Mat 267 Engineering Calculus III Updated on 04/30/2011 Dr. Firoz
Test 1 solution:
1. a) Find the center and radius of the sphere
2 2 2
4 4 4 8 16 1 0x y z x y
.
Show work here:
2 2 2
2 2 2
2 2 2
4 4 4 8 16 1 0
2 4 1/ 4 0
( 1) ( 2) 21/ 4
x y z x y
x y z x y
x y z
The center is at (1, -2, 0), and radius
21
2
b) Find the center and radius of the sphere
2 2 2
4 4 4 24 16 8 44 0x y z x y z
.
2 2 2
2 2 2
2 2 2
4 4 4 24 16 8 44 0
6 4 2 11 0
( 3) ( 2) ( 1) 25
x y z x y z
x y z x y z
x y z
The center is at (3, -2, 1), and radius
5
2. a) Given a vector
4,4,3r
. Find a vector of magnitude 10 in the direction of
the given vector.
Show work here:
10 4,4,3
41
u
b) Find x and y components of a vector of magnitude 8 and an angle of
3
with
the positive x direction.
Show work here:
8cos 4, 8sin 4 3
34
xy
3. a) Test if the following vectors are coplanar:
1,4, 7 , 2, 1, 4ab
and
0, 9,18c
. Use the concept of scalar triple product.
Show work here:
1 4 7
( ) 2 1 4 0,
0 9 18
a b c yes coplanar
b) Test if the following vectors are coplanar:
4 7 , 2 4u i j k v i j k
and
9 18w j k
. Use the concept of scalar triple product.
Mat 267 Engineering Calculus III Updated on 04/30/2011 Dr. Firoz
Show work here:
1 4 7
( ) 2 1 4 0,
0 9 18
u v w yes coplanar
4. Find orthogonal projection of
5, 6, 6P
onto
1, 4,8Q
. Remember
that the orthogonal projection of a vector
b
onto a vector
a
is defined and
denoted by
aa
orth b b proj b
Show work here:
Q2
QQ
29
Proj 1, 4, 8
81
Orth Proj 376/81, 602 /81, 254/81
PQ
PQ
Q
P P P
5. a) Find the area of the parallelogram with vertices P(0, 0, 0), Q(-5, 1, 3),
R(-5, 0, 1) and S(-10, 1, 4).
Show work here:
5, 1, 3 , 5, 0, 1PQ PR
i j k
5 1 3 i 10j 5k
5 0 1
Area=| |= 126 .
PQ PR
PQ PR sq unit
b) Find the area of the triangle with vertices P(5, 4, -5), Q(3, 1, -6), and R(2, 2, -8).
Show work here:
2, 3, 1 , 3, 2, 3PQ PR
i j k
2 3 1 7i 3j-5k
3 2 3
1
Area= | |= 83 / 2 .
2
PQ PR
PQ PR sq unit
6. For what values of x are the vectors
5, 1,ax
and
5, 1,bx
orthogonal?
Show work here: For orthogonality
2
a b = 25+1+ 0x
has no real solution.
Therefore is no such x value we have.
Mat 267 Engineering Calculus III Updated on 04/30/2011 Dr. Firoz
7. a) Find the distance of the point
( 5, 3,1)P
from the line which passes through the
two given points
(0, 2,1)Q
and
( 3,0, 1)R
.
Show work here:
i j k
5 1 0 2i+10j+13k
3 2 2
||
Distance = = 273/17
PQ QR
PQ QR
d unit
QR
b) A man started walking from a point (-3, 1) and reached at the point (-2, 5). The man
started walking in the direction of the vector
5, 2a
and changes his direction
only once, when he turns at a right angle. What are the coordinates of the point
where he makes the turn? Round your answer to two decimal places.
Show work here:
( ) 3 5 ,1 2r t t t
and the distance d is given as
2 2 2
(1 5 ) (4 2 )D d t t
has minimum at
0.4483t
(check it)
The required point where he turns at right angle is ( -0.76, 1.90)
8. A horizontal clothesline is tied between 2 poles, 18 meters apart. When a mass of
2 kilograms is tied to the middle of the clothesline, it sags a distance of 2 meters.
What is the magnitude of the tension on the ends of the clothesline?
Solution:
T | | cos i | |sin j =|T|( 9/ 85i+2/ 85j)
2
W=2 9.8=2T T 45.18
85
TT
T T
2 kg
9. Find the parametric equations of the line through the point
(0,1,2)
that is
perpendicular to the line
1 , 1 , 2x t y t z t
and parallel to the plane
20x y z
.
