**Unformatted text preview: **H OMEWORK S OLUTIONS MATH 1910
Fall 2017 Sections 6.3, 6.4, 6.5 Problem 6.3.11
Find the volume of revolution about the x-axis for f(x) = csc x over the interval [π/4, 3π/4].
S OLUTION.
Z 3π/4
π(csc)2 dx Volume =
π/4 3π/4 = π [− cot x]x=π/4
= −π(−1 − 1)
= 2π 6.3.11 Problem 6.3.16
Sketch the region enclosed by the curves y = x2 and y = 2x + 3, describe the cross section perpendicular
to the x-axis located at x, then find the volume of the solid obtained by rotating the region about the xaxis.
S E C T I O N 6.3 Volumes of Revolution 685
two curves
therefore
intersectlooks
at x = like
−1 and
x = 3. The region enclosed by the two curves is shown in
S OLUTIONThe
. The
enclosed
region
this:
the figure below. y
8 y = 2x + 3 6
4 y = x2 2
1 x 2 (b) When the region is rotated about the x-axis, each cross section is a washer with outer radius R = 2x + 3 and inner radius
r = x2.
The cross section
perpendicular
to the x-axis is a washer with outer radius 2x + 3 and inner
(c) the
The volume
of theofsolid
revolution
radius x2 , so
volume
theofsolid
of isrevolution is
!
# Z3 ! 3
3 "
Z3
2
2 2
2
4
1088π
π 2 (2x +
(4x
2 3)
2 − (x ) dx = π
2 + 12x + 9 − x )4dx
π
(2x + 3)−1 − (x ) dx = π
4x
dx =
.
6.3.16
−1 + 12x + 9 − x
15
%&3
−1
−1 $
&
4 3
1088π
1
x + 6x 2 + 9x − x 5 && =
.
3
15
5
−1 =π 17. y = 16 − x, y = 3x + 12, x = −1 solution
(a) Setting 16 − x = 3x + 12, we find that the two lines intersect at x = 1. The region enclosed by the two
curves is shown in the figure below.
y y = 16 − x
y = 3x + 12 10 1
−1 −0.5 0.5 1 x (b) When the region is rotated about the x-axis, each cross section is a washer with outer radius R = 16 − x
and inner radius r = 3x + 12.
(c) The volume of the solid of revolution is
! 1 " # ! 1 55. Find the volume of the cone obtained by rotating the region under the segment joining (0, h) and (r, 0)
about the y-axis.
solution The segment joining (0, h) and (r, 0) has the equation
h
y =− x+h
r or x = r
(h − y).
h Rotating the region under this segment about the y-axis produces a cone with volume
"h
!
"
πr2 h
πr2
(h − y)2 dx = − 2 (h − y)3 ""
2
h
3h
0
0
= Problem 6.3.56 1 2
π r h.
3 2 2 2 56. The torus (doughnut-shaped solid) in Figure 15 is obtained by rotating the circle (x − a) + y = 2b
The torus (doughnut-shaped
solid) shown below is obtained by rotating the circle (x − a) + y2 = b2
around the y-axis (assume that a > b). Show that it has volume 2π 2 ab2 . Hint:2After2simplifying it, evaluate
around the y-axis
(assuming
that
athe
> area
b).ofShow
that it has volume 2π ab .
the integral
by interpreting
it as
a circle.
y a a +b
x FIGURE 15 Torus obtained by rotating a circle about the y-axis.
2 2 2 Rotating
the region
by the circle
− a) + y = bwashers
about the y-axis
a torus
S OLUTIONsolution
. The cross
sections
ofenclosed
this volume
of(xrevolution
withproduces
outer
#
# radii given by
whose cross sections are washers with outer radius R = a + b2 − y 2 and inner radius r = a − b2 − y 2 .
the right half
of the circle and inner radii given by the left half of the circle. Solving for x, we
The volume of the torus is then
get:
'2 %
! b $%
! b& p
&
& p '2 (
2 − y2
π 2 =b
a 2+− by2 2− →
y2 x −
a−
b2 − b
y 22 − y
dy2 =→
4aπ
(x − a)
−
a
=
±
x
=−ba b±2 − y 2bdy.
−b
p
remaining
definite
integral the
is one-half
the areafor
of a the
circlerighthand
of radius b; therefore,
the the
volume
of the(our outer
Note that xNow,
= the
a+
b2 p
−
y2 gives
equation
side of
circle
torus is
2
2
radius), with x = a − b − y describing the lefthand side (our inner radius). Thus the
1
4aπ · π b2 = 2π 2 ab2 .
volume of the solid is:
2
Z
57. b Sketchp
the hypocycloid x 2/3 + y 2/3p
= 1 and find the
volume of the solid obtained by revolving it
V = about
π the x-axis.
