The document provides an overview of using integrals to calculate the areas between curves, volumes of solids, lengths of curves, and other quantities. It discusses: - Approximating the area between curves using Riemann sums and defining the integral as the limit of these sums. - Calculating areas between curves by integrating f(x) - g(x) from the limits of integration, which may be where the curves intersect. - Partitioning a region into subregions if the bounding curve formulas change, and integrating over each subregion. - Calculating areas by integrating with respect to y instead of x if the region is described in terms of y. - Combining calculus

1. CrnprpR
Applications
of Integrals
OYERVIEW Many things we want to know can be calculated with integrals: the
areasbetweencurves,the volumes and sudaceareasof solids, the lengthsof curves,
the amount of work it takes to pump liquids fiom belowground, the forces againsr
floodgates,the coordinatesof the points where solid objects will balance.We define
all of these as limits of Riemann sums of continuous functions on closed intervals.
that is, as integrals,and evaluatethese limits with calculus.
There is a pattern to how we define the integrals in applications, a pattern
that, once learned,enablesus to define new integralswhen we need them. We look
at specific applications 0rst, then examine the pattern and show how it leads to
integrals in new situations.
AreasBetweenCurves
This section shows how to lind the areas of regions in the coordinate plane bv
integrating the functions that define the regions, boundaries.
The Basic
Formula a Limitof Riemann
as Sums
Supposewe want to find the area of a region that is bounded above by the curve
-y =.l(r), below by the curve ) = g(.r), and on the left and right by the lines
.r : 4 and r : D (Fig. 5.1). The region might accidentallyhave a shapewhose area
w e c o u l d f i n L w i r h g e o m e t r l b u r i l / a n d g a r e i t r b i l r a r ) o n r i n u o ut,u n c r r o n,r
l . c e
usuaily have to flnd the area with an integral.
To see what the integml should be, we lirst approximate the region with,?
basedon a parlition p : {.r0, ... , r,,} of [ri, D] (Fig. 5.2, on
verticalrectangles rr,
the following page). The area of the tth rectangle(Fig. 5.3, on rhe followrng page;
ls
AA1 : tsjcfi1 x width : Ll (c.t) g(ci)l A4.
We then approximatethe area of the region by adding the areasof the ,?rectanqles:
A* f ^A, -Itr,,.r grr.,rlA,:,.
' The region between y: f(x) and
As ll P | -+ 0 the sums on rhe right approachthe limit
y: g ( x ) a n d t h e l i n e sx : a a n d x = b . _ g(,r)l d,r because
/j t/(r)
365

2. Chapter5: Applicationsof Integrals
356
T - s(ck)
f(t)
Stctll
::.: We approximate regionwith
the
perpendicular the x-axis.
to AAk = areaof kth rectangle,f(c*) - g(cr) = height, lxk = width
rectangles
this integral'
/and g arecontinuous. takethearea theregionto be thevalueof
of
We
That is,
lim )-l/(c,)-gtcrtlAxr - fo rf ,r,-
A- Stxt]dx
nP o-- J
Definition
with /(r) > g(x) throughout b], ihenthe area
g [4,
Ifl ancl arecontinuous
the curves : /(*) andy : g(t) ftom a to b is the
y
of the regionbetween
integral [/ - 8 | liom a to D:
o[
rb
U(x)- g(x)ldx. (1)
A=
J
To apply Eq. (1) we take the following steps.
How to Find the Area Between Two Curves
Graph the cunes and draw a tepresentative rectangle. Thts rcveals
l.
which curve is / (upper curve) and which is g (1owercurye). It also
helps find the limits of integration if you do not aheady know them.
,s'I ,=sinr 2. Find the limits of integration.
I--''-
3. Write a formula for f(x) - g(x). Simplify it if you can.
4. lntegrate [f(x) - gQ, fron a to b Tl1e number you get is the area.
H Find the area between lv = sec2 and y : 3i1;5 from 0 to z/4'
x
EXAMPLE 1
Solution
Step t; We sketch the cutves and a vertical rectangle(Fig. 5.4). The upper curve
is the gmph of /(;r) = secz;r: the lower is the graph of g(;) : sipI'
Sfep 2; The limits of integration are already given: a :0' b = r /4'
,'.]].The regionin Example'lwith a -
SteP3: f (x) - 8(x) = sec2;r sin't
rectangle.
typicalapproximating

