1. No category

LAST (family) NAME:
FIRST (given) NAME:
ID # :
MATHEMATICS 2Q04: SAMPLE FINAL EXAM
McMaster University Final Examination
Day Class
Duration of Examination: 3 hours
SAMPLE FINAL EXAM 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. (a) (8 pts.) For a certain constant A, the vector field
F(x, y, z) = (4 x3 y 2 + 8 x y z) i + (A x4 y + 4 x2 z) j + (4 x2 y + 8 z) k
is conservative. Find the constant A and, for that particular value of A, express F(x, y, z)
as the gradient of some scalar function f (x, y, z).
Solution. We have
¯
¯
¯
¯
i
j
k
¯
¯
∂
∂
∂
¯
¯ = (0) i + (0) j + (4 A x3 y − 8 x3 y) k
∇×F=¯
∂x
∂y
∂z
¯
¯ 4 x3 y 2 + 8 x y z A x 4 y + 4 x2 z 4 x2 y + 8 z ¯
For F to be conservative, we need ∇ × F = 0, i.e. A = 2. When A = 2, we can thus write
F = ∇f , for some scalar function f or, equivalently
∂f
= 4 x3 y 2 + 8 x y z
∂x
∂f
= 2 x4 y + 4 x2 z
∂y
∂f
= 4 x2 y + 8 z
∂z
(1)
(2)
(3)
Using (1), we have f (x, y, z) = x4 y 2 + 4 x2 y z + C(y, z). Using this equation together with
(2) yields
∂f
∂C
= 2 x4 y + 4 x2 z +
= 2 x4 y + 4 x2 z
∂y
∂y
= 0, i.e. C(y, z) = C(z). Thus, f (x, y, z) = x4 y 2 + 4 x2 y z + C(z) and using
and thus ∂C
∂y
this last equation together with (3), we obtain
∂f
= 4 x2 y + C 0 (z) = 4 x2 y + 8 z,
∂z
which yields C 0 (z) = 8 z or C(z) = 4 z 2 + B, B a constant. Taking B = 0, we have thus
f (x, y, z) = x4 y 2 + 4 x2 y z + 4 z 2 .
Continued. . .
Final Exam / Math 2Q04
-3-
NAME:
ID #:
(b) (4 pts.) For the conservative vector field obtained in part (a) and for the arc of a curve
C parametrized by
r(t) = (cos 2t) i + (cos 3t) j + (cos2 4t) k,
evaluate
0 ≤ t ≤ π,
Z
F · dr.
C
Solution. Since the beginning point is r(0) = (1, 1, 1) and the ending point is r(π) =
(1, −1, 1), we have, using (a),
Z
F · dr = f (1, −1, 1) − f (1, 1, 1) = 1 − 9 = −8.
C
Continued. . .
Final Exam / Math 2Q04
-4-
NAME:
ID #:
2. (a) (6 pts.) Let R be the solid region above the x, y plane and below the paraboloid
z = 9 − x2 − y 2 . Let S be the closed surface which forms the boundary of R, oriented using
the outward pointing normal. Note that S consists of two pieces, one which is part of the
paraboloid and the other one which is part of the x, y plane. Consider the vector field
F(x, y, z) = (−y) i + (x) j + (z + 1) k.
Compute the integral
ZZ
F · n dS
S
using the definition of the integral of a vector-field on a surface.
Solution. Note that the paraboloid intersect the x, y plane on the circle x2 + y 2 = 9, z = 0.
Let S1 be the part of S on the paraboloid and S2 the part of S on the x, y plane. We can
parametrize S1 by r1 (x, y) =< x, y, 9 − x2 − y 2 > where (x, y) ∈ D = {(x, y), x2 + y 2 ≤ 9}.