(a + b2 − y2 )2 − (a − b2 − y2 )2 dy
−b
solution
A sketch of the hypocycloid is shown below. =π Zb
p
p
(a2 + 2a b2 − y2 + (b2 − y2y )) − (a2 − 2a b2 − y2 + (b2 − y2 )) dy
1 b Zb p
4a b2 − y2 dy =π −1 −b Zb p
= 4aπ
b2 − y2 dy.
−b 1 x −1 ) *
2/3 3/2 For the
y = ± 1−x
. Rotating this region about the x-axis will produce a solid whose
p
Rbhypocycloid,
)
*3/2
2−
2with
Observe that
b
y
dy
describes
the
of half
of a ofcircle
ofofradius
b,will
sobewe have that
cross sections
are
disks
radius
R
=
1
−
x 2/3area
. Thus
the volume
the solid
revolution
−b
2 2 %2 3
1 +4aπ( πb ),=
the volume of the torus! is
desired.
2 2π ab −xas
9 7/3 9
2/32 3/2 π −1 (1 − x ) dx = π + x
3
7 2 '"1
"
32π
− x 5/3 + x "" =
.
5
105
−1 6.3.56 1
−a a −1 x FIGURE 16 The hyperbola with equation y 2 − x 2 = 1. solution Problem 6.3.59 √
Each cross section is a disk of radius R = 1 + x 2 , so the volume of the hyperboloid is
%
&'a
% 3
&
! a
! a "#
$2
'
1
2a + 6a
1 + x 2 dx = π
(1 + x 2 ) dx = π x + x 3 '' = π
π
3
3
−a
−a
−a A ”bead” is59.formed
byisremoving
a cylinder
of radius
r rfrom
center
a sphere
of Rradius
A “bead”
formed by removing
a cylinder
of radius
from the
the center
of aofsphere
of radius
(FigureR. Find the
volume of the
with
r =of1the
and
= 2.
17).bead
Find the
volume
beadRwith
r = 1 and R = 2.
y y h
r x R x FIGURE 17 A bead is a sphere with a cylinder removed.
2 2 2 The equation of the outer circle is x + y2 = 2 ,2and the2inner cylinder intersects the sphere
S OLUTIONsolution
. The equation
of the outer circle is x + y = 2 = #4 so we are rotating the region
√
when y =p
± 3. Each cross section of the bead is a washer with outer radius 4 − y 2 and inner radius 1, so
2 and x = 1 about the y-axis. These equations intersect at 1 = 4 − y2
enclosed√
by
x
=
4
−
y
the volume is given by
√
√
√
(
*
&2
! √washers
!to
or y = ± 3. p
Thus we have
perpendicular
from
y = − 3 to y = 3 with
" y-axis
$
3 %)
3the
√
π inner
4 − y 2 1−so
12 the
dy =
π √ 3is:
− y 2 dy = 4π 3.
√
radius
volume
outer radius 4 − y2 and
− 3
− 3
Z √3 p
2
2
Further Insights and
Challenges
2
V=π √
4 − y − 1 dy 60.
Find the volume V of−the3bead (Figure 17) in terms of r and R. Then show that V = π6 h3 , where
h is the height of the bead. ThisZ √
formula
has a surprising consequence: Since V can be expressed in terms
3
of h alone, it follows that two beads of height 12cm, one formed from a sphere the size of an orange and the
=
π
(3
− y )dy
√
other from a sphere the size of the earth,
would have the same volume! Can you explain intuitively how this
− 3
is possible?
√3
3 of the
solution The equation for the outer circle
bead is x 2 + y 2 = R 2 , and the inner cylinder intersects
y
#
√
− cross section
2 . Each
2
2
the sphere when y = ± =
R 2π− r3y
√ of the bead is a washer with outer radius R − y
3
y=−
3
and inner radius r, so the volume is
√ 3!
* √ 3! √ 2 2
&
! √ 2 2 (%) √
R −r √
3
(− 3)
2
(R3)
−+
r 2 − y 2 ) dy
3− r32 −dy = π −√3(−
)
2
2
−
3 − R −r
3
√
&' R 2 −r 2
√ 2%
√ 2 1 !
'
4
√
= (R 2 − r 2 )3/2 π.
