Alanlar Hakk nda Ek Bilgi E griler Aras ndaki Alanlar Alanlar ...kisi.deu.edu.tr/kaan.gurbuzer/matematik1009/...Alanlar Hakk nda Ek Bilgi E griler Aras ndaki Alanlar fve g’nin pozitif

Alanlar Hakkında Ek Bilgi Egriler Arasındaki Alanlar

Alanlar Hakkında Ek BilgiEgriler Arasındaki Alanlar

f ve g, [a, b] aralıgındaki her x icin f(x) ≥ g(x) esitsizligini saglayansurekli fonksiyonlar olmak uzere y = f(x), y = g(x) egrileri, x = a vex = b dusey dogruları arasındaki S bolgesini dusunelim.

More about Areas � � � � � � � � � � � � � � � �

In Chapter 5 we defined and calculated areas of regions that lie under the graphs of functions. Here we use integrals to find areas of more general regions. First we con-sider regions that lie between the graphs of two functions. Then we look at regionsenclosed by parametric curves.

Areas between Curves

Consider the region that lies between two curves and and be-tween the vertical lines and , where and are continuous functions and

for all in . (See Figure 1.)Just as we did for areas under curves in Section 5.1, we divide S into n strips of

equal width and then we approximate the ith strip by a rectangle with base andheight . (See Figure 2. If we like, we could take all of the sample pointsto be right endpoints, in which case .) The Riemann sum

is therefore an approximation to what we intuitively think of as the area of S.

This approximation appears to become better and better as . Therefore, wedefine the area of as the limiting value of the sum of the areas of these approxi-mating rectangles.

SAn l �

FIGURE 2 (b) Approximating rectangles

x

y

b0 a

(a) Typical rectangle

x

y

b0 a

f(x i*)f(x i*)-g(x i*)

_g(x i*)x i*

Îx

�n

i�1 � f �xi*� � t�xi*�� x

xi* � xi

f �xi*� � t�xi*��x

�a, b�xf �x� � t�x�tfx � bx � a

y � t�x�y � f �x�S

6.1

447

In this chapter we explore some of the applications ofthe definite integral by using it to compute areas be-tween curves, volumes of solids, lengths of curves, theaverage value of a function, the work done by a vary-ing force, the center of gravity of a plate, the force ona dam, as well as quantities of interest in biology, eco-nomics, and statistics. The common theme in most ofthese applications is the following general method,

which is similar to the one we used to find areas under curves. We break up a quantity into a largenumber of small parts. We then approximate eachsmall part by a quantity of the form and thusapproximate by a Riemann sum. Then we take thelimit and express as an integral. Finally, we evalu-ate the integral by using the Evaluation Theorem orSimpson’s Rule.

QQ

f �x i*� �x

Q

x

y

b0 a

y=©

y=ƒ

S

FIGURE 1S=s(x, y) | a¯x¯b, ©¯y¯ƒd

Guess the area of an island.Resources / Module 7

/ Areas / Start of Areas


S bolgesinin A alanı

A =

∫ b

a[f(x)− g(x)]dx. (1)

olarak tanımlanır. g(x) = 0 ozel durumunda S, f ’nin grafiginin altındakalan bolge olur.


f ve g’nin pozitif oldugu durumda, (1)’in neden dogru oldugunu sekildengorebilirsiniz.

We recognize the limit in (1) as the definite integral of . Therefore, we havethe following formula for area.

The area of the region bounded by the curves , and thelines , , where and are continuous and for all in

, is

Notice that in the special case where , is the region under the graph of and our general definition of area (1) reduces to our previous definition (Definition5.1.2).

In the case where both and are positive, you can see from Figure 3 why (2) istrue:

EXAMPLE 1 Find the area of the region bounded above by , bounded below by, and bounded on the sides by and .

SOLUTION The region is shown in Figure 4. The upper boundary curve is andthe lower boundary curve is . So we use the area formula (2) with ,

, and :

In Figure 4 we drew a typical approximating rectangle with width as a reminderof the procedure by which the area is defined in (1). In general, when we set up anintegral for an area, it’s helpful to sketch the region to identify the top curve , thebottom curve , and a typical approximating rectangle as in Figure 5. Then the areaof a typical rectangle is and the equation

summarizes the procedure of adding (in a limiting sense) the areas of all the typical rectangles.

Notice that in Figure 5 the left-hand boundary reduces to a point, whereas in Fig-ure 3 the right-hand boundary reduces to a point. In the next example both of the sideboundaries reduce to a point, so the first step is to find a and b.