3. 5.1 AreasBetweenCurves 367
Step 4:
{r' *- sinx)r/x : ftan + cos
-r x]i/a
lo'o
f'.+l,to+:+
-t
L f,
Curves
That Intersect
When a region is detemined by curves that intersect, the intersection points give
the limits of integration.
- x2
Find the area of the region enclosedby the parabola ! :2
EXAMPLE 2
: -y.
and the line y
Solution
Sfep t; Sketch the curves and a vertical rectangle (Fig. 5.5). Identifying the upper
(-{,,f(-r)
-x. The ,{-coordinates
and the lower curyes, we take /(x) :2- g(r):
xz arLd
pointsare the limits of integration.
of the intersecdon
Step 2; We find the limits of integation by solving | :2 - xz and y : -1 ri-
( 1 ,1 )
multaneously for .r.'
-x [ q r l L t/r1 r ) . u r d . q ( f r
2-xz :
x2-x-2:0 R.rir..
(x+l)(x -2):0 Fxrr.'
(a s(r))
-1,
x: x :2. sot.
The region runs from x : - 1 to.r : 2. The limits of integrationare a : - l, b : 2.
Step 3:
5.5 The regionin Example with a
2
typicalapproximatingrectangle.
Step 4
t2 f -2 -t12
- :
o: rf,.,>e(x)ldx I t z + x - x z t d x = l z r + z
- ,I
l' JJ
J-t I r
/ 4 8 / | 1
(o*r-r,l (-'*l*r/
?qq
-
-'2
At
3 2 l
Technofogy The Intersectionof Two Graphs One of the difficult and some-
times frustrating parts of integmtion applications is finding the limits of inte-
gration. To do this you often have to find the zeroes of a function or the
intenection points of two curues.
g(r) using a graphing utility, you enter
To solve the equation /(x):
yr: f@) and y2: g@)

4. Chapter Applications lntegrals
of
5:
368
you
andusethe grapherroutine to find the Pointsof intersection'Altematively'
- g(T) : 0 with a root finder'Try bothprocedures
lun solve*tJ /(:r)
quation
with
and g(x):3-x'
f (x):lnx
hidden
When points of intersectionarc not clearly revealedor you suspect
further use of calculus
behavior,additional work with the graphing utility or
may be necessary,
ISECT
=.79205996845
curves : Inx and lz:3 - x' usinga built-infundion
y1
a) The intersecting
intersection
to find the -
: x
O) Usinga built-inroot finder to find the zero of f(x) Inx 3 +
with ChangingFormulas
Boundaries
points' we partition
If the formula for a bounding curve changesat one or mole
th,gionintosubregionsthatconespondtotheformulachangesandapplyEq.
(1) to eachsubregion.
Find the areaof the region in the fint quadrantthat is bounded
EXAMPLE 3
- 2'
:
aboveby y : .r/i and below by the t-axis and the line y x
solution
is the graphof
Step t.' The sketch(Fig. 5.6) showsthat the region'supperboundary
<2to g(x) =
fromg(;r):0for0 S r
changes
f(; : JV. Theloweiboundary
-' Z tot Z : t : 4 (thereis agreement r : 2)' We partition the region at r : 2
at
rectanglefor eachsubregion'
into subregiJns ani f and tketh a representative
A
5.6 When the formulafor a boundingcurve
changes, area integralchanges match
to
the
(Example3).
'fe limits of integrationfor regionA area : O andb :,2' The lefchand
step 2:
li#t for region B rs a = 2. To find the right-hand limit, we solve
the equations

5. 5.1 AreasBetween
Curves 369
y : rG andy : x - 2 simultaneously r.'
for
EqratcI (.r)rnd
^-
- z
v^
.9(11
a:(x.-2)2:x.2-4x+4 SqLrafeboth
x 2- 5 x + 4 : o Re$'r'ite.
(.r-1)(r-4):0 FacL(r'.
-_1 -_i
Solve.
Only the value.x=4 satisfies equation
the The valuer:1is an
Ji:x-2.
extraneous introduced squaring. right-hand
root by The limit is b:4.
f (x)- S@): -0:
S t e p3 . ' F o r 05 x = 2 :
- e ( x ) : , 6E - ( x - 2q :
)
Ji - x +2
F o r2 a x a 4 : f(x)
Step 4.' We add the areaof subregions and B to find the total area:
A
f)-f4
=
T o t a fr e a | J i d x I l t J x - x + 2 t d x
a
-o - - - - /
J Jz
areaof B
arca of A
'l2
f) f1 -2 14
fil.+l1*'''-v+2,1,
/2 -
?ef,,_o_ (1,,0,',, .' 8)_
_8
r r , r{zt'''-z++1
?,t, : !.
-, ,
33
lntegrationwith Respect y
to
If a region's bounding curves are describedby functions of y, the approximating
rectanglesare horizontal insteadof vertical and the basic formula has y in place
of r.
t
1
I
L*,,,
use tJteformula
In Eq. (2),/always denotes right-hand
the
- stytldy.
e:
l.' vrrt
(2)
curve and I the left-hand curve, so
fO) - e(y) is nonnegative.