We have
¯
¯
¯i j
¯
k
¯
¯
¯
N1 (x, y) = ¯1 0 −2 x¯¯ = (2 x) i + (2 y) j + k (outward pointing)
¯0 1 −2 y ¯
and
ZZ
ZZ
< −y, x, 10 − x2 − y 2 > · < 2 x, 2 y, 1 > dx dy
ZZ
Z 2π Z 3
2
2
(10 − r2 ) r dr dθ
=
10 − x − y dx dy =
0
0
D
¸3
·
Z 3
99 π
r4
3
2
=
= 2π
10 r − r dr = 2 π 5 r −
.
4 0
2
0
F · n dS =
S1
D
Also, we can parametrize S2 ¯by r2 (x,¯y) =< x, y, 0 > where (x, y) ∈ D and D is the same as
¯ i j k¯
¯
¯
above. We have N2 (x, y) = ¯¯1 0 0 ¯¯ = k (inward pointing) and, after changing the sign of
¯0 1 0 ¯
N2 ,
ZZ
ZZ
F · n dS =
< −y, x, 1 > · < 0, 0, −1 > dx dy
S2
D
ZZ
Z
=−
Z
3
1 dx dy = −
D
Continued. . .
2π
0
0
·
r2
r dr dθ = −2π
2
¸3
= −9 π.
0
Final Exam / Math 2Q04
-5-
NAME:
ID #:
Therefore,
ZZ
ZZ
ZZ
F · n dS =
F · n dS +
S
F · n dS =
S1
S2
99 π
81 π
− 9π =
.
2
2
(b) (6 pts.) Compute the surface integral in part (a) using the divergence theorem.
Solution. By the divergence theorem,
ZZ
ZZZ
ZZZ
F · n dS =
∇ · F dV =
1 dV
S
R
R
Since R = {(x, y, z), (x, y) ∈ D, 0 ≤ z ≤ 9 − x2 − y 2 },
ZZZ
ZZ Z
ZZ
9−x2 −y 2
1 dV =
R
D 0
2π Z 3
Z
=
0
0
3
Z
= 2π
0
Continued. . .
9 − x2 − y 2 dA
1 dz dA =
D
(9 − r2 ) r dr dθ
·
r2 r4
9 r − r dr = 2 π 9 −
2
4
¸3
3
=
0
81 π
.
2
Final Exam / Math 2Q04
-6-
NAME:
ID #:
3. Let D be the region in the first quadrant of the x, y plane (i.e. the region where x, y > 0)
where x3 ≤ y ≤ 2 x3 and 1 ≤ x y ≤ 5. Consider the change of variables
½ y
=u
x3
xy = v
(a) (5 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.
Solution. The equation x3 ≤ y ≤ 2 x3 is equivalent to 1 ≤
equation 1 ≤ x y ≤ 5 is equivalent to 1 ≤ v ≤ 5. Hence,
y
x3
≤ 2 or 1 ≤ u ≤ 2. The
D∗ = {(u, v), 1 ≤ u ≤ 2, 1 ≤ v ≤ 5}.
We have y =
v
x
and thus u =
y
x3
=
v
.
x4
Hence, x4 = uv . Since x > 0 on D,
x = u−1/4 v 1/4
Continued. . .
and y =
v
= u1/4 v 3/4 .
x
Final Exam / Math 2Q04
NAME:
-7ID #:
(b) (7 pts.) Use the change of variable in part (a) to compute the area of the region D.
Solution. We have
¯ ∂x ∂x ¯ ¯ 1 −5/4 1/4 1 −1/4 −3/4 ¯
¯
¯ ¯
∂(x, y) ¯¯ ∂u
v
u
v
∂v ¯ = ¯− 4 u
4
¯
= ¯ ∂y ∂y
1 −3/4 3/4
3 1/4 −1/4 ¯
¯
¯
u
v
u
v
∂(u, v)
4
4
∂u
∂v
3 −1
1 −1
1 −1
=− u −
u =− u .
16
16
4
By the change of variables formula,
¯
ZZ
ZZ ¯
¯ ∂(x, y) ¯
¯
¯ du dv
Area(D) =
1 dx dy =
¯
¯
D
D∗ ∂(u, v)
Z 2Z 5
1 −1
=
u dv du = [ln(u)]21 = ln(2) − ln(1) = ln(2).