=3π √
(R 2 − r 2 )y3 −√y 3 '' √
3 3 − R 2 −r 2
3
=π 6 3−
3−
3
3
√
1
Now, h = 2 R 2 − r 2 = 2(R 2 −
√r 2 )1/2 , which gives h3 = π8(R 2 − r 2 )3/2 and finally (R 2 − r 2 )3/2 = 8 h3 .
Substituting into the expression
= 4π for3.the volume gives V = 6 h3 . The beads may have the same volume but
π R −r √ R 2 −r 2 2 R 2=− π
y2 clearly the wall of the earth-sized bead must be extremely thin while the orange-sized bead would be thicker. 3 6.3.59 18. f (x) = √ 1
x2 +1 [0, 2], , about x = 0 solution A sketch of the solid is shown below. Each shell has radius x and height √
of the solid is S E C T I O N 6.4 1
x2 +1 , so the volume The Method of Cylindrical Shells 705 #
! 2 "
$%
& ''2
√
1 −2
2+
2
11. y = (x 2 + 1)−2 , 2π
y = 2x− (x
1)
,
x
=
2
dx = 2π
x + 1 '' = 2π( 5 − 1).
√
x2 + 1
0
0
solution
region enclosed by y = (x 2 + 1)−2 , y = 2 − (x 2 + 1)−2 and x = 2 is shown below. When
ProblemThe
6.4.11
y
−2 =
rotating this region about the y-axis, each shell has radius x and height 2 − (x 2 + 1)−2 − (x 22 + 1)−2
Use the2 Shell−2
Method to compute the volume obtained
by rotating the region enclosed by y = (x +1) ,
1
2 − 2(x + 1)
. The volume of the resulting solid is
y = 2 − (x2 + 1)−2 and x = 2 around the y-axis.
0.8
#
"
#$2
! 2 " 0.6
! 2
$
2x
1
2
−2
2
$ = 32 π.
2x − 2
dx = 2π x + 2
x(2
− 2(x
+ 1)looks
) dxlike
= 2π
2π
S OLUTION
. The
region
this:
2
5
x + 1 $0
0
0
0.2 (x + 1)
−2y 1
−1
y = 2 − (x 2 + 1)−2 2.0 19. f (x) = a − x 2 x [0, a],1.5 about x = −1 with a > 0, 1.0 below. Each shell has radius x − (−1) = x + 1 and height a − x,
solution A sketch of the solid is shown
y = (x 2 + 1) 2
0.5
so the volume of the solid is
! a
! a$
&
x
2π
(x + 1) (a − x) dx = 2π0.5 1.0a +1.5
(a −2.0
1)x − x 2 dx
− 0 0 "
#'a
'
12. y = 1 − |x − 1|, y = 0
a − 1 2 2 1 3 −2
Shells centered at the y-axis have radius
2 −x2(x− +x1) '' . Thus the volume is given
= x2πandaxheight
+
3 is shown
0
solution
The region enclosed by y = 1 − |x − 1| and2the x-axis
below. When rotating this
by
# [0, 1],2 the shell has radius x and
region about the y-axis,Z two different shells are "
generated.
For
each
x
∈
2
3
− 1) a
a (a + 3)
aZ 2(a
2
2
and
2xvolume
= 2π
+height
=
π.
height x; for each x ∈ [1, 2], the shell has
2 − x.−The
of the resulting
solid is
2 radius
−2x a
2πx 2 − 2(x + 1)
= 2π
dx3
22x − 23
2
(x
+
1)
! 1
! 2
!0
!
0
1
2 2
2π
x(x) dx + 2π
x(2 − x) dx = 2π a 2(x 2 ) dx
+
2π
(2x − x 2 ) dx
1
= 2π x0 + 2
0
1
1
x + 1 x=0
"
#$
"
#$
1 3 $$1
1 3 $$2
2
32
=
2π
+
2π
x
−
x
x
π.
=
$
$ = 2π. 6.4.11
3
0
1
5 3
−2 − ay 20.Problem
f (x) = 1 6.4.20
− x 2 , [−1, 1], x=c a −2 −1 1.0 with cy =>
x 1 y=2−x 0.5 below. Each shell has radius c − x and height 1 − x 2 , so the
solution A sketch of the solid is shown
Sketch the solid obtained by rotating the region underneath the graph of f(x) = 1 − x2 over [−1, 1] about
volume of the solid is
x method.
the axis x = c for c > 1, then calculate its volume using the Shell
"
#'1
! 1
! 1$ 0
0.5
1.0
1.5
$
&
& 2.0
'
1 4 c 3 1 2
2
3
2
' = 8cπ .
