φ=π
=
φ= 23π
¸ρ=− cos1 φ
dφ
ρ=0
π
15 π
(−1 + 24 ) =
5
5
207 π
192 π 15 π
+
=
.
5
5
5
This integral can also be computed using cylindrical coordinates since R can be expressed
√
in cylindrical coordinates by the inequalities 0 ≤ θ ≤ 2 π, −1 ≤ z ≤ 2 and 0 ≤ r ≤ 4 − z 2
ZZZ
Z
2
2
2π
2
Z
2
√
4−z 2
Z
2 (x + y + z ) dV =
R
Z
2
√
Z
0
4−z 2
= 4π
Z
−1
2
=π
0
r3 + z 2 r dr dz = 4 π
0
2 2
2
2
Z
2
−1
Z 2
(4 − z ) + 2 z (4 − z ) dz = π
−1
207 π
=
5
Continued. . .
−1
−1
·
2 (r2 + z 2 ) r dr dz dθ
r4 z 2 r2
+
4
2
¸√4−z2
dz
0
·
z5
16 − z dz = π 16 z −
5
¸2
4
−1
Final Exam / Math 2Q04
-10-
NAME:
ID #:
5. Let S be the surface parametrized by the vector function
r(u, v) = hv cos u, v sin u, v 2 i,
for 0 ≤ u ≤ 2 π, 0 ≤ v ≤ 2.
(a) (6 pts.) Compute the surface area of S.
Solution. We have
ru = hv (− sin u), v (cos u), 0i
and
rv = hcos u, sin u, 2 vi.
Thus,
¯
¯
¯
¯
i
j
k
¯
¯
ru × rv = ¯¯v (− sin u) v (cos u) 0 ¯¯ = h2 v 2 (cos u), 2 v 2 (sin u), −(sin2 u + cos2 u) vi
¯ cos u
sin u
2 v¯
= v h2 v (cos u), 2 v (sin u), −1i
and
kru × rv k = v kh2 v (cos u), 2 v (sin u), −1ik
p
= v 4 v 2 cos2 u + 4 v 2 sin2 u + 1
√
= v 1 + 4 v2
Letting D = {(u, v), 0 ≤ u ≤ 2 π, 0 ≤ v ≤ 2}, the surface area A(S) of S is then given by
the integral
ZZ
Z 2 Z 2π √
kru × rv k du dv =
v 1 + 4 v 2 du dv
D
0
·
0
(1 + 4 v 2 )3/2
= 2π
12
3/2
= π
Continued. . .
(17)
6
−1
.
¸2
= 2π
0
(17)3/2 − 1
12
Final Exam / Math 2Q04
-11-
NAME:
ID #:
(b) (6 pts.) If F(x, y, z) = y 2 i + x2 j + z 2 k, compute the surface integral
ZZ
F · n dS
S
where the unit normal n corresponds to the given parametrization for S (i. e. n has the same
direction as ru × rv ).
Solution. We have
ZZ
ZZ
F · n dS =
F(r(u, v)) · ru × rv du dv
D
ZS Z
=
hv 2 sin2 u, v 2 cos2 u, v 4 i · h2 v 2 (cos u), 2 v 2 (sin u), −vi du dv
ZDZ
=−
(2 v 4 ) (sin2 u) (cos u) + (2 v 4 ) (cos2 u) (sin u) − v 5 du dv
Z 2 D
Z 2π
Z 2
Z 2π
Z
4
2
4
2
=
2 v dv
sin u cos u du +
2 v dv
cos u sin u du −
0
Since
Z 2π
0
0
0
·
sin3 u
sin u cos u du =
3
¸2 π
2
we have
ZZ
Z
2
F · dS = −
S
Continued. . .
Z
2π
0
Z
2π
5
0
Z
·
2π
5
v dv
0
= 0 and
0
v dv
0
0
2
1 du
0
·
¸2 π
cos3 u
cos u sin u du = −
= 0,
3
0
v6
1 du = −2 π
6
2
µ
¸2
= −2 π
0
32
3
¶
= −
64 π
.