A � limn l �

�n

i�1 �yT � yB� �x � y

b

a �yT � yB� dx

�yT � yB� �xyB

yT

�x

� e �12 � 1 � e � 1.5

A � y1

0 �ex � x� dx � ex �

12 x 2]1

0

b � 1a � 0, t�x� � xf �x� � exy � x

y � ex

x � 1x � 0y � xy � ex

� yb

a f �x� dx � y

b

a t�x� dx � y

b

a � f �x� � t�x�� dx

A � �area under y � f �x�� area under y � t�x��

tf

fSt�x� � 0

A � yb

a � f �x� � t�x�� dx


y � f �x�, y � t�x�A2

f � t

A � limn l �

�n

i�1 � f �xi*� � t�xi*�� x1

448 � CHAPTER 6 APPLICATIONS OF INTEGRATION

FIGURE 3

A=j ƒ dx-j © dxa

b

a

b

0 x

y

a b

y=ƒ

y=©

S

0 x

y

1

y=´

y=x Îx

x=1

1

FIGURE 4

0 x

y

a b

yT

yB

yT-yB

Îx

FIGURE 5

A = [y = f(x)’in altındaki alan]− [y = g(x)’in altındaki alan]

=

∫ b

af(x) dx−

∫ b

ag(x) dx =

∫ b

a[f(x)− g(x)] dx.


Ornek 1Ustten y = ex, alttan y = x ve kenarlardan x = 0 ve x = 1 ile sınırlı olanbolgenin alanını hesaplayınız.

Cozum.

Bolge Sekilde gosterilmistir.

We recognize the limit in (1) as the definite integral of . Therefore, we havethe following formula for area.

The area of the region bounded by the curves , and thelines , , where and are continuous and for all in

, is

Notice that in the special case where , is the region under the graph of and our general definition of area (1) reduces to our previous definition (Definition5.1.2).

In the case where both and are positive, you can see from Figure 3 why (2) istrue:

EXAMPLE 1 Find the area of the region bounded above by , bounded below by, and bounded on the sides by and .

SOLUTION The region is shown in Figure 4. The upper boundary curve is andthe lower boundary curve is . So we use the area formula (2) with ,

, and :

In Figure 4 we drew a typical approximating rectangle with width as a reminderof the procedure by which the area is defined in (1). In general, when we set up anintegral for an area, it’s helpful to sketch the region to identify the top curve , thebottom curve , and a typical approximating rectangle as in Figure 5. Then the areaof a typical rectangle is and the equation

summarizes the procedure of adding (in a limiting sense) the areas of all the typical rectangles.

Notice that in Figure 5 the left-hand boundary reduces to a point, whereas in Fig-ure 3 the right-hand boundary reduces to a point. In the next example both of the sideboundaries reduce to a point, so the first step is to find a and b.

A � limn l �

�n

i�1 �yT � yB� �x � y

b

a �yT � yB� dx

�yT � yB� �xyB

yT

�x

� e �12 � 1 � e � 1.5

A � y1

0 �ex � x� dx � ex �

12 x 2]1

0

b � 1a � 0, t�x� � xf �x� � exy � x

y � ex

x � 1x � 0y � xy � ex

� yb

a f �x� dx � y

b

a t�x� dx � y

b

a � f �x� � t�x�� dx

A � �area under y � f �x�� area under y � t�x��

tf

fSt�x� � 0

A � yb

a � f �x� � t�x�� dx


y � f �x�, y � t�x�A2

f � t

A � limn l �

�n

i�1 � f �xi*� � t�xi*�� x1


FIGURE 3

A=j ƒ dx-j © dxa

b

a

b

0 x

y

a b

y=ƒ

y=©

S

0 x

y

1

y=´

y=x Îx

x=1

1

FIGURE 4

0 x

y

a b

yT

yB

yT-yB

Îx

FIGURE 5

Ust sınır egrisi y = ex ve alt sınıregrisi y = x’tir. Dolayısıyla, (1)’dekiformulde f(x) = ex, g(x) = x, a = 0,ve b = 1 kullanırsak

A =

∫ 1

0(ex − x)dx =

[ex − 1

2x2]10

= e− 1

2− 1 = e− 1.5

elde ederiz.


Bazı bolgelerle calısmak icin x degiskenini y’nin fonksiyonu olarakdusunmek gerekir. f ve g surekli ve her c ≤ y ≤ d icin f(y) ≥ g(y)olmak uzere x = f(y), x = g(y), y = c ve y = d denklemleriyle sınırlıolan bolgenin alanı

A =

∫ d

c[f(y)− g(y)]dy

olur.

EXAMPLE 4 Figure 8 shows velocity curves for two cars, A and B, that start side byside and move along the same road. What does the area between the curves repre-sent? Use Simpson’s Rule to estimate it.