2πS OLUTION
(c − x). A1 sketch
− x dx
=
2π
x
−
cx
−
x
+
c
dx
=
2π
−
−
+
cx
x
x
x
'
of
the
solid
is
shown
below.
4
3
2
3
−1
−1
−1
In Exercises 13 and 14, use a graphing utility to find the points of intersection of the curves numerically and
y
then compute the volume of rotation of the enclosed
region about the y-axis.
1 13. y = 12 x 2 , y = sin(x 2 ), x≥0 solution The region enclosed by y = 12 x 2 and y = sin x 2 is shown below. When rotating this region
about the y-axis, each shell has radius x and height sin x 2 − 12 x 2 .x Using a computer algebra system, we find
−1
1 on cthe 2c
− 1 2cis+ 1x = 1.376769504. Thus, the volume of the
that the x-coordinate of the point of intersection
right
resulting solid of revolution is
"
# 2 Method to calculate the volume of rotation
InEach
Exercises
21–26, shell
sketchhas
the!radius
enclosed
region
use 1the
Shell
1.376769504
cylindrical
c−
x andand
height
1−
x , so the volume of the solid is
2
2
x sin x − x dx = 1.321975576.
2π
about the x-axis.
Z1
Z
1 2
0
8cπ
21. x = y, y = 0, x2π(c
= 1− x)(1 − x2 ) dx = 2π
x3 − cx2 − x + c dx =
.
6.4.20
y
3
−1
−1
solution When the region shown below
is rotated about the x-axis, each shell has radius y and height
1
1 − y. The volume of the resulting solid is
#'
"
! 1 y = sin x 2
! 1
1 2 1 3 ''1
π
x2
= = 2π
y(1 − y) dy = 2π
(y − y 2 ) ydy
−
y
y
2π
2
' = 3.
2
3
0
0
0
x
4 1
0
14. y = 1 − x4, y = x, x≥0 solution The region enclosed by y = 1 − x 4 , y = x, and the y-axis is shown below. Using the shell
method, each shell has radius x and height 1 − x 4 − x. Using a computer algebra system, we find that the 1
0 26. y = x 1/3 − 2, y = 0, 0.5 1.0 1.5 2.0 x x = 27 solution When the region shown below is rotated about the x-axis, each shell has radius y and height
27 − (y + 2)3 . The volume of the resulting solid is Problem 6.4.26
! ! 1#
$
$
#
1
2
3
2π enclosed
dy
y · 27 −
2)3 dy
y 4 calculate
Sketch the region
by(y
the+curves
y=
= 2π
x1/3 − 19y
2, y −=12y
0, x −=6y27,−and
its volume of
0
revolution about the
x-axis using the shell method. 0
%
&'
19 2
38π
3 4 1 5 ''1
3
=
2π
=
y
−
4y
−
.
y
−
y
'
S OLUTION. The region looks like this:
2
5
2
5
0
y 1.0
0.8
0.6 y= 3 5 10 15 20 25 30 x−2 0.4
0.2
0 x 3
27.
Determine
which ofatthethe
following
is the
appropriate
needed
to 2)
determine
ofis
the solid
Each
shell centered
x-axis has
radius
y andintegrand
height 27
− (y +
. Thus the
thevolume
volume
obtained by rotating around the vertical axis given by x = −1 the area that is between the curves y = f (x)
Z 1 over the interval [a, b], where aZ ≥
1 0 and f (x) ≥ g(x) over that
and y = g(x)
interval.
38π
.
6.4.26
2πy 27 − (y + 2)3 dy = 2π
19y − 12y2 − 6y3 − y4 dy =
(a) x(f (x) 0− g(x))
5
0
(b) (x + 1)(f (x) − g(x)) (c) x((f (x) − 1) − (g(x) − 1)) (d) (x − 1)(f (x) − g(x)) Problem 6.4.57 (e) x(f (x + 1) − g(x + 1))
Use the Shell Method to find the volume of the torus obtained by rotating the circle (x − a)2 + y2 = b2
solution
When(assume
the region
= fproblem
(x) and6.3.56.)
y = g(x) is rotated around the vertical axis
about the y-axis
a >between
b). (Seethe
thecurves
picturey for
x = −1, each shell has radius x − (−1) = x + 1 and height f (x) − g(x). The appropriate integrand is
therefore
(b): .(xWe
+ 1)(f
− g(x)).with radius x going from x = a − b to x = a + b. Each of these
S OLUTION
have(x)
cylinders
p
(b
cylinders has height 2 b2 − (x − a)2 so we have volume:
28. Let y = f (x) be a decreasing function on [0, b], such that f (b) = 0. Explain why 2π 0 xf (x) dx =
( f (0)
Z a+b h denotes
Z a+b q
q the inverse of f .