6. Chapter 5: Applicationsof lntegrals
370
Find the area of the region in Example 3 by integrating with
EXAMPLE 4
rcspect to ).
Solution
(c()),
).)
x:y+2 Step t; We sketchthe region and a typical horiz,ontalrectanglebasedon a panition
of an interval ofy values(Fig. 5.7). The region's right-handboundary is the line x :
y f 2, so l (y) : y + 2. The left-hand boundary is the curve ir : y2, so g(y) : y2.
Step 2: The lower limit of integration is l = 0. We find the upper lirnit by solving
I : y *2 and r: )2 simultaneously y; for
5.7 lt takestwo integrations find the l+t:l
to
areaof this regionif we integrate with
2=o
,t-)
respectto x It takes only one if we
with respect y (Example
integrate to 4). ( ) + 1 ) ( ) 2 ): 0 Iurt,,
1'-
_t ) S ,i l .
The upperlimit of integmtion D : 2. (The valuey : -1 givesa point of inter-
is
sectlor
belowthe r-axis.)
Step3:
=)+2 y2=2iy )?
/())-s())
Step 4:
v,rt- :
e:
1,,'' srt,rat 1 2 + y - y ' ) l d y
l,'
f r,2 rt l'
: +; 1l
L2Y 3 ln
8 l0
- o.+ t -
4
3
3
This is the result of Example 3, fbund with less work.
Integrals
with Formulas
from Geometry
Combining
andgeometry.
The fastest
way to find an areamay be to combinecalculus
The Area of the Region in Example 3 Found the Fastesi
EXAMPLE5
way
Find the area of the region in Example 3.
So/ution The area we want is the areabetweenthe curye I : Ji,O = x S 4, and
1:rf the x-axis, minus the area of a triangle with base 2 and height 2 (Fig. 5.8):
2 lr2rr2r
/ x r u = f or t a ,
.2
'.2
2 ,11
J .lo
5.8 The areaof the blue regionis the
:3ttl-o -'z:+'
area underthe parabola = 1& minus
,
y
the areaof the triangle. JJ
-

7. Exercises
5.1 371
Moral of Examples3-5 It is sometimes easierto lind the areabetween two
with respect y instead ,r. Also, it may help to combine
curves integrating
by to of
geometry calculus.
and After sketching region,takea moment determine
the to the
bestwav to oroceed.
Exercises
5.1
Find the areasof the shadedregions in Exercises 1-8.
l. 2.

8. Chapter Applications Integrals
of
5:
372
16. y : x 2 - 2 x and Y=r
area.
In Exercises findthetotalshaded
9-12,
and y = -tt2 14x
t7. ) : 12
10.
9,
] 18. y = l - l x ' ano ]:x-14
(-3,5) 19. y = x 4 - 4 x 2 + 4 and Y=a2
I and y:0
20, y = r n @ = 7 , a>0,
_,_,2
5y = x * 6 (How many intersection points are
21. y = {fl nd
there?)
2x3 - (x212)+4
5x 22,y:lx2-41'and ):
2
I
Find the areasof the regions enclosedby the lines and curves
23-30.
Exercises
-3) : r : 0 , a n d) : 3
x:2y2,
23.
24. x = y2 u1t4 a: y 12
and' 4x - Y = 16
25. y2 - 4x:4
26, x - y2 :0 arld x +2Y2 =3
^nd r+3Y2:2
2 7 ,x * y 2 : O
and ,r*)a:2
2 8 .x - y 2 / t : 0
(-2,4
y
29.x =y2-l and x='yJl
3 0 . r = ) l 3- y 2 a n d x = 2 y
-2 -l Find the areasof the regions enclosedby the curves in Exercises
I 31'34.
and .xa-Y:l
31.4x2+y:4
I a n d 3 x 2- y : 4
y :O
3 2 .x 3
I, (3,5) for xZ0
and :r*)a:1,
3 3 .x * 4 y 2 : 4
and 4.t*y2:0
3 4 ,x a y z : 3
12.
Find the arcas of the regions enclosedby the lines and curves in
)
I+.
35-42.
Exercises
(3,6)
6
and y = sin2.r, 0:.x57t
3 5 .y = 2 s 1 n x
and y:5e92.r, -7t13=x Sr13
v 36.):8cos.r
.t2
and 1:1
3 7 .y : c o s ( r x / 2 )
38. y = sin(nx/2) and 1t:r
(3,1) and'x:tt/4
39, l:sec2.r, y=tarfx, x:-tr/4,
and .x:-tan2), -7r14=r 3n/4
4O. :ta#y
x
r and -x=0, 0=y<7t/2
4 1 .r : 3 s i n l / i o s y
-'
4 2 .y : s s s 2 1 1 t a J 3 1 r r d y : r 1 / 1 , - l = i r : l
3) a
rcgion enclosed the curve
by
43. Find the areaof the propeller-shaped
Find the areasof the regions enclosedby the lines and curves in - y3 : 0 andthe line.r - Y : 0.
t
13-22.
Exercises region enclosedby the
44. Find the area of the propeller-shaped
1 3 . y : a 2- 2 clruesx -yll3 : 0 andr - )l/5 = 0.
ar'd y:2
14. y =2x - x2 arLd Y: -J 45, Find the areaof the region in the first quadrantboundedby the
line I :1, the line,t :2, the curve1 : 1/-r'?, the -t-aris.
and
15. y :1+ and ) :8r