SOLUTION We know from Section 5.3 that the area under the velocity curve A repre-sents the distance traveled by car A during the first 16 seconds. Similarly, the areaunder curve B is the distance traveled by car B during that time period. So the areabetween these curves, which is the difference of the areas under the curves, is thedistance between the cars after 16 seconds. We read the velocities from the graphand convert them to feet per second .

Using Simpson’s Rule with intervals, so that , we estimate the dis-tance between the cars after 16 seconds:

Some regions are best treated by regarding x as a function of y. If a region isbounded by curves with equations , , , and , where and

are continuous and for (see Figure 9), then its area is

If we write for the right boundary and for the left boundary, then, as Fig-ure 10 illustrates, we have

Here a typical approximating rectangle has dimensions and .�yxR � xL

A � yd

c �xR � xL� dy

xLxR

x

c

d

y

0

y=dx=g(y)

x=f(y)

y=c

Îy

FIGURE 9

0 x

y

c

dxRxL

xR-xL

Îy

FIGURE 10

A � yd

c � f �y� � t�y�� dy

c � y � df �y� � t�y�t

fy � dy � cx � t�y�x � f �y�

� 367 ft

� 23 �0 � 4�13� � 2�20� � 4�23� � 2�25� � 4�28� � 2�29� � 4�29� � 30�

y16

0 �vA � vB� dt

�t � 2n � 8

�1 mih � 52803600 fts�


FIGURE 8

0

10

20

30

40

50

60A

B

2 4 6 8 10 12 14 16 t(seconds)

√ (mi/h)

t 0 2 4 6 8 10 12 14 16

0 34 54 67 76 84 89 92 95

0 21 34 44 51 56 60 63 65

0 13 20 23 25 28 29 29 30vA � vB

vB

vA


Ornek 2y = x− 1 dogrusu ve y2 = 2x+ 6 paraboluyle sınırlı olan bolgenin alanınıbulunuz.

Cozum.

Iki denklemi ortak cozersek kesisim noktalarını (−1,−2) ve (5, 4) olarakbuluruz.

EXAMPLE 5 Find the area enclosed by the line and the parabola.

SOLUTION By solving the two equations we find that the points of intersection areand . We solve the equation of the parabola for x and notice from

Figure 11 that the left and right boundary curves are

We must integrate between the appropriate -values, and . Thus

We could have found the area in Example 5 by integrating with respect to x insteadof y, but the calculation is much more involved. It would have meant splitting theregion in two and computing the areas labeled and in Figure 12. The method weused in Example 5 is much easier.

Areas Enclosed by Parametric Curves

We know that the area under a curve from to is , where. If the curve is given by the parametric equations and ,

, then we can calculate an area formula by using the Substitution Rule forDefinite Integrals as follows:

EXAMPLE 6 Find the area under one arch of the cycloid

(See Figure 13.)

SOLUTION One arch of the cycloid is given by . Using the SubstitutionRule with and , we have

� r 2( 32 � 2�) � 3�r 2

� r 2[ 32 � � 2 sin � �

14 sin 2�]0

2�

� r 2 y2�

0 [1 � 2 cos � �

12 �1 � cos 2��] d�

� r 2 y2�

0 �1 � cos ��2 d� � r 2 y

2�

0 �1 � 2 cos � � cos2�� d�

A � y2�r

0 y dx � y

2�

0 r�1 � cos ��r�1 � cos �� d�

dx � r�1 � cos �� d�y � r�1 � cos ��0 � � � 2�

y � r�1 � cos ��x � r�� sin ��

�or y

� t�t� f ��t� dt�A � y

b

a y dx � y

�

t�t� f ��t� dt

� t � �y � t�t�x � f �t�F�x� � 0

A � xba F�x� dxbay � F�x�

A2A1

� �16 �64� � 8 � 16 � ( 4

3 � 2 � 8) � 18

� �1

2 y 3

3 � � y 2

2� 4y�

�2

4

� y4

�2 (�1

2 y 2 � y � 4) dy

� y4

�2 [�y � 1� � ( 1

2 y 2 � 3)] dy

A � y4

�2 �xR � xL � dy

y � 4y � �2y

xR � y � 1xL � 12 y 2 � 3

�5, 4��1, �2�

y 2 � 2x � 6y � x � 1

SECTION 6.1 MORE ABOUT AREAS � 451

x

y

_2

4

0

(_1, _2)

(5, 4)

xR=y+1

12xL= ¥-3

FIGURE 11

0 x

y

�3

(5, 4)

(_1, _2)

y=x-1

A¡

y=_ 2x+6

A™

y= 2x+6œ„„„„„

œ„„„„„

FIGURE 12

� The limits of integration for are foundas usual with the Substitution Rule. When

, is either . When , is the remaining value.t

x � b or �tx � a

t

0

y

x2πr

FIGURE 13

� The result of Example 6 says that thearea under one arch of the cycloid isthree times the area of the rolling circlethat generates the cycloid (see Example7 in Section 1.7). Galileo guessed thisresult but it was first proved by theFrench mathematician Roberval and theItalian mathematician Torricelli.