3
Final Exam / Math 2Q04
-12-
NAME:
ID #:
PART II: Multiple choice part. Circle the letter corresponding to the correct answer for
each of the questions 6 to 15 in the box below. Indicate your choice very clearly. Ambiguous
answers will be marked as wrong. There is only one correct choice for each question and an
incorrect answer scores 0 marks.
QUESTION #
V1
Continued. . .
ANSWER:
6.
A
B
C
D
E
7.
A
B
C
D
E
8.
A
B
C
D
E
9.
A
B
C
D
E
10.
A
B
C
D
E
11.
A
B
C
D
E
12.
A
B
C
D
E
13.
A
B
C
D
E
14.
A
B
C
D
E
15.
A
B
C
D
E
Final Exam / Math 2Q04
NAME:
-13ID #:
6. (4 pts.) Compute the directional derivative of the function
f (x, y, z) = sin(x + y 2 + z 3 )
at the point P = (−1, 3, −2) in the direction where f decreases the fastest at P .
(A) −
34
3
√
→ (B) − 181
√
(C) − 163
√
96
(D)
5
√
67
(E) −
3
(ENTER YOUR ANSWERS ON THE CHART ON PAGE 12.)
Solution. We have
∇f (x, y, z) = hcos(x + y 2 + z 3 ), 2 y cos(x + y 2 + z 3 ), 3 z 2 cos(x + y 2 + z 3 )i
and thus
∇f (−1, 3, −2) = h1, 6, 12i = h1, 6, 12i
The function f decreases the fastest at P = (−1, 3, −2) in the direction of −∇f (−1, 3, −2)) =
h1, 6, 12i and the directional derivative of f in that direction is given by
p
√
−k∇f (−1, 3, −2))k = − 12 + (6)2 + (12)2 = − 181 .
Continued. . .
Final Exam / Math 2Q04
-14-
NAME:
ID #:
7. (4 pts.) Let C be the closed curve parametrized by
x = cos(t),
y = sin t +
1
cos(2 t),
2
0 ≤ t ≤ 2π
and let D be the region enclosed by the curve C (see the picture below). Use Green’s theorem
to compute the area A of the region D.
→ (A) A = π
(B) A =
3π
2
(C) A =
2π
3
(D) A =
7π
6
(E) A =
4π
3
0.5
K
1.0
K
0
0.5
0.5
1.0
K
0.5
K
1.0
K
1.5
Solution. Using Green’s theorem with P (x, y) = 0 and Q(x, y) = x, we obtain
I
ZZ
x dy =
1 dA = A.
C
On the other hand,
I
Z
x dy =
C
Z
2π
Z
0
2π
0
= π.
2π
cos2 t − 2 cos2 t sin t dt
0
·
¸2 π
1 cos(2 t)
t sin(2 t)
cos3 t
2
+
− 2 cos t sin t dt =
+
+2
2
2
2
4
3 0
cos t (cos t − sin(2 t)) dt =
=
Continued. . .
D
Final Exam / Math 2Q04
-15-
NAME:
ID #:
8. (4 pts.) Evaluate the line integral
Z
I=
C
x
ds
1 + 2y
where C is the curve parametrized by
x = et , y = e2 t , z = 2
(A) I =
e3 − e
3
(B) I =
e2 − 1
4
e3 t
,
3
0 ≤ t ≤ 2.
(C) I = e − 1
(D) I =
e2 − e3
2
→ (E) I =
e4 − 1
2
Solution. We have
r(t) = het , e2 t , 2
e3 t
i
3
and
r0 (t) = het , 2 e2 t , 2 e3 t i.