9. Exercises
5.1 373
Suppose area of the region betweenthe graph of a positive
46. Find the area of the tdangular region in the first quadrant the
continuousfunction / and the .x-axisfrom x : 4 to x : b is 4
boundedon the left by the y-axis and on the right by the curves
units. Find the areabetween curves)r: /(.r) and
the
) = sin.rand): cos,r. square
y-2f(x)ftomx:atox-b.
belowby the pambola : ,r2 andabove by
47, The regionbounded )
Which of the following integals, if either calculates areaof
the
: 4 is to be parlitioned into two subsections equal
of
the line )
region shownhere?Give reasonsfor your answer.
areaby cutting acrossit with the horizontal line ), = c. the shaded
a) Sketchthe region and draw a line ) : c acrossit that looks fL f)
-l-x)d:(=
I (x 2xdx
al |
about right. In terms of c, what are the coordinates the
of J
J-t I
points where the line and parabolaintersect?Add them to '
rot tl
-zxax
bt / r-x-(r))dx-
your figure. |
JI
J-t
b) Find c by integratingwith respectto ),. (This puts c in the
limits of integration.)
to.x.(Thisputsc into the
c) Find c by integntingwith respect
integrand well.)
as
''2 and
Find the areaof the regionbetween curve )):3
the
the line y = 1 by integratingwith respectto (a) t, (b) ).
49. Find the areaof the region in the first quadrantboundedon the
left by the ),-axis,below by the line y = t /4, aboveleft by the
curve y = I * .,[, and aboveright by he (i.jtr..te : 2/ Ji,
y
50. Find the area of the region in the first quadrantboundedon the
left by the y-axis, below by the curve r : 2/t, aboveleft by
right by the line i :3 - ).
the curyer : () - l)'?,andabove
54. True, sometimestue, or never tue? The area of the region
between graphsof the continuous functions): /(i) and
the
: g(r) andthe verticallinesr : a andx: b (a < b) rs
)
rb
I lf @) - s(x)ld)c.
J
Give reasonsfor your answer.
S CASExplorationsand Proiects
In Exercises 55-58, you will find the area between curves in the plane
when you cannot find their points of intersectionusing simple algebra.
Use a CAS to perform the following steps:
The figure here shows triangle AOC inscribed in the region cut
a) Plot the curves together to see what they look like and how many
ftom the parabola ) - .r2 by the line y: a2. Find the limit of
(hey hae.
points of inrersection
the latio of the area of the triangle to the area of the parabolic
b) Use the numerical equation solver in your CAS to find all the
region as 4 approaches zerc.
points of intersection.
- g(r)l over consecutive pairs of intersection
c) Integate
f(r)
values.
d) Sum together the integrals found in part (c).
^- ^ -2, I . g 1 x .:1 - J
x
5 5 . '/3 2 3 :
1x1
t4
1 0 .g ( . r r : 8 - l 2 r
5 6 ./ t x t : i - t r ' -
57. /(r) =x *sin(2tc), g(x) : 13
58. /(.x) : -r2sqsir, g(tc) : x1 - x

10. Chapter 5: Applicationsof Integrals
374
Volumes Slicing
by
Finding
From the areasof regions with curved boundaries,we can calculatethe volumes of
cylinders with curved basesby multiplying base area by height. From the volumes
of such cylinders, we can calculate the volumes of other solids.
5licing
Supposewe want to find the volume of a solid like the one shown in Fig. 5.9. At
Cross sectionR(jr). Its area is A(ir).
each point r in the closed inteNal la, rl the cross section of the solid is a region
R(x) whose area is A(r). This makesA a real-valuedfunction of .r' If it is also a
continuous lunction of .r, we can use it to define and calculate the volume of the
solid as an integral in the following way.
We partition the interval [4, ]l along the r-a{is in the usual manner and slice
the solid, as we would a loaf of bread, by planes petpendicular to the -r-axis at
the partition points. The tth slice, the one between the planes at xk r and tn, has
approximatelythe same volume as the cylinder betweenthesetwo planes basedon
the region R(,rr) (Fig. 5.10). The volume of this cylinder is
Vr : base area x height
sedion
lf the areaA(x) of the cross
: A(rr) x (distancebetween the planes at -rr r and xr)
functionof x, we can
R(x)is a continuous
find the volumeof the solid bY : A(xr)Axr.
A(x) from a to b.
integrating
The volume of the solid is therelbre approximatedby the cylinder volume sum
- r,. r,r-.
?-,'^'^
This is a Riemann sum for the function A(r) on [c, Dl. We expectthe approxlmahons
from these sums to improve as the norm of the partition of la, bl goes to zero, so
we define their limiting integral to be the volume of the solid.
Approxrmatlng
Plane at.r[ I cllinderbased p l a n en r r ,
on]((rt]
-'i.---------'
-,
The cylinder's base
is the region R(,v*).
NOTTO SCAI-E
view of the sliceof the solidbetweenthe
. Enlarged
cylinder'
planes xk r and xk and its approximating
at