Cozum (devamı).

Parabolun denklemini x icin cozeriz ve Sekilden sag ve sol sınır egrilerini

xL =1

2y2 − 3 ve xR = y + 1

olarak buluruz. Integrali, uygun y degerleri y = −2 ve y = 4 arasındahesaplarsak

A =

∫ 4

−2(xR − xL) dy

=

∫ 4

−2

[(y + 1)− (

1

2y2 − 3)

]dy =

∫ 4

−2

(−1

2y2 + y + 4

)dy

= −1

2

(y3

3

)+y2

2+ 4y

]4−2

= −1

6(64) + 8 + 16− (

4

3+ 2− 8)

= 18 buluruz.


Not

EXAMPLE 5 Find the area enclosed by the line and the parabola.

SOLUTION By solving the two equations we find that the points of intersection areand . We solve the equation of the parabola for x and notice from

Figure 11 that the left and right boundary curves are

We must integrate between the appropriate -values, and . Thus

We could have found the area in Example 5 by integrating with respect to x insteadof y, but the calculation is much more involved. It would have meant splitting theregion in two and computing the areas labeled and in Figure 12. The method weused in Example 5 is much easier.

Areas Enclosed by Parametric Curves

We know that the area under a curve from to is , where. If the curve is given by the parametric equations and ,

, then we can calculate an area formula by using the Substitution Rule forDefinite Integrals as follows:

EXAMPLE 6 Find the area under one arch of the cycloid

(See Figure 13.)

SOLUTION One arch of the cycloid is given by . Using the SubstitutionRule with and , we have

� r 2( 32 � 2�) � 3�r 2

� r 2[ 32 � � 2 sin � �

14 sin 2�]0

2�

� r 2 y2�

0 [1 � 2 cos � �

12 �1 � cos 2��] d�

� r 2 y2�

0 �1 � cos ��2 d� � r 2 y

2�

0 �1 � 2 cos � � cos2�� d�

A � y2�r

0 y dx � y

2�

0 r�1 � cos ��r�1 � cos �� d�

dx � r�1 � cos �� d�y � r�1 � cos ��0 � � � 2�

y � r�1 � cos ��x � r�� sin ��

�or y

� t�t� f ��t� dt�A � y

b

a y dx � y

�

t�t� f ��t� dt

� t � �y � t�t�x � f �t�F�x� � 0

A � xba F�x� dxbay � F�x�

A2A1

� �16 �64� � 8 � 16 � ( 4

3 � 2 � 8) � 18

� �1

2 y 3

3 � � y 2

2� 4y�

�2

4

� y4

�2 (�1

2 y 2 � y � 4) dy

� y4

�2 [�y � 1� � ( 1

2 y 2 � 3)] dy

A � y4

�2 �xR � xL � dy

y � 4y � �2y

xR � y � 1xL � 12 y 2 � 3

�5, 4��1, �2�

y 2 � 2x � 6y � x � 1

SECTION 6.1 MORE ABOUT AREAS � 451

x

y

_2

4

0

(_1, _2)

(5, 4)

xR=y+1

12xL= ¥-3

FIGURE 11

0 x

y

�3

(5, 4)

(_1, _2)

y=x-1

A¡

y=_ 2x+6

A™

y= 2x+6œ„„„„„

œ„„„„„

FIGURE 12

� The limits of integration for are foundas usual with the Substitution Rule. When

, is either . When , is the remaining value.t

x � b or �tx � a

t

0

y

x2πr

FIGURE 13

� The result of Example 6 says that thearea under one arch of the cycloid isthree times the area of the rolling circlethat generates the cycloid (see Example7 in Section 1.7). Galileo guessed thisresult but it was first proved by theFrench mathematician Roberval and theItalian mathematician Torricelli.

Ornekdeki alanı, y yerine x e gore integral alarak da bulabilirdik ama budurumda hesaplamalar daha karmasık olurdu. Bolgeyi, Sekilde goruldugugibi, A1 ve A2 diye ikiye ayırmamız gerekirdi. Ornekde kullandıgımızyontem, cok daha basittir.

