MATH 172
EXAM 1
Section 502
Fall 1998
Solutions
P. Yasskin
Multiple Choice: (4 points each)
;
1. Evaluate
a. 2 2=
b. 1 =
c. 1 2=
d. = " 1
2
0
4 " x2
1 dx
by interpretating it as an area.
e. 2 =T correctchoice
3
This is a quarter circle on top of
a rectangle. So
2
;
1
2
4 " x2
0
0
1
1 dx
1 =Ÿ2 2
4
1 =r 2
4
LW
21
=2
2
x and y 1 x 2 for 0 t x t 6
using 3 rectangles with equal widths and with heights given by the function values at
the left endpoints.
a. 17
b. 29
c. 34T correctchoice
d. 46
e. 58
2. Approximate the area between the curves y
30
The widths are each 2.
The heights a gotten by subtracting
20
the function values at 0, 2 and 4:
10
0
A
2
4
2Ÿ1 " 0 2Ÿ5 " 2 2Ÿ17 " 4 34
6
1
3. A girl walks(runs) in a straight line with acceleration aŸt velocity is vŸ0 3, find her velocity at t 2.
a. 12 " cos 2T correctchoice
b. 12 cos 2
c. 10 " cos 2
d. 10 cos 2
e. 8 sin 2
aŸt 4t sin t
vŸt ; 4t sin t dt 2t 2 " cos t C
C
4
vŸt 4. Compute:
2t 2 " cos t 4
vŸ2 vŸ0 4t sin t. If her initial
"1 C
3
12 " cos 2
1
; x 3 / 7 dx
0
a. " 7
4
b. " 4
7
c. 7
3
d. 3
7
e. 7 T correctchoice
10
1
1
10/7
7
3/7
; x dx 7x
10
10
0
0
; x x 2 " 1x dx
2x 7/2 2x "3/2 C
3
7
2x 7/2 " 2x 1/2 CT correctchoice
7
2x 37/2 2x "3/2 C
3
3
3
3/2
x
x
" lnx 2x Ÿx 2 " lnx C
3
3
2x 3/2 x 3 " lnx C
3
3
7/2
1
2
x " x dx ;Ÿx 5/2 " x "1/2 dx 2x " 2x 1/2 C
7
5. Compute:
a.
b.
c.
d.
e.
; x
2
2
; x 4 " x 2 dx
6. Compute:
0
a.
b.
c.
d.
e.
u
2 2
3
8 T correctchoice
3
24
32
3
6
4 " x2
du
2
0
a.
b.
c.
d.
e.
u
;
1/4
0
" 1 du
"2x dx
2
0
; x 4 " x 2 dx
" 1 ; u 1/2 du
2
4
;
7. Compute:
1/4
0
x dx
2u 3/2
3
"1
2
0
4
x
2 at u
0
x
0 at u
4
" 1 ¡0 ¢ 1
2
8
3
sinŸ=t dt
1
= 2 "1
1
= 1" 2
1 " 1
= 2 =
1 " 1 T correctchoice
= = 2
" 1
= 2
1
=t
du = dt
= du dt
sinŸ=t dt
2
2Ÿ4 3/2
3
1
= ; sin u du
" =1 cos =
4
1 ¡" cos u ¢
=
1 ¡cos 0 ¢
=
1 "
=
" =1 cosŸ=t 1/4
0
1
= 2
3
8. The mass density of a 3 cm bar is
mass of the bar.
a. 4 gm
b. 10 gm
c. 12 gmT correctchoice
d. 18 gm
e. 30 gm
3
3
M ; 1 x 2 dx x x
3
0
12
>
>
1 x 2 gm
cm
for 0 t x t 3. Find the total
3
39
0
9. The mass density of a 3 cm bar is
density of the bar.
gm
a. 4 cm T correctchoice
gm
b. 10 cm
gm
c. 12 cm
gm
d. 10 cm
3
gm
e. 13 cm
4
3
> ave 1 ; 1 x 2 dx M
3 0
3
>
12
3
1 x 2 gm
cm
for 0 t x t 3. Find the average
4
1 x 2 gm
cm for 0 t x t 3. Find the
x-coordinate of the center of mass of the bar.
(If you prefer, you may think of this as a plate of uniform density > 1 between
y 1 x 2 and the x-axis for 0 t x t 3.)
a. 3
2
b. 2
c. 7
3
d. 33 T correctchoice
16
e. 99
4
3
3
3
2
4
st
1 mom ; xŸ1 x 2 dx ; x x 3 dx x x
2
4 0
0
0
st
9
81
18
81
99
x 1 mom 99 1 33
M
2
4
4
4
4 12
16
10. The mass density of a 3 cm bar is
4
11.
(15 points) The area between the curve x
16 " y 4
and the y-axis is rotated about the y-axis. Find the
volume of the solid swept out.
This is a y-integral. The limits occur when x 0. So y 4 16 or y o2. Cutting
up the y-axis, drawing a rectangle and rotating, gives a disk of radius x 16 " y 4 .
So the volume is
2
2
2
y5 2
dy ; =Ÿ16 " y 4 dy = 16y "
V ; = 16 " y 4
5 "2
"2
"2
5
Ÿ2 1 256=
2= 16 2 "
64= 1 "
5
5
5
12.
(15 points) The area between the curve y
4x " x 2
and the x-axis is rotated about the y-axis. Find the
volume of the solid swept out.
This is an x-integral. The limits occur when y 0. So xŸ4 " x 0 or x 0, 4.
Cutting up the x-axis, drawing a rectangle and rotating, gives a cylinder of radius
x and height y 4x " x 2 . So the volume is
4
4
4
3
4
V ; 2=xŸ4x " x 2 dx ; 2=Ÿ4x 2 " x 3 dx 2= 4x " x
3
4 0
0
0
128=
256
1
1
256
1
2=
512=
512=
"
"
3
3
12
4
4
3
5
13. (15 points) Find the arc length of the parametric curve x
t 0 and t 1.
HINT:
t 2a t 2ab t a 1 t b
dy
dx 3t 5
3
4t
dt
dt
2
dx 2 dy
5 2
3 2
Ÿ3t Ÿ4t dt
dt
L
u
L
;
1
dx
dt
0
9t
4
2
16
du
2
dy
dt
9t 10
16t 6
1 t6, y
2
t 3 9t 4
t 4 between
16
1
dt
; t 3 9t 4 16 dt
0
3
36t dt
1 du
36
1
1 ; 1 u 1/2 du 1 2u 3/2
36
3
36 t0
t 0
2
3/2
2
1
" Ÿ16 3/2
Ÿ25 3
36 3
t 3 dt
1
36
2Ÿ9t 4
16 3/2
3
1 ¡125 " 64 ¢
54
61
54
1
t 0
(15 points) A bowl is formed by rotating the curve
14.
y
4
x 2 for 0 t x t 2 about the y-axis. This bowl
is full of water. How much work is done in pumping
2
-2
0
-2
the water out the top of the bowl? Leave the density
2
as > and the acceleration of gravity as g.
2
This is a y-integral for 0 t y t 4. A slice perpenducular to the y-axis at height y is
2
a thin disk of radius x y . So its volume is dV = y dy, it mass is
dm > dV >=y dy and its weight is dF dm g >g=y dy. This weight must be lifted
a distance D 4 " y. So the work done is
4
4
4
4
y3
W ; D dF ; >g=yŸ4 " y dy >g= ; 4y " y 2 dy >g= 2y 2 "
3 0
0
0
0
4
4
32>g=
64
2
>g= 32 "
32>g= 1 "
3 0
3 0
3
6