11. 5 . 2 F i n d i n g o l u m eb y S l i c i n g 3 7 5
V s
Definition
The volume of a solid of known integrable cross-sectionarea A(x) from
x = a to x: D is the integralofA from a to r.
v : fo e61a,. (j)
J
T o a p p l l E q . { l ) . w e r a k et h e f o l l o w i n g: r e p . .
How to FindVolumesby the Method of Slicing
1 Sketch the solid and a typical cross section.
2. Find a formula for A(r).
3. Find the limits of integration.
4. Integrate A(x) to find the volume.
EXAMPLE 7 A pyramid 3 m high has a square base that is 3 mon a side.
The cross section of the pyramid perpendicularto the altitude r m down from the
ve ex is a squarer m on a side. Find the volume of the pyramid.
Solution
Step 1: A sketcll. We draw the pyramid with its altitude along the r-axis and its
vertexat the origin and includea typical crosssection(Fig. 5.11).
Typical cfoss
section
--r$€t
-=_,,.,
sections the pyramidin Example
The cross of 1
are 5quare5.
Step 2: A.fonnula for A(x). The cross section at r is a squarex meters on a side,
so lts area ls
A(r) : 12'
Step 3: Tlle lintits of integration. The squaresgo from r :0 to;r :3
Step 4: The volwne.
'l'
lb t' - r.
-,
At^trlt -
v=I | t:dr r |=9
J.. J,, .l ,
The volume is 9 mr.

12. Chapter Applications Integrals
of
376 5:
wedge cut from a cylinderof radius3 by two planes.
is
2 A curved
EXAMPLE
Bonaventura Cavalieri ('l 598-1647) planecrosses
One planeis perpendicular the axis of the cylinder.The second
to
Cavalied, a studentof Galileo's, discovered the nrst plane at a 45' angle at the centerof the cylinder. Find the volume of the
that if two plane rcgions can be afiangedto wedge.
lie over the sameinterval of the i-axis 1n
sucha way that they haveidentical vertical Solution
crosssectionsat every point, then the rcgions perpen-
Step 1: A sketch. draw the wedgeand sketcha typical crosssection
We
havethe sarnearea.The theorem(and a letter
dicularto the:r-axis(Fig. 5.12).
from Galileo) were
of recommendation
enoughto win Cavalied a chai at the
Univelsityof Bolognain 1629.The solid
2lt -7
geometryversionin Example 3, which
Cavalieri neverproved,was given his name
by later geometen.
have
Crcsssections
the samelongthat
everypoint in [d, b]
5-t2 The wedgeof Example slicedperpendicular
2.
sections redangles.
are
to the x-axis.
The cross
Step 2: Theformula for A(x). The cross section at r is a rectangleof area
A(') : (heishtxwidth) (zJe - 'r)
: (r)
: z x t t: Y - x ' .
^,
Step 3: The linxits of integration. The rectangles from r :0 to.r :3.
run
Step 4: The volune.
?b ^3
Ax)dx: | 2tcJ9x2dx
-
v:l
Ja JO
113
: -;e - xr3/'
)o
I
,f : o+?e),/' 1! = lf r1r. nncsfalc-
J . L n d s u b s t i l U r eb r c k .
tr
- 19
Cavalieri's Theorerrr
EXAMPLE 3
Cavalieri's theoremsaysthat solids with equal altitudesand identical parallel cross-
theorem. Thesesolids
5.13 Cavalieri's
section areashave the samevolume (Fig. 5.13). We can seethis imrnediately from
havethe samevolume.You can illustrate
Eq. (1) becauset}Ie cross-sectionarea function A(.r) is the same in each case. f
this yourself
with stacks coins.
of