Hacimler

Hacimler

S’yi bir duzlemle kesip S’nin kesiti dedigimiz duzlemsel bolgeyi eldeederek baslayacagız. a ≤ x ≤ b olmak uzere x-eksenine dik ve xnoktasından gecen Px duzlemindeki S’nin kesitinin alanı A(x) olsun(S’yi, sekilde gosterildigi gibi x’ten gecen bir bıcakla dilimledigimizi ve budilimin alanını hesapladıgımız dusununuz). x, a’dan b’ye arttıkca kesitinalanı A(x) degisecektir.

called the base, and a congruent region in a parallel plane. The cylinder consists ofall points on line segments perpendicular to the base that join to . If the area ofthe base is and the height of the cylinder (the distance from to ) is , then thevolume of the cylinder is defined as

In particular, if the base is a circle with radius , then the cylinder is a circular cylin-der with volume [see Figure 1(b)], and if the base is a rectangle with length

and width , then the cylinder is a rectangular box (also called a rectangular paral-lelepiped) with volume [see Figure 1(c)].

For a solid S that isn’t a cylinder we first “cut” S into pieces and approximate eachpiece by a cylinder. We estimate the volume of S by adding the volumes of the cylin-ders. We arrive at the exact volume of S though a limiting process in which the num-ber of pieces becomes large.

We start by intersecting S with a plane and obtaining a plane region that is called across-section of Let be the area of the cross-section of in a plane perpen-dicular to the -axis and passing through the point , where . (See Figure 2.Think of slicing with a knife through and computing the area of this slice.) Thecross-sectional area will vary as increases from to .

Let’s divide S into n “slabs” of equal width by using the planes , , . . . toslice the solid. (Think of slicing a loaf of bread.) If we choose sample points in

, we can approximate the slab (the part of that lies between the planesand ) by a cylinder with base area and “height” . (See Figure 3.)�xA�xi*�PxiPxi�1

SSiith�xi�1, xi�xi*

Px2Px1�x

FIGURE 2

y

0 xba x

S

Px

A(a)A(b)

A(x)

baxA�x�xS

a � x � bxxPxSA�x�S.

V � lwhwl

V � �r 2hr

V � Ah

VhB2B1A

B2B1

B2

FIGURE 1

h

B¡

B™

h

r

h

l

(a) CylinderV=Ah

(b) Circular cylinderV=πr@h

(c) Rectangular boxV=lwh

w


Watch an animation of Figure 2.Resources / Module 7

/ Volumes / Volumes by Cross-Section

Hacimler

Tanım 3 (Hacmin Tanımı)

S, x = a ve x = b arasında uzanan bir cisim olsun. A surekli birfonksiyon olmak uzere x’ten gecen ve x−eksenine dik olan Px

duzlemindeki S’nin kesitinin alanı A(x) ise S’nin hacmi

V =

∫ b

aA(x)dx

olarak tanımlanır.

V =

∫ b

aA(x)dx formulunu kullandıgımız zaman hatırlamamız gereken

onemli nokta A(x)’in, x’ten gecen ve x−eksenine dik dilimlemeyle eldeedilen kesitin alanı olmasıdır.

Hacimler

Ornek 4

Yarıcapı r olan bir kurenin hacminin V =4

3πr3 oldugunu gosteriniz.

Cozum.

Kureyi, sekildeki gibi, merkezi baslangıc noktasında olacak sekildeyerlestirirsek Px duzlemiyle kurenin kesisimi, (Pisagor teoreminden)yarıcapı y =

√r2 − x2 olan bir cember olur.

The volume of this cylinder is , so an approximation to our intuitive con-ception of the volume of the th slab is

Adding the volumes of these slabs, we get an approximation to the total volume (thatis, what we think of intuitively as the volume):

This approximation appears to become better and better as . (Think of the slicesas becoming thinner and thinner.) Therefore, we define the volume as the limit of thesesums as . But we recognize the limit of Riemann sums as a definite integral andso we have the following definition.

Definition of Volume Let be a solid that lies between and . If thecross-sectional area of in the plane , through x and perpendicular to the x-axis, is , where is a continuous function, then the volume of is

When we use the volume formula it is important to remember that is the area of a moving cross-section obtained by slicing through perpendicu-

lar to the -axis.

EXAMPLE 1 Show that the volume of a sphere of radius is

SOLUTION If we place the sphere so that its center is at the origin (see Figure 4), thenthe plane intersects the sphere in a circle whose radius (from the PythagoreanTheorem) is . So the cross-sectional area is

A�x� � �y 2 � ��r 2 � x 2 �

y � sr 2 � x 2

Px

V � 43 �r 3

r

xxA�x�

V � xba A�x� dx

V � limn l �

�n

i�1 A�xi*� �x � y

b

a A�x� dx

SAA�x�PxS

x � bx � aS

n l �

n l �

V � �n

i�1 A�xi*� �x

V�Si � � A�xi*� �x

SiiA�xi*� �x

y

0 xx¶=ba=x¸ ⁄ ¤ ‹ x¢ x∞ xß

FIGURE 3

xi-1 xi

y

0 xba x*i

Îx

S

SECTION 6.2 VOLUMES � 455

FIGURE 4

x

y

0 x r

r

_r

y

Hacimler

Cozum (devamı).