Show work here:
1,1,1 1, 1,2 3, 1, 2u
The equation of the line is
0,1,2 3, 1, 2
3 , 1 , 2 2
Lt
x t y t z t
Mat 267 Engineering Calculus III Updated on 04/30/2011 Dr. Firoz
10. Consider the path
2
( ) (18 , 9 , 9ln ), 0r t t t t t
. Find the length of the curve
between the points
(72,144,9ln4)P
and
(18,9,0)Q
Solution:
4
1
( ) 18,18 ,9/ , | ( )| 9(2 1/ )
9(2 1/ ) 9(15 ln 4) 147.48
r t t t r t t t
L t t dt
Test 2 solution
1. A) Let
( , ) xy
f x y xe
.
a) Find
(1,1)f
and b)
(1,1)
u
Df
in the direction of
3v i j
, where
||
v
uv
Show work here:
2
( , ) (1 ), , (1,1) 2 , , 3 / 2, 1/ 2
3 1/ 2
xy xy
u
f x y e xy x e f e e u
D f e e
B) Let
( , ) xy
f x y xe
.
a) Find
(1,1)f
and b)
(1,1)
u
Df
in the direction of
3v i j
, where
||
v
uv
Answer: a)
(1,1) 2 ,f e e
b)
(1,1) ( 3 / 2 1)
u
D f e
2. A) Find the equation of the tangent plane to
22
( , ) xy
z f x y e
at the point
(1, 1,1)
.
And approximate the value of
(1.1, 0.98)f
using linear approximation.
Show work here:
( , ) 2 2 1, (1.1, 0.98) 2(1.1) 2( 0.987) 1 1.24f x y x y f
B) Find the equation of the tangent plane to
22
( , ) xy
z f x y e
at the point
(1, 1,1)
. And approximate the value of
(0.97, 1.2)f
using linear approximation.
Show work here:
22
22
2 (1, 1) 2
2 (1, 1) 2
1 2( 1) 2( 1) 2 2 1
(0.97, 1.2) 2(0.97) 2( 1.2) 1 0.54
xy
xx
xy
yy
f xe f
f ye f
z f x y x y
f
Mat 267 Engineering Calculus III Updated on 04/30/2011 Dr. Firoz
3. a) Suppose that
22
cos , ,
r
z e r st s t
. Find
(1,1)
z
t
,
(1,1)
z
s
round
your answer to two decimal places.
Solution:
22
( , ) cos ( sin )
(1,1) 1.47
rr
zt
s t se e
tst
z
t
22
(1,1) cos ( sin ) 1.47
rr
zs
te e
sst
b) Let
( , ) ( ( , ), ( , )),W s t F u s t v s t
where
,Wu
and v are differentiable,
(1,0) 2, (1,0) 2, (1,0) 6, (1,0) 3, (1,0) 5, (1,0) 4
s t s t
u u u v v v
. Further
given that
(2,3) 1, (2,3) 10
uv
FF
. Find
(1,0)
s
W
,
(1,0)
t
W
.
Solution:
(1,0) 1( 2) 10(5) 52
W W u W v
s u s v s
(1,0) 1(6) 10(4) 34
t
WW
t
4. A) Given
22
( , ) 3f x y x xy y
. Find the differential
dz
in terms of x and y.
Show work here:
2 3 3 2 (2 3 ) (3 2 )dz xdx ydx xdy ydy x y dx x y dy
B) Given
22
( , ) 3f x y x xy y
. Find the differential
dz
in terms of x and y.
2 3 3 2 (2 3 ) ( 3 2 )dz xdx ydx xdy ydy x y dx x y dy
5. The dimensions of a closed rectangular box are measured as 90 cm, 50 cm,
and 60 cm respectively with the error in each measurement at most 0.2 cm.
Use differential to estimate the maximum error in calculating the surface area
of the box. Do not forget the unit.
Solution:
2 2 2
2 2 2 2 2 2
2(90 50 90 60 50 60)0.2 160 .
S lw lh wh
dS ldw wdl ldh hdl wdh hdw
sq cm
Mat 267 Engineering Calculus III Updated on 04/30/2011 Dr. Firoz
6. Verify if the function
( , ) sin( )u x t x at
satisfies the wave equation
22
2
22
0
uu
a
tx
.