π 0 (h(x))2 dx, where
2πx(2 b2 − (x − a)2 )dx = 4π
x b2 − (x − a)2 dx
a−b a−b Substituting u = x − a and du = dx we get
Zb
4π
−b √ Zb p
Zb p
p
u b2 − u2 du + 4πa
b2 − u2 du
(u + a) b2 − u2 du = 4π
−b −b √ p
Note that u b2 − u2 is an odd function (u b2 − u2 = −((−u) b2 − (−u)2 ) so it is 0 integrated about the symmetric interval [-b,b] (or use substitution u = bsin(θ), du = bcos(θ)dθ to
show this fact).
Rb √
Thus the volume is 4πa −b b2 − u2 du. The integral here is equal to half of the area of a circle
2 2
2
of radius b, we have that the volume is 4πa( πb
2 ) = 2π ab . 5 6.4.57 π "y m . 2 Thus the weight of one layer is
9800π ! y "2
2 "y N. The layer must be lifted 12 − y meters, so the total work needed to empty the tank is
# 10
! y "2
9800π
(12 − y) dy = π(3.675 × 106 ) J ≈ 1.155 × 107 J.
2
0 Problem 6.5.21 21.Calculate
Horizontal
Figurerequired
11; water
from
a small
hole at
Evaluate
the cylinder
work (ininjoules)
toexits
pump
all of
the water
outthe
of top.
the Hint:
cylindrical
tankthe
of integral
radius rby
interpreting
of it as
the area of a circle.
and lengthpart
l shown
below.
Water exits here. r FIGURE 11 S OLUTION. Consider a horizontal “slice” of water of thickness dy at a distancepy above vertical
solution Place the origin along the axis of the cylinder. At location y, the layer of water
is a rectangular
$ allow y to be negative.) The volume of the slice is 2` r$2 − y2 dy m3 ,
center of the tank. (We
p volume of the layer is 2ℓ r 2 − y 2 "y, and
slab of length ℓ, width 2 r 2 − y 2 and thickness
"y.
Thus,
the
$ is 2 · 9.8 · 1000` r2 − y2 dy N. The slice must be lifted a
and so the force needed to lift the slice
thedistance
force needed
to
lift
the
layer
is
19600ℓ
r 2tank,
− y 2 "y.
Thetotal
layerwork
must done
be lifted
of r − y to get to the top of the
so the
is a distance r − y, so the total
work needed to empty the tank is given by
Zr #
r p
# rZ%
# r Z r% p
%p
r
2
2
2
2
2
2 · 9.8
· 1000`r 2 −
r y−2 (r
y−
(ry)
−dy
y) dy
= 19600`r r 2 −r y−
y −
dr19600ℓ
− 19600` y y
y2 dr.
2 dy
2 dy.
19600ℓ
= 19600ℓr
r 2 −r y −
−r
−r
−r
−r −r −r The first integral on the right is equal to half of the area of a semicircle of radius r, and the
Now,
second integral is zero since the integrand
# r % is odd (you can also use substitution to find this).
Thus the work done is
y r 2 −y 2 du = 0
−r 1
19600`r
πr2 = 9800`πr3 J.
6.5.21
2
because the integrand is an odd function and the integration interval is symmetric with respect to zero.
Moreover, the other integral is one-half the area of a circle of radius r; thus,
# r %
1
r 2 − y 2 dy = π r 2 .
Problem 6.5.32
2
−r
A 500-kg
wrecking
hangs
from athe
12-meter
Finally,
the total
work ball
needed
to empty
tank is cable of density 15 kg/m attached to a crane. Calculate
the work done if the crane lifts the ball&from ground
level to 12 meters in the air by drawing in the cable.
'
1 2
19600ℓr
π r − 19600ℓ(0) = 9800ℓπ r 3 J.
S OLUTION. We will treat the cable2and the ball separately. Consider a segment of cable of
length dy that must be lifted y feet. The work needed to lift the cable is W(y) = (15 dy)9.8y J,
22.soTrough
in Figure
12;lift
water
the total
work to
the exits
cableby
is pouring over the sides.
Z1
215 · 9.8y dy ≈ 10, 584 J.