Dolayısıyla, bu kesitin alanı

A(x) = πy2 = π(r2 − x2)

olur. a = −r ve b = r alarak hacim tanımını kullanırsak

V =

∫ r

−rA(x)dx =

∫ r

−rπ(r2 − x2)dx

= 2π

∫ r

0(r2 − x2)dx

= 2π

[r2x− x3

3

]r0

= 2π

[r3 − r3

3

]=

4

3πr3

buluruz.

Hacimler

Ornek 5Tabanı, bir kenarı L olan bir kare ve yuksekligi h olan piramidin hacminibulunuz.

Cozum.

Sekillerden goruldugu gibi, baslangıc noktası Q’yu piramidin tepenoktasına ve x-eksenini piramidin merkez eksenine yerlestiririz.

Since lies on the circle, we have and so the base of the triangleis . Since the triangle is equilateral, we see from Figure 13(c)

that its height is . The cross-sectional area is therefore

and the volume of the solid is

EXAMPLE 8 Find the volume of a pyramid whose base is a square with side andwhose height is .

SOLUTION We place the origin at the vertex of the pyramid and the -axis along itscentral axis as in Figure 14. Any plane that passes through and is perpendicularto the -axis intersects the pyramid in a square with side of length , say. We canexpress in terms of by observing from the similar triangles in Figure 15 that

and so . [Another method is to observe that the line has slope and so its equation is .] Thus, the cross-sectional area is

The pyramid lies between and , so its volume is

NOTE � We didn’t need to place the vertex of the pyramid at the origin in Example 8.We did so merely to make the equations simple. If, instead, we had placed the centerof the base at the origin and the vertex on the positive -axis, as in Figure 16, you can y

�L2

h 2 x 3

3 �0

h

�L2h

3

V � yh

0 A�x� dx � y

h

0 L2

h 2 x 2 dx

x � hx � 0

x

y

O

x h

FIGURE 14

sx

h

Lx

y

O

P

FIGURE 15

A�x� � s 2 �L2

h 2 x 2

y � Lx�2h�L�2h�OPs � Lxh

x

h�

s2

L2�

s

L

xssx

xPx

xO

hL

� 2 y1

0 s3 �1 � x 2 � dx � 2s3�x �

x 3

3 �0

1

�4s3

3

V � y1

�1 A�x� dx � y

1

�1 s3 �1 � x 2 � dx

A�x� � 12 � 2s1 � x 2 � s3s1 � x 2 � s3 �1 � x 2 �

s3 y � s3s1 � x 2� AB � � 2s1 � x 2ABC

y � s1 � x 2B


Resources / Module 7/ Volumes

/ Start of Mystery of theTopless Pyramid

x

y

h

0

y

FIGURE 16

Since lies on the circle, we have and so the base of the triangleis . Since the triangle is equilateral, we see from Figure 13(c)

that its height is . The cross-sectional area is therefore

and the volume of the solid is

EXAMPLE 8 Find the volume of a pyramid whose base is a square with side andwhose height is .

SOLUTION We place the origin at the vertex of the pyramid and the -axis along itscentral axis as in Figure 14. Any plane that passes through and is perpendicularto the -axis intersects the pyramid in a square with side of length , say. We canexpress in terms of by observing from the similar triangles in Figure 15 that

and so . [Another method is to observe that the line has slope and so its equation is .] Thus, the cross-sectional area is

The pyramid lies between and , so its volume is

NOTE � We didn’t need to place the vertex of the pyramid at the origin in Example 8.We did so merely to make the equations simple. If, instead, we had placed the centerof the base at the origin and the vertex on the positive -axis, as in Figure 16, you can y

�L2

h 2 x 3

3 �0

h

�L2h

3

V � yh

0 A�x� dx � y

h

0 L2

h 2 x 2 dx

x � hx � 0

x

y

O

x h

FIGURE 14

sx

h

Lx

y

O

P

FIGURE 15

A�x� � s 2 �L2

h 2 x 2

y � Lx�2h�L�2h�OPs � Lxh

x

h�

s2

L2�

s

L

xssx

xPx

xO

hL

� 2 y1

0 s3 �1 � x 2 � dx � 2s3�x �

x 3

3 �0

1

�4s3

3

V � y1

�1 A�x� dx � y

1

�1 s3 �1 � x 2 � dx

A�x� � 12 � 2s1 � x 2 � s3s1 � x 2 � s3 �1 � x 2 �

s3 y � s3s1 � x 2� AB � � 2s1 � x 2ABC

y � s1 � x 2B


Resources / Module 7/ Volumes

/ Start of Mystery of theTopless Pyramid

x

y

h

0

y

FIGURE 16

Boylece, x’ten gecen ve x-eksenine dik olanher Px duzleminin piramitlekesisimi bir karedir. Bu karenin bir kenarına s diyelim.