13. Exercises
5.2 377
Exercises
5.2
Areas
Cross-Section
In Exercises and 2, find a fomula for the arcaA(r) of the cross
I
sectionsof the solid perpendicular the l-axrs.
to
1. The solid lies betweenplanesperpendicular the r-axis at i =
to
-l and,r:1. In eachcase,the crosssections perpendicular
to the r-axis betweentheseplanesrun from the semicircley =
-^4 - ,r2 to the semicircley : 4[ - ,2.
a) The cross sectionsme circular disks with diametersin the
xy-plane.
planesperpendicular ther-axis at.r : 0
The solidliesbetween to
andl : 4. The crosssectionsperpendicular the-t-axisbetween
to
these planes run from the parabola y = -aE to the parabola
r: Ji.
a) The cross sectionsare circular disks with diametersin the
r)-plane.
b) The crcsssections squares
are with bases ther),-plane.
in
b) The crosssectionsare squareswith basesin the r)-plane.
c) The cross sectionsarc squarcswith diagonalsin the xy-
plane. (The length of a square'sdiagonalis .,4 times the
lengthof its sides.)
c) The crosssections squares
are with diagonals ther)-plane.
in
d) The crosssectionsare equilateraltriangleswith basesin the
rJ-plane.
Volumes Slicing
by
Find the volumesof the solids in Exercises3-12.
3. The solid lies betweenplanesperpendicular the,-axis at .r : 0
to
and r :4. The crosssections perpendicular the axis on the
to
The crosssectionsarc equilateraltriangleswith bases the
in interval 0 5 i :4 are squareswhose diagonalsrun ftom the
d)
parabola1 : -.vE to the parabolaf : Jtr.
ry-plane.

14. 374 chapter5: Applications Integrals
of
4, The solid lies between planes perpendicularto the r-axis at -x :
- 1 and r : I . The cross sections perpendicular to the ir axis are
run from lhe parabolay - x7 to
circulardisks whosediamerers
the Darabola :2 - t2.
t
5. The solid lies betweenplanesperpendicular the r-axis at
to Cavalieri's
Theorem
r : - I andr : 1. The crosssections
perpendicular the axis
to
ll. A twisted solid. A square sidelengths lies in a planeper-
of
betweentheseplanesare vertical squares whosebase edges
pendicular a line L Onevertex the square on t. As this
to of lies
run from the semicircle - - 1(1- rt lo rhe semicircle -
y
J
squ&emovesa distance alongl, the square
l? tums one revo-
!7-7.
lution aboutZ to generatea corkscrewlike column with square
perpendicular ther-axis at r =
6. The solidlies between planes to crosssections.
I and n: l. The crosssections perpendicular the ,r-axis
to
a) Find the volume of the column.
planes squares
between these are whosediagonals frcm the
run
b) What will the volumebe if the squaretums twice instead
semicircle : -{ -7 to the semicircle
y: JT=V. Qn
) Givercasons your answer
of once? for
lengthof a square's
diagonal r/Z timesthelengthof its sides.)
is
berween curve) : 2 vfi; planes
12. A solid lies between
7. t he base thesolidis thereSion
of lhe
The qoss sections perpendicular tbe -{-axisat
and the interval[0, r] on the.r-axis. perpen- to
-r = 0 ard .r : 12.The cross
dicularto the r-axis are
sections planes perpendicular
by
to ther-axis arecirculardisks
whosediametersrun from the line
) : -r/2 to the line ) : -r.
Explain why the solid has the
same volumeas a right circular
conewith baseradius3 and
height12.
13. Cavalieri'soriginal theorem. ProveCavalieri's
original theo
veftical equilateral triangles with bases running from the
a)
rem (marginalnote, page 376), assuming
that each region is
I-axrs to me culve:
boundedabove belowby thegraphs continuous
and of functions.
b) vertical squareswith basesrunning from the r-axis to the
t4. The volume of a hemisphere(a classical
cutve. application af Cav-
alieri's theorem). Derive the formula V : (2/3)zR3 for the
8. The solid lies between planes pe.pendicularto the r-axis at -x =
volumeof a hemisphere radiusR by comparing crosssec-
of its
-r /3 and x : xr/3. The crcss sectionsperpendicularto the
r-
tionswith the crosssections a solid right circularcylinderof
of
axis are
mdiusR andheightR from which a solidright circularconeof
circular disks with diameters running from the curve y :
a)
baseradius andheishtR hasbeenremoved.
R
tan,r to the curve ): secr;
verlical squares whose base edges run from the curve y :
b)
tanr to the cuNe ): SeCr.
9. The solid lies betweenplanesperpendicularto the y-axis at y : 0
and ) : 2. The cross sectionsperpendicularto the ]-axis are cir-
cular disks with diametersrunning from they-axis to the parabola
10. The base of the solid is the disk j'? + ),2 5 1. The cross sections
I and y - I
by planesperpendicularto the )-axis between], -
are isoscelesdght triangles with one leg in the disk.