1
1 4
Continued. . .
Final Exam / Math 2Q04
-8-
NAME:
ID #:
4. (12 pts.) Let S be the hemisphere given by x ≥ 0, x2 + y 2 + z 2 = 9, oriented so that the
normal points away from (0, 0, 0).
Let C be the boundary of S, positively oriented relative to S. For the vector field F(x, y, z) =
(−y 2 z) j + (y z 2 ) k, use Stokes’ theorem to calculate the line integral
I
F · dr.
C
Hint: Is there a simpler looking surface with the same boundary C?
Warning: Do not waste your time calculating the line integral directly. No credit will be
given for that.
Solution. We have
¯
¯
¯i
¯
j
k
¯∂
¯
∂
∂ ¯
2
2
¯
∇ × F = ¯ ∂x
∂y
∂z ¯ = (y + z ) i.
¯ 0 −y 2 z y z 2 ¯
The boundary of S is the circle y 2 + z 2 = 9 on the y, z plane (x = 0) and the positive
orientation corresponds to counterclockwise motion in the y, z plane. That same circle is
also the boundary of the surface S1 consisting of the part of the y, z plane inside C, oriented
so that the normal point in the direction of the positive x-axis. We can parametrize S1 by
r1 (y, z) =< 0, y, z > where (y, z) ∈ D = {(y, z), y 2 + z 2 ≤ 9}. We have
¯
¯
¯ i j k¯
¯
¯
N1 (x, y) = ¯¯0 1 0 ¯¯ = i.
¯0 0 1 ¯
Using Stokes’theorem we have thus
I
ZZ
ZZ
F · dr =
(∇ × F) · n dS =
< y 2 + z 2 , 0, 0 > · < 1, 0, 0 > dy dz
C
S
D
ZZ 1
Z 2π Z 3
Z 2π Z 3
2
2
2
=
y + z dy dz =
r r dr dθ =
r3 dr dθ
D
0
0
0
0
· 4 ¸3
4
81 π
r
3
=
.
= 2π
= 2π
4 0
4
2
Continued. . .
Final Exam / Math 2Q04
NAME:
-9ID #:
If the original
p surface S is chosen instead for the computation, we can parametrize it with
r(y, z) =< 9 − y 2 − z 2 , y, z > where (y, z) ∈ D and D is as above. We have
¯
¯
¯
i
j k¯
¯
¯ −y
y
z
¯ √ 2 2 1 0¯
N(x, y) = ¯ 9−y −z
) j + (p
) k.
¯ = i + (p
2
2
¯
¯ √ −z
9−y −z
9 − y2 − z2
0
1
¯
¯
2
2
9−y −z
Using Stokes’theorem we have thus
I
ZZ
F · dr =
(∇ × F) · n dS
C
S1
ZZ
z
y
=
< y 2 + z 2 , 0, 0 > · < 1, p
,p
> dy dz
9 − y2 − z2
9 − y2 − z2
D
ZZ
81 π
=
y 2 + z 2 dy dz =
,
2
D
as before.
Continued. . .
Final Exam / Math 2Q04
-10-
NAME:
ID #:
5. (12 pts.) Find the solution u(x, y) of the Laplace equation
∂ 2u ∂2u
+
= 0 on the rectangle
∂x2 ∂y 2
0 ≤ x ≤ 3, 0 ≤ y ≤ 2 subject to the boundary conditions
½
u(0, y) = 0, u(3, y) = sin( 5 π2 y ),
0 ≤ y ≤ 2,
u(x, 0) = 0, u(x, 2) = 0,
0 ≤ x ≤ 3.
∂ 2u ∂ 2u
+
= 0 that are of the
∂x2 ∂y 2
form u(x, y) = X(x) Y (y) and satisfy the boundary conditions
Solution. We first look for non-trivial solutions u(x, y) of
u(x, 0) = 0, u(x, 2) = 0,
0 ≤ x ≤ 3.