Hacimler

Cozum (devamı).

s’yi x cinsinden yazmak icin Sekildeki benzer ucgenleri kullanırsak

x

h=s/2

L/2=s

L

elde ederiz. Buradan s = LX/h. buluruz. Boylece; kesit alanı,

A(x) = s2 =L2

h2x2

olur. Piramit, x = 0 ve x = h arasında kaldıgı icin hacmi

V =

∫ h

0A(x)dx =

∫ h

0

L2

h2x2dx

=L2

h2x3

3

∣∣∣∣h0

=L2h

3olur.

Hacimler

Ornek 6y =√x egrisi, x-ekseni ve x = 1 dogrusuyla sınırlanan bolgeyi x−ekseni

cevresinde dondurmekle elde edilen cismin hacmini bulunuz.

Cozum.

EXAMPLE 3 Find the volume of the solid obtained by rotating the region bounded by, , and about the -axis.

SOLUTION The region is shown in Figure 7(a) and the resulting solid is shown in Fig-ure 7(b). Because the region is rotated about the y-axis, it makes sense to slice thesolid perpendicular to the y-axis and therefore to integrate with respect to y. If weslice at height y, we get a circular disk with radius x, where . So the area of across-section through y is

and the volume of the approximating cylinder pictured in Figure 7(b) is

Since the solid lies between y � 0 and y � 8, its volume is

FIGURE 7

x

y

0

y=8

x=0y=˛

orx=Œ„y

x

y

0

8

y x(x, y)Îy

(a) (b)

� �[ 35 y 53 ]0

8�

96�

5

V � y8

0 A�y� dy � y

8

0 �y 23 dy

A�y� �y � �y 23 �y

A�y� � �x 2 � � (s3 y)2 � �y 23

x � s3 y

yx � 0y � 8y � x 3

FIGURE 6 (a) (b)

x0 x

yy=œ„x

1

œ„x

Îx

0 x

y

1


See a volume of revolution being formed.Resources / Module 7

/ Volumes / Volumes of Revolution

Hacimler

Cozum (devamı).

Bolge ilk Sekilde gosterilmistir. Eger bolgeyi x−ekseni cevresindedondurursek ikinci Sekildeki cismi elde ederiz. x’ten gecen kesit, yarıcapı√x olan bir cemberdir. Bu kesitin alanı

A(x) = π(√x)2

= πx

olur ve yaklasımı veren silindirin (kalınlıgı ∆x olan bir disk) hacmi ise

A(x)∆x = πx∆x

olarak bulunur. Bu cisim x = 0 ile x = 1 arasındadır. Dolayısıyla; hacim,

V =

∫ 1

0A(x)dx =

∫ 1

0πxdx = π

x2

2

∣∣∣∣10

=π

2

olur.

Hacimler

Ornek 7y = x3, y = 8 ve x = 0 ile sınırlı olan bolgeyi y−ekseni cevresindedondurerek elde edilen cismin hacmini bulunuz.

Cozum.

EXAMPLE 3 Find the volume of the solid obtained by rotating the region bounded by, , and about the -axis.

SOLUTION The region is shown in Figure 7(a) and the resulting solid is shown in Fig-ure 7(b). Because the region is rotated about the y-axis, it makes sense to slice thesolid perpendicular to the y-axis and therefore to integrate with respect to y. If weslice at height y, we get a circular disk with radius x, where . So the area of across-section through y is

and the volume of the approximating cylinder pictured in Figure 7(b) is

Since the solid lies between y � 0 and y � 8, its volume is

FIGURE 7

x

y

0

y=8

x=0y=˛

orx=Œ„y

x

y

0

8

y x(x, y)Îy

(a) (b)

� �[ 35 y 53 ]0

8�

96�

5

V � y8

0 A�y� dy � y

8

0 �y 23 dy

A�y� �y � �y 23 �y

A�y� � �x 2 � � (s3 y)2 � �y 23

x � s3 y

yx � 0y � 8y � x 3

FIGURE 6 (a) (b)

x0 x

yy=œ„x

1

œ„x

Îx

0 x

y

1


See a volume of revolution being formed.Resources / Module 7

/ Volumes / Volumes of Revolution

Hacimler

Cozum (devamı).