Thus,
p
√
√
(et )2 + (2 e2 t )2 + (2 e3 t )2 = e2 t + 4 e4 t + 4 e6 t = et 1 + 4 e2 t + 4 e4 t
p
= et (1 + 2 e2 t )2 = et (1 + 2 e2 t ).
kr0 (t)k =
Therefore,
Z
C
x
ds =
1 + 2y
Z
Z
2
0
2
=
0
Continued. . .
et
et (1 + 2 e2 t ) dt
2
t
1 + 2e
· 2 t ¸2
e
e4 − 1
2t
e dt =
=
.
2 0
2
Final Exam / Math 2Q04
-16-
NAME:
ID #:
9. (4 pts.) A particle moves on a curve C parametrized by
√
r(t) = h2 + 3 t, 3 − 4 t, 1 + 75 ti,
starting at time t = 0. What is the value of the x-coordinate of the particle after it has
covered a distance of s = 20 units.
(A) x = 3
(B) x = 5
→ (C) x = 8
(D) x = 10
(E) x = 13
√
Solution. We have r0 (t) = h3, −4, 75i for all t and
q
√
√
0
kr (t)k = 32 + 42 + ( 75)2 = 100 = 10.
The distance covered by the particle in the interval of time from 0 to t is
Z t
Z t
0
s=
kr (u)k du =
10 du = 10 t.
0
0
Thus, if s = 20, we have t = 2 and the x-coordinate of the particle is then x = 2+(3) (2) = 8.
Continued. . .
Final Exam / Math 2Q04
-17-
NAME:
ID #:
10. (4 pts.) A parametric surface S is parametrized by
r(u, v) = hu2 − v 2 , u + v, u vi,
−∞ < u, v < ∞.
The equation for the plane tangent to S at the point (−3, 3, 2) is:
(A)
x+y+z =2
→ (B)
(C)
x + 10 y − 6 z = 15
2 x + 3 y − 4 z = −5
(D) 4 x + y + 5 z = 1
(E) −2 x + 5 y + z = 23
Solution. We have
ru = h2 u, 1, vi
and
rv = h−2 v, 1, ui
Thus,
¯
¯
¯ i
¯
j
k
¯
¯
ru × rv = ¯¯ 2 u 1 v ¯¯ = hu − v, −2 (u2 + v 2 ), 2 (u + v)i.
¯−2 v 1 u¯
Since h−3, 3, 2i = r(1, 2), a vector normal to the surface S at the point (−3, 3, 2) is given
by the vector
(ru × rv ) (1, 2) = h−1, −10, 6i.
The equation for the plane tangent to S at the point (−3, 3, 2) is thus
− (x + 3) − 10 (y − 3) + 6 (z − 2) = 0
or, equivalently,
x + 10 y − 6 z = 15 .
Continued. . .
Final Exam / Math 2Q04
-18-
NAME:
ID #:
11. (4 pts.) Suppose that the function f (x, y, z) = x z + ex y + y 2 z 2 is a potential function
for the vector field F(x, y, z) (i. e. ∇f = F). Let C be the path parametrized by the vector
function
­
®
r(t) = 3 − t, 2 + t, t et−1 , 0 ≤ t ≤ 1.
The value of the path integral
Z
F · dr
C
is which of the following:
(A) 2
(B) 5
(C) e3 − 2
→ (D) 11
(E) 3 e + 2
Solution. Note that the beginning point of the path C is (3, 2, 0) while its ending point is
(2, 3, 1). By the fundamental theorem of line integrals
Z
F · dr = f (2, 3, 1) − f (3, 2, 0) = (2 + e6 + 9) − (0 + e6 + 0) = 11 .
C
Continued. . .
Final Exam / Math 2Q04
-19-
NAME:
ID #:
12. (4 pts.) Let V be the solid region bounded by the surfaces z = 2 − x2 and z = y 2 − 2.