(4)
This yields the equations
X 00 (x) Y (y) + X(x) Y 00 (y) = 0,
and, dividing both sides by X(x) Y (y), we obtain
X 00 (x) Y 00 (y)
+
= 0 or
X(x)
Y (y)
X 00 (x)
Y 00 (y)
=−
X(x)
Y (y)
Since the left-hand side of the previous equation is a function of x only while the right-hand
side is a function of y only, we deduce that they must both be equal to the same constant
C. We have thus
X 00 (x)
Y 00 (y)
=−
=C
X(x)
Y (y)
The boundary conditions (4) are equivalent to X(x) Y (0) = X(x) Y (2) = 0, for 0 ≤ x ≤ 3,
and show that Y (0) = Y (2) = 0. We thus need to solve the boundary-value problem
½ 00
Y (y) + CY (y) = 0,
Y (0) = Y (2) = 0.
Non-trivial solutions only exist when C = ( nπ
)2 and are of the form Yn (y) = Cn sin( nπy
),
2
2
2
for each integer n ≥ 1. The equation X 00 (x) − ( nπ
)
X(x)
=
0
has,
for
each
n
≥
1,
solutions
2
of the form
nπx
nπx
) + Bn sinh(
)
Xn (x) = An cosh(
2
2
Continued. . .
Final Exam / Math 2Q04
-11-
NAME:
ID #:
We have thus obtained particular solutions of the PDE of the form
n
nπx
nπx o
nπy
un (x, y) = An cosh(
) + Bn sinh(
) sin(
)
2
2
2
and that satisfy also (4). Using the superposition principle, the series
u(x, y) =
∞ n
X
An cosh(
n=1
nπx
nπx o
nπy
) + Bn sinh(
) sin(
)
2
2
2
is also a solution of the PDE that satisfy (4). The condition u(0, y) = 0 for 0 ≤ y ≤ 2 yields
then
∞
X
nπy
), 0 ≤ y ≤ 2,
0=
An sin(
2
n=1
which shows that An = 0, for all n ≥ 1 and thus that
u(x, y) =
∞
X
n=1
Bn sinh(
nπx
nπy
) sin(
)
2
2
The condition u(3, y) = sin( 5 π2 y ) for 0 ≤ y ≤ 2 yields then
∞
X
5πy
nπ3
nπy
sin(
)=
Bn sinh(
) sin(
),
2
2
2
n=1
0 ≤ y ≤ 2.
By inspection (or, by an explicit computation of the coefficient in the sine Fourier series on
15π
the right-hand
we
£ side),
¤ find that Bn = 0, for n ≥ 1 and n 6= 5 and that B5 sinh( 2 ) = 1,
15π
i.e. B5 = 1/ sinh( 2 ) .
The solution is thus given by:
) sin( 5πy
)
sinh( 5πx
2
2
u(x, y) =
.
sinh( 5π3
)
2
Continued. . .
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 #
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
-13-
NAME:
ID #:
6. (4 pts.) Compute the directional derivative of the function
f (x, y, z) = 3 x ey
2
z
−
2z
xy
at the point ( 12 , 2, 0) in the direction in which f (x, y, z) increases the fastest at that point.
Its value is
(A) 1
→ (B) 5
(C) 7
(D) 9
(E) 11
(ENTER YOUR ANSWERS ON THE CHART ON PAGE 12.)
Solution. We have
∇f (x, y, z) = (3 ey
2
z
+
2z
2z
2
2
2
) i + (6 x y z ey z +
) j + (3 x y 2 ey z −
)k
2
2
x y
xy
xy
and
∇f (1/2, 2, 0) = 3 i + 4 k.
The directional derivative in the direction of fastest increase at (1/2, 2, 0) is
√
√
k∇f (1/2, 2, 0)k = 32 + 42 = 25 = 5.
Continued. . .