Show work here:
22
2
2
2
cos( ), sin( )
cos( ), sin( )
uu
a x at a x at
tt
uu
x at x at
xx
22
2
22
0
uu
a
tx
, yes, verifies.
7. Find all the relative maximum and minimum and saddle points on the given
surface
44
( , ) 4 1z f x y x y xy
. Do not forget to check the sign of the
discriminant at each stationary point. Write your answer in the box below.
Answer:
33
2 2 2
4 4 0, 4 4 0,
12 , 12 , 4, [ ]
xy
xx yy xy xx yy xy
f x y f y x
f x f y f D f f f
We have the solution (0, 0), (1, 1) and (-1, -1) as stationary points
Check that saddle point at (0, 0, 1), min at (1, 1) and (-1, -1) and no max.
8. Evaluate the integral
( 2 )
D
x y dA
, where D is the region bounded by the
parabolas
22
2 , 1 .y x y x
You may round your answer to two decimal
places. Put your answer in the box below
Solution:
2
2
11
12
( 2 ) 2 2.13
x
Dx
x y dA x ydydx
9. A) Evaluate the integral by reversing the order of the integral
2
13
03
x
ye dxdy
.
Write your answer in the box below.
Solution:
2 2 2
3 /3 3
13
03 0 0 0
1350.35
3
x
x x x
y
x
e dxdy e dydx e dx
Mat 267 Engineering Calculus III Updated on 04/30/2011 Dr. Firoz
B)
3
4 4 4
3
22
82 3
00 0 0
2221527.38
x
x x x
ye dxdy e dydx x e dx
10. A) Find
2
11
22
10
1
1
xdydx
xy
using polar coordinates. Write answer
in the box below.
22
1xy
Solution:
2/2 4cos
11 2
22
10 00
1128/9 14.22
1
xdydx r drd
xy
22
( 2) 4xy
B)
21
44 22
2
00 00
11.3
1
xx x y dydx rdrd
r
Test3 solution:
Write down the double integral and also triple integral to find the volume of the
tetrahedron bounded by the coordinate planes and the plane
4 4 2z x y
.
Evaluate both the integrals.
2 2
D
22yx
E
x y
x
Complete the limit:
1 2 2
00
(4 4 2 ) 4/3 1.33
x
DD
zdA zdxdy x y dydx
Complete the limit:
4 4 2
1 2 2
0 0 0
4/3 1.33
xy
x
EE
dV dzdydx dzdydx
2. Find the volume of the solid that lies within the sphere
2 2 2 4x y z
, above
the xy-plane, and below the cone
22
z x y
Mat 267 Engineering Calculus III Updated on 04/30/2011 Dr. Firoz
Solution: Draw the picture and identify the region.
/2 2 2 2
/4 0 0
82
sin 11.85
3
E
V dV d d d
3. A sphere has center at (0, 0, 0) and radius equal to m. Use spherical coordinate to
evaluate the volume of the sphere
Solution: the sphere has the equation
2 2 2 2
x y z m
23
23
0 0 0
4
sin 4.19
3
mm
V d d d m
4. Evaluate the line integral
sin cos
C
xdx ydy
, where C is the arc consisting of the
top half of the circle
22
1xy
from (1, 0) to (-1, 0) and the line segment from
(-1, 0) to (-2, 3). Write exact answer, no decimals.
Solution: Check that on
1
C
curve:
cos , sin , 0x t y t t
and
2
C
1
C
on
2
C
curve:
1 , 3 , 0 1x t y t t
12
1
00
sin cos sin cos sin cos
sin(cos )( sin ) cos(sin )(cos ) sin( 1 )( ) cos(3 )3
cos1 cos2 sin3
C C C
xdx ydy xdx ydy xdx ydy
t t dt t t dt t dt t dt
5. a) Show that the line integral
(1 )
xx
C
ye dx e dy
is independent of path, where
C is any path from (0, 1) to (1, 2).
b) Use line integral with parametric representation for C to evaluate integral.
Solution:
1,
x x x
QP
P ye Q e e
xy
, the path C is independent.
1
0
(1 ) (1 (1 ) ) 2/
x x t t
C
ye dx e dy t e e dt e
Where C has the parametric representation:
( ) 0,1 (1,1) , 1r t t x t y t
Mat 267 Engineering Calculus III Updated on 04/30/2011 Dr. Firoz
6. Use Greens’ theorem to evaluate the line integral
4
C
x dx xydy
, where C is the
triangular path consisting of the line segments from (0, 0) to (1, 0), from (1, 0) to
(0, 1), and from (0, 1) to (0, 0).
