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(MATH102)[2007](s)midterm~4660^_65263.pdf
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HKUST MATH 102
Third Midterm Examination

Multivariable and Vector Calculus
12 April 2007
Answer all .ve questions Time allowed C 120 minutes
Directions C This is a closed book examination. No talking or whispering are allowed. Work must be shown to receive points. An answer alone is not enough. Please write neatly. Answers which are illegible for the grader cannot be given credit.
Note that you can work on both sides of the paper and do not detach pages from this exam packet or unstaple the packet.
Student Name:
Student Number:

Tutorial Session:

Question No. Marks 1 /20 2 /20 3 /20 4 /20 5 /20 Total /100
22 2 y2 y
The intersection of the two surfaces (elliptic cylinders) x+ =1and z+ = 1 consists of
22
two curves.

(a)
Parameterize each curve in the form r(t)=(x(t),y(t),z(t)).

(b)
What is the arc length of one of the curves?

(c)
Find the volume bounded by the two elliptic cylinders above.


2 2 y

(d) Find the surface area of the part of the elliptic cylinder z + = 1 that lies within the elliptic
2
2
2 y

cylinder x+ =1.
2

Solution:
2 y2

(a) Fix .rst x(t), y(t) to satisfy x+ = 1. i.e.
2
x = cos t
y = 2 sin t
z = cos t
. r(t) = (cos t, 2 sin t, cos t), where 0 . t< 2.
or r(t)=(t, 2(1 . t2), t), where . 1 . t< 1.


(b) r.(t)=(. sin t, 2 cos t, sin t)
2.
s = r .(t) dt
0
2. . = sin2 t + 2 cos2 t + sin2 t dt 0
2.
=2 dt
0
=2. 2.

(c)
21.y2 /21.y2 /2
V =8 dz dxdy
00 0
21.y2 /2
y2 =8 1 . dx dy 00 2
22
y
=8 1 . dy 0 2

Problem 1 (20 points) Your Score:
.. 2
3
y
=8 y .
6
0
=16 2/3=7.54.

(d) The required surface of the elliptic cylinder above the xy-plane is
12
2
y
z =1 .
2
zx =0
12
..
12
..
2 2
1

11

y

y

1 .


. 2y = . y 1 .

zy =
22 2

2

2

22
y4 . y
. 1+ z 2 + z 2 =1+ = .
xy
4 . 2y2 4 . 2y2 Therefore, the surface area above the xy-plane (with x,y > 0) is


4 . y2
S = dA
4 . 2y2
R
21.y2 /2
4 . y2
= dx dy
4 . 2y2
00
2
4 . y2 y2
=1 . dy
4 . 2y2
0 2
1 2
=4 . y2 dy
2 0
Let y = 2 sin ., then
.1 .

12 .
S =4 . 4 sin2 . 2 cos . d., where .1 = sin.1 =
2 0 24
.1
= 2 cos 2 . d.
0
.1
= (1+cos2.) d.
0
...1
1
= . + sin 2.
2
0
1
= .1 + sin 2.1
2
. 1

=+
42 Therefore, the required total surface is 8S =2. + 4. C3C

2
82 2 x
ye

(a) Evaluate dx dy by reverse the order of integration.
x8
0 y1/3

(b) If Q =[.1, 1] [0, 2], evaluate the double integral |y . x2| dx dy, given that it exists.
Q
(Hint: reverse the order of integration)

Solution:
(a) This type II integral cannot be computed as it is. We write it as a type I integral: from the
1/33

boundary relation x = ywe obtain y = xand y = 8 corresponds to x = 2:
2
2 x 3 2x
yedy dx x8
00
29 x 2
xe
= dx
x8
0 3
2
x 2
x
= e dx
0 3

2
2
1 .
x
= e
6
0
e4 . 1
= .
6
2

(b) If we integ