Final Exam / Math 2Q04
-14-
NAME:
ID #:
7. (4 pts.) Let C1 and C2 denote the circles x2 + y 2 = 1 and x2 + y 2 = 4, respectively, both
being described in the counterclockwise direction. Let
P (x, y) = y 5 + 4 y x4
and Q(x, y) = x5 + 6 x y 4 .
Use Green’s theorem to compute
Z
Z
P dx + Q dy −
C2
P dx + Q dy.
C1
The answer is:
(A)
0
(B) −
6π
13
(C)
21 π
3
(D)
9π
7
→ (E)
63 π
4
Solution. By Green’s theorem,
Z
Z
ZZ
∂Q ∂P
−
dx dy
P dx + Q dy −
P dx + Q dy =
∂y
C2
C1
D ∂x
ZZ
ZZ
4
4
4
4
=
(5 x + 6 y ) − (5 y + 4 x ) dx dy =
x4 + y 4 dx dy,
D
D
where D = {(x, y), 1 ≤ x2 + y 2 ≤ 4}. Passing to polar coordinates
ZZ
Z 2π Z 2
4
4
x + y dx dy =
r4 (cos4 θ + sin4 θ) r dr dθ
D
0
1
µZ 2π
¶ · 6 ¸2
2
(1 + cos(2 θ))
(1 − cos(2 θ))2
r
=
+
dθ
4
4
6 1
0
¶
µ
¶
µZ 2π
Z
2π
21
3 1
21
63 π
(1 + cos2 (2 θ))
dθ
=
+ cos(4 θ) dθ
=
.
=
2
2
4 4
2
4
0
0
Continued. . .
Final Exam / Math 2Q04
-15-
NAME:
ID #:
8. (4 pts.) A curve in the x, y, z space is given parametrically by
x = t + 1, y = 3 t2 + 2, z = 6 t3 + 3.
The length of its arc from the point (1, 2, 3) to the point (2, 5, 9) is:
(A)
389
5
(B) 18
√
(C) 46
→ (D) 7
(E) 11
Solution. Let r(t) =< t + 1, 3 t2 + 2, 6 t3 + 3 >. Then,
r0 (t) =< 1, 6 t, 18 t2 >
and
p
p
1 + (6 t)2 + (18 t2 )2 = 1 + 2 (18 t2 ) + (18 t2 )2
p
= (1 + 18 t2 )2 = 1 + 18 t2 .
kr0 (t)k =
Since (1, 2, 3) = r(0) and (2, 5, 9) = r(1), the length of the arc of curve from the point (1, 2, 3)
to the point (2, 5, 9) is
Z 1
Z 1
£
¤1
0
kr (t)k dt =
1 + 18 t2 dt = t + 6 t3 0 = 7.
0
Continued. . .
0
Final Exam / Math 2Q04
-16-
NAME:
ID #:
9. (4 pts.) The function z(u, v) is given as the composition
z=
x
, where x = u cos v and y = u sin v.
x + 2y
A correct expression for the partial derivative
→ (A)
∂z
is:
∂v
2x
2y
(−u sin v) −
(u cos v)
2
(x + 2 y)
(x + 2 y)2
(B)
cos v
cos v + 2 sin v
(C)
x
x
(−u sin v) +
(u cos v)
x + 2y
x + 2y
(D)
−2
(cos v + 2 sin v)3
(E) none of the above
Solution. We have, by the chain rule,
∂z
∂z ∂x ∂z ∂y
=
+
∂v
∂x ∂v ∂y ∂v
2y
2x
=
(−u sin v) −
(u cos v) .
2
(x + 2 y)
(x + 2 y)2
Continued. . .
Final Exam / Math 2Q04
-17-
NAME:
ID #:
10. (4 pts.) A surface S is given by the equation
x2 + z y 3 + z 4 = 6.