(0, 1)
Solution:
11
4
00
1/6
x
CD
QP
x dx xydy dA ydydx
xy
1yx
Where
4,QP
P x Q xy y
xy
(0,0) (1,0)
7. Show that the vector field
2 3 3
( , , ) (2 3i )j k
zz
F x y z y xy e ye
is conservative.
Also find the scalar function
( , , )f x y z
such that
grad f F
.
Solution:
2 3 3
23
zz
i j k
Curl F F o
x y z
y xy e ye
the field is conservative
For scalar function we use the following results;
2 3 3
2 , 3
zz
x y z
f y f xy e f ye
2 3 3
( , ) 2 ( , ) 2 , ( , ) ( )
zz
yy
f xy h y z f xy h y z xy e h y z ye g z
Now
2 3 3
( ) 3 ( ) ( )
zz
z
f xy ye g z f ye g z g z k
Thus the scalar function is
23
( , , ) z
f x y z xy ye C
8. Find the flux of the vector field
( , , ) i j kF x y z z y x
across the unit sphere
center at the origin.
Solution: Use
sin cos , sin sin , cos ,0 ,0 2x y z
and
( , ) sin cos sin sin cos ,r xi yj zk i j k
22
sin cos sin sin sin cosr r i j k
( ( , ) cos sin sin sin cosF r i j k
Now
22 3 2
00
( , , )
( ) (2sin cos cos sin sin )
4 /3 4.19
SD
D
F x y z dS F ndS
F r r dA d d
Check the integral.
Mat 267 Engineering Calculus III Updated on 04/30/2011 Dr. Firoz
Quiz 1: February 11, 2011
1. Name
2. Find the equation of the largest sphere center at (2, 4, 6) contained in the first
octant.
Solution: The sphere contained in the first octant must have radius 2 units. The
equation of the sphere is
2 2 2
( 2) ( 4) ( 6) 4x y z
3. A man walks due east on a deck of a ship at 5 miles per hour. The ship is moving
north at a speed of 11 miles per hour. Find the speed and direction (in radians) of
the man relative to the surface of the water.
Solution:
5,0 , 0,11uv
The resultant velocity is
5,0 0,11 5,11w u v
(0,11)
The speed
22
| | 5 11 146w
The direction with north:
2
1
2
11
cos 0.43
146 11
(5,0)
Quiz 2: March 25, 2011
Set A
1. Evaluate
2 3 2 3
22
22
0 3 0 3
9ln5
11
xy x
dydx dx y dy
xx
2. Evaluate
22
3
D
x y dA
, where
{( , )| 0, 1, , 0}D x y x y y x z
1
2 2 2 2
00
3 ( 3 ) 5/6
y
D
x y dA x y dxdy
y = x
3. Evaluate
3
11
0
x
y
e dxdy
by changing the order of integration.
2
33
1 1 1
0 0 0
1
3
x
xx
y
e
e dxdy e dydx
4. Evaluate by using polar coordinate:
2
33 22
00
6
xx x y dydx
, compare with
problem number 10(B) in test 2.
Mat 267 Engineering Calculus III Updated on 04/30/2011 Dr. Firoz
Set B
1. Evaluate
2 3 3 2
22
22
0 3 3 0
9ln5
11
x y y
dxdy x dx dy
yy
2. Evaluate
3
11
0
y
x
e dydx
by changing the order of integration.
3
11
0
1
3
y
x
e
e dydx
compare with set A.
3. Evaluate
22
3
D
x y dA
, where
{( , )| 0, 1, , 0}D x y x y y x z
1
2 2 2 2
00
3 ( 3 ) 5/6
y
D
x y dA x y dxdy
4. Evaluate by using polar coordinate:
2
22 22
00
xx x y dydx
2
2 2 /2 2cos
2 2 2
0 0 0 0
16/9
xx x y dydx r drd
Quiz 3: April 22, 2011
1. Given that
( , , ) i j kr x y z x y z
, use spherical coordinate for a unit sphere
center at origin to find the normal vector
rr
.
