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(math100)[2009](f)final~kwwongaa^_10410.pdf
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Math100 Introduction to Multivariable Calculus
Fall 2009
C Final Examination C

Name:
Student ID:

Lecture and Tutorial Section:

.
There are FIVE questions in this midterm examination.

.
Answer all questions.

.
You may write on both sides of the paper if necessary.

.
You may use a HKEA approved calculator. Calculators with symbolic calculus functions are not allowed.

.
The full mark is 100.




Question Points
Q1
Q2
Q3
Q4
Q5

Total
1. Let R be the circular disk: x2 + y2 1 which lies on the xy-plane, and let S be the portion of the paraboloid z =1+ x2 + y2 that lies above the circular disk R. Find the area of S by evaluating a suitable surface integral.
[20 points]
z
yx
Solution:
The equation of the surface is z = f(x, y)=1+ x 2 + y 2 . Then the desired surface area is given by
.f .f
S = ()2 +( )2 +1 dA =4x2 +4y2 +1 dA.
R .x .y R
We may convert to polar coordinates by letting x = r cos and y = r sin .Since R is a circular region lying on the xy-plane, its polar representation is 0 r 1and 0 2.Then we have
2 1 . 2r=1
1
S = 1+4r2 rdr d = (1+4r 2)3/2 d,
12
00 0 r=0

2 55 . 1
= d,
12
0

(5 5 . 1)
= 5.33.
6
2. Evaluate the triple integral xy dxdydz,where R is the solid region in the .rst octant (i.e,
R
x 0, y 0and z 0) which is enclosed by the planes x = y, x =0, z = 0, and the ellipsoid surface 4x2 +4y2 + z2 = 16. [20 points]
Solution z
V = xy dxdydz,
R
x= 2 y= 4.x2 z=2 4.x2.y2
= xy dz dy dx,
x=0 y=xz=0
y
. 2
x=2 y=4.x
=2xy 4 . x2 . y2 dydx, x=0 y=x
x

. 2
x=2 y=4.x
2
2 23/2
= x . (4 . x . y )dx,
3
x=0 y=x

. x=2 2
2 2)3/2 1 2)5/2 32
= x(4 . 2x dx = . (4 . 2x =
3 15 15
x=0 0
Or by polar cooridnates: x = r cos , y = r sin ,

. 2.y
. x=2 y=4.x2 . z=2 4.x2
V = xy dxdydz, = xy dz dy dx,
Rx=0 y=xz=0

. 2
x=2 y=4.x
=2xy 4 . x2 . y2 dydx,
x=0 y=x

r=2 =/2 . r=2 =/2 . = 2cos sin r3 4 . r2 d dr , = sin2 r 3 4 . r2 dr r=0 =/4 r=0 =/4
r=2 . 2 u=0
1 3 u=4.r 1 1/2
= r 4 . r2 dr = . (4 . u)u du
24
r=0 u=4
. u=0
21 5 32
3/2
= . u + u 2 =
3 10 15
u=4
3. Consider the following vector .elds
x + y
(i) .F(x, y)= .i +(y +tan.1 x).j ;
1+ x2
.2 .
(ii) G(x, y, z)= zx.i +(xy + z).j + yzk .
d 1
(Recall that tan.1 x = 2 .)
dx 1+x
(a)
Determine whether the given vector .elds are conservative vector .elds or not. Must show your reasoning for full credit. [8 points]

(b)
Evaluate the line integral .F d.r, where the oriented curve C is parametrized by


C
.
t+cos t2
.r(t)= e ,t +sin(t2) , 0 t .

which runs from .r(0) to .r(). [12 points]
Solution
(i) is conservative, since
.x + y 1

=
.y 1+ x2 1+ x2
. 1
.1
y +tan x =
2
.x 1+ xare equal. [2 pts for knowing what to check, 2 pts for correct checking]
(ii) is not