The equation for the plane tangent to S at the point (−2, 1, 1) is:
x+2
y−1
z−1
=
=
−4
3
5
(A)
→ (B)
(C)
−4 x + 3 y + 5 z = 16
x = −2 + 4 t, y = 1 − 3 t, z = 1 − 5 t
(D) −4 x + 3 y + 5 z = 0
(E) 4 (x − 2) − 3 (y + 1) − 5 (z + 1) = 0
Solution. Let f (x, y, z) = x2 + z y 3 + z 4 . We have
∇f (x, y, z) = (2 x) i + (3 z y 2 ) j + (y 3 + 4 z 3 ) k
and
∇f (−2, 1, 1) = (−4) i + (3) j + (5) k.
The equation for the plane tangent to S at the point (−2, 1, 1) is thus
(−4)(x + 2) + 3 (y − 1) + 5 (z − 1) = 0
or
−4 x + 3 y + 5 z = 16.
Continued. . .
Final Exam / Math 2Q04
-18-
NAME:
ID #:
11. (4 pts.) Let f (x, y) be a continuous function. After a correct change of order of
integration, the double integral
Z
1
Z
ln(1+x)
f (x, y) dy dx
0
becomes:
Z ln(1+x) Z
(A)
x ln(2)
Z
1
f (x, y) dx dy
0
Z
ln 2
x ln(2)
1
(B)
f (x, y) dx dy
0
0
Z
ln 2
Z
y/ ln 2
→ (C)
f (x, y) dx dy
ey −1
0
Z
1
Z
ln(1+y)
f (x, y) dx dy
(D)
0
y ln(2)
(E) none of the above
Solution. The domain of integration in the x, y plane is the region
D = {(x, y), 0 ≤ x ≤ 1, x ln(2) ≤ y ≤ ln(1 + x)}
= {(x, y), 0 ≤ y ≤ ln(2), ey − 1 ≤ x ≤ y/ ln(2)}.
Therefore,
Z
1
Z
Z
ln(1+x)
ln 2
Z
y/ ln 2
f (x, y) dy dx =
0
Continued. . .
x ln(2)
f (x, y) dx dy.
0
ey −1
Final Exam / Math 2Q04
-19-
NAME:
ID #:
12. (4 pts.) Let V be the solid region bounded by the surfaces
z = 18 − 2 [(x − 1)2 + y 2 ] and z = 6 + (x − 1)2 + y 2 .
The volume of V can be obtained by computing which one of the following double integrals:
Z
Z √4−(x−1)2
3
→ (A)
−1
Z
(B)
0
−
(C)
√
−1
−
(D)
−1
Z
4
−
0
√
−
√
12 − 2(x − 1)2 − 4 y 2 dy dx
4−(x−1)2
Z √6+(x−1)2
(E)
12 − 3(x − 1)2 − 3 y 2 dy dx
2−(x−1)2
Z √4−(x−1)2
3
12 − 3(x − 1)2 − 3 y 2 dy dx
4−(x−1)2
Z √5−(x−1)2
3
Z
√
12 − 3(x − 1)2 − 3 y 2 dy dx
4−(x−1)2
Z √4−(x−1)2
4
Z
−
√
12 − 3(x − 1)2 − 3 y 2 dy dx
18−(x−1)2
Solution. The two surfaces intersect when
18 − 2 [(x − 1)2 + y 2 ] = 6 + (x − 1)2 + y 2
i.e. on the circle (x − 1)2 + y 2 = 4 in the plane z = 10. Letting
D = {(x, y), (x − 1)2 + y 2 ≤ 4}
p
p
= {(x, y), −1 ≤ x ≤ 3, − 4 − (x − 1)2 ≤ y ≤ 4 − (x − 1)2 },
we can express V as
V = {(x, y, z), (x, y) ∈ D, 6 + (x − 1)2 + y 2 } ≤ z ≤ 18 − 2 [(x − 1)2 + y 2 ]
and thus
ZZZ
vol(V ) =
Z
3
=
−1
Continued. . .
ZZ
1 dV =
V
Z √4−(x−1)2
√
−
4−(x−1)2
12 − 3 [(x − 1)2 + y 2 ] dA
D
12 − 3(x − 1)2 − 3 y 2 dy dx.
