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(MATH100)final-1999.pdf
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Math 100 -Introduction to Multivariable Calculus
FINAL EXAMINATION

Fall Semester, 1999
Time Allowed: 2.5 Hours. Total Marks: 100

Student Name:
Student Number:

1. (10 marks) Locate all relative maxima, relative minima and saddle points of the function
f(x, y)= x 3 + y 3 . 3x . 3y.
2. (10 marks) Evaluate the double integral .. 1
dxdy,
D (1 + x2 + y2)2
where

2
D = .x + y 2 R2 .
x 0,y 0. .
3. (15 marks) Evaluate the line integral
. xy
(e 2 + y 2)dx +(e2 + x 2)dy
C
where C is the boundary of the region between y =2x2 and y =2x and is oriented counterclockwise.
4. (15 marks) Evaluate the line integral
(sin x + sin y)dx + (1 + x cos y)dy,
C
where C is the line segment from (0, 0) to (, ).
1
5. (15 marks) Find the .ux of F.across the surface of conical solid bounded by
z = x2 + y2 and z =1,
where

.2.2.2.
F (x, y, z)= xi + yj + z k.
222 2
6.
(15 marks) Find the volume lying inside both the sphere x+y+z= aand the cylinder x2 + y2 = ax, where a is a positive constant.

7.
(20 marks) Use the Divergence Theorem to evaluate the surface integral


F (x, y, z) .n dS,

where
F.(x, y, z)=2x 2.i +(y + z . 4xy).j + .k

and is the surface of the upper-hemisphere
2
z = .1 . x2 . y, z> 0
without the bottom disk (i. e., not including the disk x2 + y2 1 on the xy-plane) and is oriented upward.
1. Locate all relative maxima, relative minima and saddle points of the function f(x, y)= x 3 + y 3 . 3x . 3y. Solution The critical points: fx =3x 2 . 3=0=. x = 1, fx =3y 2 . 3=0=. y = 1. Hence, there are four critical points: (1, 1), (1, .1), (.1, 1), (.1, .1),
Also, fxx =6x, fyy =6y, fxy =0=. D = fxxfyy xy = 36xy.
. f2
Thus,
at (1, 1),D > 0& fxx > 0=. relative minimum,
at (1, .1),D < 0=. saddle point,
at (.1, 1),D < 0=. saddle point,
at (.1, .1),D > 0& fxx < 0=. relative maximum.

2. Evaluate the double integral
.. 1

dxdy,
D (1 + x2 + y2)2
where

2
D = .x + y 2 R2 .
x 0,y 0. .
Solution
.. 1
dxdy
D (1 + x2 + y2)2
. 2 . R 1
= rdrd
00 (1 + r2)2
.11 ..
r=R
= .
2
2 21+ r
r=0
R2
= .
4(1 + R2)
3. Evaluate the line integral
. xy
(e 2 + y 2)dx +(e2 + x 2)dy
C
where C is the boundary of the region between y =2x2 and y =2x and is oriented counterclockwise. Solution By Greens Theorem, we have
. xy
(e 2 + y 2)dx +(e2 + x 2)dy
C
= (2x . 2y)dxdy
R
. 1 . 2x 1
= (2x . 2y)dydx = . .
2
02x5
4. Evaluate the line integral
(sin x + sin y)dx + (1 + x cos y)dy,
C
where C is the line segment from (0, 0) to (, ).
Solution Since

..
(sin x + sin y)= (1+ x cos y),
.y .x
we know that the vector .eld
(sin x + sin y).i + (1 + x cos y).j
is conservative. The potential function satis.es

.
. .x = sin x + sin y, . = 1+ x cos y.
.y
Solving the .rst equation, we have
= . cos x + x sin y + h(y).
Substituting this into the second equation for to have

x cos y +