15. 5.3 Volumesof Solidsof Revolution-Disksand Washers 379
rn::rt sotids Revorution-Disks
of
E n
The most common applicarion of the method of slicing is to solids of revolution.
Solids of revolution are solids whose shapescan be generatedby revoJvingplane
regions about axes. Thread spools are solids of revolution; so are hand weights
and billiard balls. Solids of revolutign sometimeshave volumes we can find with
formulas liom geometry, as in the case of a billiard ball. But when we want to
find the volume of a blimp or to predict the weight of a part we are going to have
turned on a lathe, formulas from geometry are of little help and we tum to calculus
lor lhe un:* ers.
6
revolution
Axisof
revolution
s
#
If we can arange for the region to be the region betweenthe graph of a contin-
uous function I = R(r), a 5 x < ,, and the x-axis, and for the axis of revolution
to be the x-axis (Fig. 5.14), we can find the solid's volume in the following way.
The typical cross section of the solid perpendicularto the axis of revolution is
a disk of radius R(x) and area
A(.r) : 71136iut1: rlR(x)12.
:
The solid's volume, being the integral ofA from.r - a to x : b, is the integral of
Cross sectionperpendjcularto
zfR(x)1'zfrom a to D.
the axis at r is a disk ofarea
,4(r) = 7r(radius)2 tt (R(,r))2
=
Volume of a Solid of Revolution (Rotation About the .r-axis)
The volume of the solid generatedby revolving about the.{-axis the region
betweenthe -r-axis and the graph of the continuousfunction ) : R(r), a S
:r 5 b, is
v = [ , ftua;ur1t : [' r l R 1 x ) l 2 l r .
a* (1)
z
J. J,,
EXAMPLE I The region betweenthe curye y :.,rE,O <.{ S 4, and the r-axis
and the x-axisfrom a to b about the
x-axis. is revolved about the .x axis to generatea solid. Find its volume.

16. 380 Chapter Applications lntegrals
5: of
:l,
Solution We draw ligures showing the region, a typical radius, and the generated
solid (Fig. 5.15).The volumeis
fb
v : I trlR(x)'dx F - qr l r
f1
: I rlrtTax /l1ft i
Jo-
(4t2
f4
- u . ,t 2 l a
=tr
lo x d x LJo =z-:8n.
1
l
J
,, .f
How to Find Volumes Using Eq. (1)
1, Draw the regionandidentifythe radiusfunctionR(x).
2. Square R(r) andmultiply by z.
3. Intesrate find the volume.
to
The axis of revolution the next example not the .x-axis, the rule for
in is but
calculating volumeis thesame:
the Integmtelr(radius)2
betweenappropriate
limits.
(b)
5.15 The region(a) and solid(b) in EXAMPLE2 Find the volumeof the solid generated revolvingthe region
by
Exampl1.e
bounded y : .rf andthe lines) : l, x:4 about line y : 1.
by the
Solution We drawfiguresshowing region,a typicalradius,andthe generated
the
solid (Fig. 5.16).The volumeis
F
f( lrr
V: I r l 'R ( x ) 1 d r
z
J,
r1
: I n l' J x _ t ) - d x
Jt
f4
: o I 'l , z E+ r ) a ,
Jt
^ .4
^ tt tn
lx- - + x l : -I
:Tl 2':x
^
J o
Lz lr -j
To find the volume of a solid generated by revolving a region between the
)-axis and a curve x : R()), c < l : d, about the y-axis, we use Eq. (i) with r
replaced by ).
Volume of a Solid of Revolution (Rotation About the y-axis)
(b)
5.16 The region(a) and solid(b) in
Example2.
-

17. 5.3 Volumesof Solidsof Revolution_Disksand Washers 391
EXAMPLE 3 Find the volume of the solid generatedby revolving Lnereglon
betweenthe y-axis and the curve x:2/y,I S l,:4, aboutthe y-axis.
Solution We draw figures showing the region, a typical radius, and the generated
solid (Fig. 5.17).The volumeis
v: :zln6)1'z
ay l_ l (lr
r
l,o
= ;) o,
t4 /'r2
J, /11 I
r
r^ 4 T r'1.
l3l
I - tlt : qr | -: | : 4n r . t
-t
Jt v L )lr L4J
.l
EXAMPLE 4 Find the volume of the solid generatedby revolving the region
betweenthe parabola :12 + I and the line I :3 aboutthe line; :3.
I
so/ut on We draw ligures showing the region, a typical radius, and the generated
solid (Fig. 5.18). The volume is
( N1
,: l.r r: )
J nrlR(y)'?dy
pD
/11r:1 rf -ll
rl2 - y212
: dy
I
(b)
/1
:r
5.17 The region(a) and solid(b) in L4-4y2+y4ldy
I
Example3.
.,,51f
: 7 r | 4 y - ; - y '+ a I
4
l J )l^
I
64rJ2
t5
R(])=3-()'?+ 1)
)
z
0
-.'t
r:t-+r
I
(a.) (b)
5.78 The region(a) and solid(b) in
Example4.
l