Final Exam / Math 2Q04
NAME:
-20ID #:
13. (4 pts.) Use spherical coordinates to evaluate the triple integral
ZZZ
z
dV,
2
2
2 2
V 1 + (x + y + z )
where V is the solid region above the x, y plane between the spheres x2 + y 2 + z 2 = 1 and
x2 + y 2 + z 2 = 4. The answer is:
(A)
π
6
→ (B)
π
ln
4
µ
17
2
¶
(C) 0
¶
1
(D)
1 + (34)2
µ
¶
5
π
(E) −
6 (1 + (25)2 )2
π
8
µ
Solution. In spherical coordinates, V can be expressed at the region where
0 ≤ θ ≤ 2 π,
We have thus
ZZZ
0 ≤ φ ≤ π/2 1 ≤ ρ ≤ 2.
Z 2 π Z π/2 Z 2
z
ρ cos(φ) 2
dV =
ρ sin(φ) dρ dφ dθ
2
2
2
2
1 + ρ4
V 1 + (x + y + z )
0
0
1
ÃZ
! µZ
¶
π/2
2
ρ3
= 2π
sin(φ) cos(φ) dφ
dρ
4
0
1 1+ρ
·
¸π/2 ·
¸2
ln(1 + ρ4 )
cos(2 φ)
= 2π −
4
4
0
1
1 ln(17) − ln(2)
= 2π( )(
)
2µ ¶ 4
π
17
= ln
4
2
Continued. . .
Final Exam / Math 2Q04
-21-
NAME:
ID #:
14. (4 pts.) Let S be the part of the cylinder x2 + y 2 = 1 above the x, y plane between the
planes z = 0 and z = y. Compute the surface area of S using a suitable parametrization for
S. The answer is:
(A) 4 π
(B)
π
2
(C)
3π
4
→ (D) 2
√
(E)
2
Solution.
The cylinder can be parametrized by r(θ, z) =< cos θ, sin θ, z >, where
(θ, z) ∈ D = {(θ, z), 0 ≤ θ ≤ π, 0 ≤ z ≤ sin θ}.
We have
¯
¯
¯ i
j
k¯¯
¯
N(θ, z) = ¯¯− sin θ cos θ 0 ¯¯ = cos θ i + sin θ j
¯ 0
0
1¯
and thus |N(θ, z)| = 1
Z
Area(S) =
Z
1 dS =
Z π Z sin θ
S
=
0
Continued. . .
0
|N(θ, z)| dθ dz
Z π
1 dz dθ =
sin θ dθ = [− cos θ]π0 = 2.
D
0
Final Exam / Math 2Q04
-22-
NAME:
ID #:
15. (4 pts.) Let V be the solid region defined by the inequality
z 2 + (x2 + y 2 )2 ≤ 1.
(Note that V is not a ball !). If the mass density at a point (x, y, z) in V is given by
ρ(x, y, z) = x2 + y 2 ,
compute the total mass of V .
Hint: Use cylindrical coordinates.
The answer is:
(A) π
(B)
π
2
→ (C)
(D)
3π
4
(E)
5π
4
2π
3
Solution. In cylindrical coordinates, the region can be described by the inequality z 2 +r4 ≤ 1
or
√
√
0 ≤ θ ≤ 2 π, 0 ≤ r ≤ 1, − 1 − r4 ≤ z ≤ 1 − r4 ,
Thus
ZZZ
Z
2
mass(V ) =
V
0
1
= 2π
2r
0
Continued. . .
Z
1
x + y dV =
Z
2π
=
3
2π
2
3
√
0
·
Z
√
1−r4
√
− 1−r4
r2 r dz dr dθ
(1 − r4 )3/2
4
1 − r dr = 2π −
3
¸1
0
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
+
an cos
,
f (x) =
2
L
n=1
∞
f (x) =
∞
X
n=1
*** END ***
bn sin
³ nπx ´
L
∇ · F dV
T
,