Solution:
22
( , , ) i j k
( , ) sin cos sin sin cos
( , ) cos cos cos sin sin
( , ) sin sin sin cos
sin cos sin sin sin cos
r x y z x y z
r i j k
r i j k
r i j
r r i j k
2. Evaluate
S
ydS
, where S is the surface
3/2 3/ 2
2( ), 0 1, 0 1
3
z x y x y
.
Round your answer correct to three decimal places.
Solution:
11
00
1 1 0.733
SD
ydS y y x dA y x ydxdy
Mat 267 Engineering Calculus III Updated on 04/30/2011 Dr. Firoz
3. Find the flux of the vector field
F( , , ) i j kx y z z y x
across the unit sphere
center at the origin.
Use
sin cos , sin sin , cos ,0 ,0 2x y z
and
( , ) sin cos sin sin cos ,r xi yj zk i j k
22
sin cos sin sin sin cosr r i j k
( ( , ) cos sin sin sin cosF r i j k
Now
22 3 2
00
( , , )
( ) (2sin cos cos sin sin )
4 /3
SD
D
F x y z dS F ndS
F r r dA d d
Check the integral.
Pop Quiz
Answers to selected homework problems:
Section: 12.5
10.
4
3
14.
2
15
16.
27
8
18. 144 20.
128
32.
Look at the domains: y z z
2
2z y y
1yx
2
1zx
x x y
A B C
From diagram A: Domain D is xy plane
22
1
1 1 1 1 1
0 0 0 0 0 0
[ ( , , ) ] [ ( , , ) ]
y
x x x
f x y z dz dydx f x y z dz dxdy
From diagram B: Domain D is xz plane
2
1 1 1 1 1 1
0 0 0 0 0 0
[ ( , , ) ] [ ( , , ) ]
x x z x
f x y z dy dzdx f x y z dy dxdz
From diagram C: It has two separate regions for domain D
Mat 267 Engineering Calculus III Updated on 04/30/2011 Dr. Firoz
2
2
1
1 1 1 1 1 1
0 0 0 0 0
11
21
1 1 1 1
0 0 0 0 0
2
[ ( , , ) ] [ ( , , ) ]
[ ( , , ) ] [ ( , , ) ]
y
zz
z
y y y z
yy
f x y z dx dydz f x y z dx dydz
f x y z dx dzdy f x y z dx dzdy
Section: 12.6
8. A hyperboloid of one sheet with axis the z axis.
12. A cone opening upward in the first octant. 16.
7
18.
2
35
20.
65
4
22.
3/2
4(8 3 )
3
28.
162
5
Section: 12.7
8. the surface of a sphere of radius 1 center at (0, 0, 1)
10. a)
2
b)
sec2 cot csc
24.
3
(5 2)
2e
26.
82
3
36. 0
Chapter 13
Section: 13.1
4. Choose different x and y values and the length of the vector, then plot them (look at
example 1 in your text book)
24.
2
cos sin sin
2
y x y xy y
z z z z
26.
11
( ) ( )
22
f x y x y
, plot yourself
Section: 13.2
6.
cos1 cos2 cos3
10.
3/ 2
1(14 1)
6
14. 35/3
16. On C2 negative, on C1 positive 20. 34.
1(15 cos1 cos4)
2
Section: 13.3
2.
6
8.
2
( , ) lnf x y x y x x K
16. a)
( , , ) yz
f x y z e x ze
b) 2e 18. 2/e
Mat 267 Engineering Calculus III Updated on 04/30/2011 Dr. Firoz
Section: 13.4
2.
2
10. 0 14. -16 16. 18.
12
Section: 13.5
4. a)
(sin cos ) cos sinx xz xy y xy z xz
b) 0
8. a) greater than zero b) is zero 10. Try yourself
16.
( , , ) sin cosf x y z xy z K
Section: 13.6
2.Portion of elliptic cylinder
22
1,0 2
49
xy z
16.
22
, , 1 2 4x x y y z x y
22.
cos , sin , 3 cos , 0 1; 0 2x s y s z s s
30.
20x y z
38.
3/2 3/2
1(26 10 )
24
Section: 13.7
6.
4
3
8.
6
6
14.
(8 2)
4
22.
3
24. 0 34
108 2
.
Concept:
1. Evaluation of double integrals and triple integrals
2. Cylindrical and spherical coordinates
3. Vector fields, scalar functions, line integral and fundamental theorem of line
integral, conservative vector field,
4. Greens’ theorem
5. Curl and divergence
6. Parametric surfaces and surface integrals