The volume of V equals
(A) 6 π
(B) 2 π
(C) 5 π
(D) 10 π
→ (E) 8 π
Solution. Note that the two surfaces intersect when z = 2 − x2 = y 2 − 2. This yields
x2 + y 2 = 4. The solid region V can thus be described as
V = {(x, y, z), x2 + y 2 ≤ 4, y 2 − 2 ≤ z ≤ 2 − x2 }.
Letting D = {(x, y), x2 + y 2 ≤ 4}, we have thus
Z Z (Z
ZZZ
Vol(V ) =
1 dV =
1 dz
V
ZZ
)
2−x2
dx dy
y 2 −2
D
4 − (x2 + y 2 ) dx dy.
=
D
Passing to polar coordinates, this last integral become
Z
2π
Z
2
Z
2
2
(4 − r ) r dr dθ = 2 π
0
0
0
·
¸2
r4
2
4 r − r dr = 2 π 2 r −
= 8π .
4 0
3
2
1
0
−1
−2
−1
−2
2
Continued. . .
1
0
0
1
−1
2
−2
Final Exam / Math 2Q04
-20-
NAME:
ID #:
13. (4 pts.) If spherical coordinates are used to evaluate the triple integral
ZZZ
z dV,
V
p
where V is the solid region bounded below by the cone z = x2 + y 2 and above by the
sphere of radius 1 centered at (0, 0, 1), by which of the following iterated integrals is the
original integral replaced ?
Z
2π
Z
→ (A)
0
Z
2π
0
Z
π
3
(B)
0
2π
Z
2 cos φ
Z
cos φ
ρ3 sin φ cos φ dρ dφ dθ
0
π
Z
cos(2 φ)
(C)
0
0
Z
2π
Z
0
Z
2π
cos φ
2
ρ3 sin φ cos φ dρ dφ dθ
0
π
4
(E)
0
Z
0
Z
ρ3 sin φ cos φ dρ dφ dθ
0
π
6
(D)
ρ3 sin φ cos φ dρ dφ dθ
0
0
Z
Z
π
4
0
Z
2 sin φ
ρ3 sin φ cos φ dρ dφ dθ
0
Solution. The equation of the sphere of radius 1 centered at (0, 0, 1) is x2 + y 2 + (z − 1)2 = 1
or x2 + y 2 + z 2p= 2 z, which becomes ρ2 = 2 ρ cos φ or ρ = 2 cos φ in spherical coordinates.
The cone z = x2 + y 2 has equation φ = π/4 in spherical coordinates. Thus the region V
is expressed in polar coordinates as the region where 0 ≤ θ ≤ 2 π, and 0 ≤ φ ≤ π/4 and
0 ≤ ρ ≤ 2 cos φ. Therefore,
ZZZ
Z
2π
Z
π/4
Z
2 cos φ
z dV ==
V
0
0
ρ3 cos φ sin φ dρ dφ dθ .
0
2.0
1.5
1.0
0.5
0.0
−1.0
Continued. . .
−0.5
0.0
−1.0
−0.5
0.0
0.5
1.0
1.0
0.5
Final Exam / Math 2Q04
-21-
NAME:
ID #:
14. (4 pts.) Compute the volume V of the solid region in the first octant (x, y, z ≥ 0)
bounded by the coordinate planes and the plane 2 x + y + 3 z = 6.
(A) V = 3
(B) V = 4
(C) V = 5
→ (D) V = 6
(E) V = 7
Solution. The region R above can be expressed at
R = {(x, y, z), 0 ≤ x ≤ 3, 0 ≤ y ≤ 6 − 2 x, 0 ≤ z ≤ 2 −
y 2x
−
}.
3
3
Therefore,
ZZZ
Z
V =
3
Z
6−2 x
Z
2− y3 − 23x
1 dV =
R
Z
3
=
Z
0
3
=
0
·
0
0
0
¸6−2 x
Z
3
Z
6−2 x
1 dz dy dx =
2−
0
3
0
y 2x
−
dy dx
3
3
2x
2x
y
(6 − 2 x)2
(2 −
dy =
(2 −
)y −
) (6 − 2 x) −
dx
3
6 y=0
3
6
0
·
¸x=3
−(6 − 2 x)3
(6 − 2 x)2
= 6 .
dx =
6
36
x=0
Continued. . .