18. Chapter 5: Applicationsof lntegrals
382
The WasherMethod
If the region we revolve to generate a solid does not border on or cross the axis of
revolution, the solid has a hole in it (Fig. 5.19). The cross sectionsperpendicular
to the axis of revolution ate washersinstead of disks. The dimensionsof a typical
washer are
R(-r)
Outerradius:
, rx )
Inner radius:
areais
The washer's
A(x) : TtlR(;)lz rVG)12: o (tn{)l' - tt(r)lt).
-
The Washer Fonnula for Finding Volumes
fb
.lrtx)f)dx
v: I (3)
ra ( [ R r x ) | ' : |
|
tl
outer inner
radius
mdius
squarcd squarcd
) = R(-r)
Notice that the function integrated in Eq. (3) is t(Rz - r2), not T (R - r)'z. Also
notice that Eq. (3) gives the disk method fomula if /(jr) is zero throughout [a, r].
Thus. the disk method is a soecial case of the washer method.
The region boundedby the curve y : xz + | and the line y :
EXAMPLE 5
-x t 3 is revolved about the -r-axis to generatea solid. Find the volume of the
solid.
5.?9 The crosssectionsof the solid of
revolution generated here are washers, Solution
not disks, the integralf Ak) dx leads
so
Sfep t; Draw the region and sketcha line segmentacrossit perpendicularto the
to a slightlydifferentformula.
(the red segment Fig. 5.20).
in
a;dsof revolution
Step 2,' Find the limits of integrationby finding the.r-coordinates the intersection
of
points.
x2+l:-x+3
,'+*-2:o
(x+2)(x-l):0
, -_1 .-l
Step 3; Find the outer and inner radii of the washerthat would be swept out by
the line segmentif it were revolved about the x-axis along with the region. ftVe
drewthe washer Fig. 5.21,but in your own work you neednot do that.)These
in
radii are the distances the endsof the line sesmentfrom the axis of revolution.
of
R(x) : -1 -1-
3
Outer radius:
r(x):a241
Inner radius:

19. 5.3 Volumes Solids Revolution-Disks Washers 383
of of and
+-1
(1,2)
lntegratlon
j . . i T h er e g i o ni n E x a m p l 5 s p a n n e d y
e b
a line segmentperpendicular the axis to
of revolution. when the regionis 5.21 The inner and outer radii of the
cmss section
Washer
about the x-axis, line
revolved the Outerradius: = r+3 washersweptout by the line segmentin
R(r)
will generatea washer.
segment r(.r) = 12 + I
Innerradius: Fi9.5.20.
Step 4; Evaluate the volume integral.
rb
-
,= ([Rrx)l' lr(x)]')
dx F ( t 1 - il
J,,
lt ^
= ({ x | 3)' tx' I lt'tdt
I ltn!1 l
/_rtr
fl
= / z(S 6x x2 xa;dx
J-2
r' r'l' tllr
-'l*t^
i sl, s LI
How to FindVolumesby the WasherMethod
Draw the region and sketch a line segmentacross it perpend.icularto
l.
the q,is of revolution. When the region is revolved, this segmentwill
generatea typical washer cross section of the generatedsolid.
2. Find the limils of integration.
Find the outer qnd inner rqdii of the washer swept out by the line
segment.
4. Integrate to flnd the volume.
To find the volume of a solid generatedby revolving a region about the )-axis,
we use the stepslisted above but integratewith respectto l instead of ,r.
The regionboundedby the parabola : x2 and the line y : 2y
EXAMPLE 6 I
quadrantis revolved about the )-axis to generatea solid. Find the volume
in the first
of the solid.

20. 384 Chapter Applications Integrals
5: of
Solution
Step t.' Draw the region and sketch a line segment across it perpendicular to the
axis of revolution, in this case the ),-axis (Fig. 5.22).
Step 2: The line and parabola intersect at ) : 0 and l : 4, so the limits of inte-
grationare c :0 andd :4.
Step 3.' The radii of the washer swept out by the line segment are R(y) : a/ry,
r(y) : y/2 (Figs.5.22 and 5.23).
?
t22 The region,limitsof integration, 523 The washersweptout by the line
and radii in Example
6. segmentin Fig.5.22.
Step 4:
Eq. (-l ) 'irh i
-
,: , (tnci)l, tr(y)t2)
dy
1
: l,^ -ltrJ')r,
(ltl' sleps I .rnd -l
:L'(,-';)*:l+-i]:, 8
3
EXAMPLE 7 The region in the first quadrantenclosedby the parabola y :2e2,
the y-axis, and the line ) : I is revolved about the line,r : 3/2 to generatea solid.
)=*2or*:f
t
5ol
Find the volume of the solid.
4!.=;-ti
Solution
Step 1.' Draw the region and sketch a line segment across it perpendicular to the
Er
axis of revolution, in this case the line ; :3/2 (Fig. 5.24).
Step 2: Tbe limits of integration are ) : 0 to l : 1.
Step 3.' The radii of the washer swept out by the line segment arc R(y) :3/2,
524 The region,limitsof integration,
r(y) : (3/2) - -/1, (Fiss. 5.24 and 5.25).
and radii in Example
7.