0 Lifting the 500 kg ball to a height of 12 bmeters requires and additional
c 500 · 12 · 9.8
h = 58, 800 J,
so the total work done is 69,384 Joules. a FIGURE 12 6 6.5.32 0 9.8ρ(x) dx = 9.8 0 8x − 2x dx = 9.8 4x − x
3 & = 176.4 N
&
0 another two meters requires an additional 352.8 J of work. The total work is therefore 661.5 J.
Exercises 35–37: The gravitational force between two objects of mass m and M, separated by a distance r,
has magnitude GMm/r 2 , where G = 6.67 × 10−11 m3 kg−1 s−1 .
35. Show that if two objects of mass M and m are separated by a distance r1 , then the work required to
increase the separation to a distance r2 is equal to W = GMm(r1−1 − r2−1 ). solution The work required to increase the separation from a distance r1 to a distance r2 is
&
! r2
GMm
GMm &&r2
dr
=
−
= GMm(r1−1 − r2−1 ).
r &r1
r2
r1 Problem 6.5.35 The gravitational
between
two35objects
of mass
m and
M, toseparated
by a satellite
distance
r, has
36. Use force
the result
of Exercise
to calculate
the work
required
place a 2000-kg
in an
orbit magnitude
GMm
−11 3
−1 −1
6 m and
where
G
=
6.67
∗
10
m
kg
s
.
1200
km
above
the
surface
of
the
earth.
Assume
that
the
earth
is
a
sphere
of
radius
R
=
6.37
×
10
e
r2
mass Me = 5.98 × 1024 kg. Treat the satellite as a point mass. Show that if two objects of mass M and m are separated by a distance r1 , then the work required to
solution The satellite will move from a distance r1 = Re to a distance
Re + 1200000. Thus, from
−1 r2 = −1
increase theExercise
separation
35, to a distance r2 is equal to W = GMm(r1 − r2 ).
S OLUTION. W = (6.67
10−11 )(5.98 × 1024 )(2000)
From
the×equation $ 1
1
−
6
6
6.37
Z × 10 6.37 × 10 + 1200000
b % ≈ 1.99 × 1010 J. W the
= workF(x)dx
37. Use the result of Exercise 35 to compute
required to move a 1500-kg satellite from an orbit
a
1000 to an orbit 1500 km above the surface of the earth.
we obtain solution The satellite will move from a distance r1 = Re + 1000000 to a distance r2 = Re + 1500000.
from Exercise 35,
r2
Z r2
ZThus,
r2
%
$
1
1
GMm
−1
dr
=−11
GMm
dr =× GMm −1
W=
=
GMm(r1−1
24
1 − r2 ).
W =
(6.67
×
10
−
)(5.98
×
10
)(1500)
2
2
6
6
r
r 1000000
6.37 × 10 +
6.37 × 10 + 1500000
r1 r
r1
r=r1 6.5.35 ≈ 5.16 × 109 J. 38. The pressure P and volume V of the gas in a cylinder of length 0.8 m and radius 0.2 m, with a movable
piston, are related by P V 1.4 = k, where k is a constant (Figure 14). When the piston is fully extended, the
gas pressure is 2000 kilopascals (kPa; 1 kilopascal is 103 newtons per square meter).
k.
The pressure(a)PCalculate
and volume
V of the gas in a cylinder of length 0.8 m and radius 0.2 m, with a movable
(b) The force on the
piston is P A, where A is the piston’s area. Calculate the force as a function of the length
1.4
piston, are xrelated
by PV
of the column
of gas. = k, where k is a constant. When the piston is fully extended, the gas
3
pressure is 2000
kilopascals
kPa isto10
newtons
square
meter).
(c) Calculate
the work(1
required
compress
the gas per
column
from 0.8
m to 0.5 m. Problem 6.5.38 0.2
x FIGURE 14 Gas in a cylinder with a piston. a) Calculate k.
b) The force on the piston is PA, where A is the piston’s area. Calculate the force as a function of the
length x of the column of gas.
c) Calculate the work required to compress the gas column from 0.8 m to 0.5 m.
S OLUTION. a) We have P = 2 × 106 Pa and V = π · 0.22 · 0.8 = 0.032π m3 ,
k = 2 × 106 · (0.032π)1.4 ≈ 80, 213.9.
b) The are of the pistion is A = 0.04π, and the volume of the column of gas is V(x) = Ax. From
P = k/V 1.4 we obtain
F = PA = k
· A = k(0.04π)−0....

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