2
Z
Final Exam / Math 2Q04
-22-
NAME:
ID #:
15. (4 pts.) Let C be the intersection of the cylinder x2 + y 2 = 1 with the plane y + z = 2.
Then, the curvature of C at the point (0, 1, 1) is:
(A) 1
(B)
1
2
→ (C)
√
(D)
√
2
2
2
(E) 2
Solution. The curve C can be parametrized by the vector function
r(t) = hcos t, sin t, 2 − sin ti,
0 ≤ t ≤ 2 π.
Note that r( π2 ) = h0, 1, 1i.
r0 (t) = h− sin t, cos t, − cos ti with r0
and
³π ´
2
= h−1, 0, 0i
³π ´
= h0, −1, 1i
r00 (t) = h− cos t, − sin t, sin ti with r00
2
√
¡ ¢
¡ ¢
¡ ¢
¡ ¢
¡ ¢
Thus r0 π2 × r00 π2 = h0, 1, 1i. This yield kr0 π2 k = 1 and kr0 π2 × r00 π2 k = 2.
Therefore, the curvature κ of C at the point (0, 1, 1) is
¡ ¢
¡ ¢
√
kr0 π2 × r00 π2 k
2 √
¡π¢
κ=
=
= 2.
13
kr0 2 k3
Continued. . .
Final Exam / Math 2Q04
NAME:
SCRATCH
Continued. . .
-23ID #:
Final Exam / Math 2Q04
-24-
NAME:
ID #:
Some formulas you may use:
T(t) =
r0 (t)
,
kr0 (t)k
N(t) =
T0 (t)
,
kT0 (t)k
v·a
r0 (t) · r00 (t)
aT =
=
,
v
kr0 (t)k
κ(t) =
kT0 (t)k
kr0 (t) × r00 (t)k
=
.
kr0 (t)k
kr0 (t)k3
kv × ak
kr0 (t) × r00 (t)k
aN = κv =
=
v
kr0 (t)k
2
projb a =
a·b
b
kbk2
d
[u(t) r(t)] = u(t) r0 (t) + u0 (t) r(t),
dt
d
d
[r1 · r2 ] = r01 (t) · r2 (t) + r1 (t) · r02 (t),
[r1 × r2 ] = r01 (t) × r2 (t) + r1 (t) × r02 (t),
dt
dt
d
d
(cos t) = − sin t,
(sin t) = cos t.
dt
dt
1 1
1 1
cos2 t = + cos(2t),
sin2 t = − cos(2t).
2 2
2 2
2 cos α cos β = cos(α − β) + cos(α + β), 2 sin α sin β = cos(α − β) − cos(α + β),
et + e−t
et − e−t
2 sin α cos β = sin(α + β) + sin(α − β) cosh t =
sinh t =
2
2
p
ρ = x2 + y 2 + z 2 ,
x = ρ cos θ sin φ, y = ρ sin θ sin φ, z = ρ cos φ
p
r = x2 + y 2 = ρ sin φ
I
ZZ
∂Q ∂P
P dx + Q dy =
−
dA
∂y
C
D ∂x
I
ZZ
F · dr =
(∇ × F) · n dS
C
S
ZZ
ZZZ
F · n dS =
S
Z L
³ nπx ´
2
an =
f (x) cos
dx, 0 < x < L.
L 0
L
Z L
³ nπx ´
2
bn =
f (x) sin
dx, 0 < x < L.
L 0
L
³ nπx ´
a0 X
f (x) =
+
an cos
,
2
L
n=1
∞
f (x) =
∞
X
n=1
*** END ***
bn sin
³ nπx ´
L
∇ · F dV
T
,