Bolge, birinci Sekilde, cisim ise ikinci Sekilde gosterilmistir. Bolgey−ekseni cevresinde donduruldugu icin y−eksenine dik bicimdedilimlemek ve integrali y’ye gore almak daha mantıklı olur. O halde; yyuksekligindeki kesit, yarıcapı x olan cembersel bir disktir. x = 3

√y

oldugu icin y’den gecen kesitin alanı

A(y) = πx2 = π(

3√x)2

= πy2/3

ve ikinci Sekilde gosterilen yaklasım silindirinin hacmi

A(y)∆y = πy2/3∆y

olur. Cisim, y = 0 ve y = 8 arasında kaldıgı icin hacmi

V =

∫ 8

0A(y)dy =

∫ 8

0πy2/3dy =

[3

5y5/3

]80

=96π

5dir.

Hacimler

Ornek 8y = x ve y = x2 egrileriyle cevrili olan R bolgesi, x-ekseni cevresindedondurulmustur. Olusan cismin hacmini bulunuz.

Cozum.

EXAMPLE 4 The region enclosed by the curves and is rotated aboutthe -axis. Find the volume of the resulting solid.

SOLUTION The curves and intersect at the points and . Theregion between them, the solid of rotation, and a cross-section perpendicular to the -axis are shown in Figure 8. A cross-section in the plane has the shape of a

washer (an annular ring) with inner radius and outer radius , so we find thecross-sectional area by subtracting the area of the inner circle from the area of theouter circle:

Therefore, we have

EXAMPLE 5 Find the volume of the solid obtained by rotating the region in Example 4about the line .

SOLUTION The solid and a cross-section are shown in Figure 9. Again a cross-sectionis a washer, but this time the inner radius is and the outer radius is .The cross-sectional area is

and so the volume of is

� �� x 5

5� 5

x 3

3� 4

x 2

2 �0

1

�8�

15

� � y1

0 �x 4 � 5x 2 � 4x� dx

V � y1

0 A�x� dx � � y

1

0 ��2 � x 2 �2 � �2 � x�2 � dx

S

A�x� � ��2 � x 2 �2 � ��2 � x�2

2 � x 22 � x

y � 2

FIGURE 8 (b)

x

y

(0, 0)

(1, 1)

y=≈y=x

�

x

y

0 x

≈

A(x)

(a) (c)

� �� x 3

3�

x 5

5 �0

1

�2�

15

V � y1

0 A�x� dx � y

1

0 ��x 2 � x 4 � dx

A�x� � �x 2 � ��x 2 �2 � ��x 2 � x 4 �

xx 2Pxx

�1, 1��0, 0�y � x 2y � x

xy � x 2y � x�


Hacimler

Cozum (devamı).

y = x ve y = x2 egrileri, (0, 0) ve (1, 1) noktalarında kesisir. Aralarındakibolge, donel cisim ve x-eksenine dik olan kesit Sekilde gosterilmistir. Px

duzlemindeki kesit, ic yarıcapı x2 ve dıs yarıcapı x olan bir halkaseklindedir. Dolayısıyla, alanını bulmak icin buyuk cemberin alanındankucuk cemberin alanını cıkarırsak

A(x) = πx2 − π(x2)2 = π(x2 − x4

)buluruz. Bu durumda,

V =

∫ 1

0A(x)dx =

∫ 1

0π(x2 − x4

)dx =

[x3

3− x5

5

]10

=2π

15

elde ederiz.

Hacimler

Donel CisimlerOnceki orneklerdeki cisimler donel cisimlerdir, cunku bir bolgenin birdogru cevresinde dondurulmesiyle elde edilmislerdir. Genelde donel bircismin hacmini

V =

∫ b

aA(x)dx veya V =

∫ d

cA(y)dy

formulunu kullanarak hesaplarız. Bu formuldeki kesit alanı olan A(x)veya A(y) yi de asagıdaki yollardan birini kullanarak buluruz:

Hacimler

Donel Cisimler (devamı)

• Eger kesit bir diskse, diskin yarıcapını (x veya y cinsinden) bulur ve

A = π(radius)2

formulunu kullanırız.

• Eger kesit bir halkaysa, ic yarıcap ric ve dıs yarıcap rdıs ı sekildenbulur ve dıstaki diskin alanından icteki diskin alanını cıkararakhalkanın alanını buluruz:

A = π(dıs yarıcap)2 − π(ic yarıcap)2.

The solids in Examples 1–5 are all called solids of revolution because they areobtained by revolving a region about a line. In general, we calculate the volume of asolid of revolution by using the basic defining formula

and we find the cross-sectional area or in one of the following ways:

� If the cross-section is a disk (as in Examples 1–3), we find the radius of thedisk (in terms of x or y) and use

� If the cross-section is a washer (as in Examples 4 and 5), we find the innerradius and outer radius from a sketch (as in Figures 9 and 10) andcompute the area of the washer by subtracting the area of the inner diskfrom the area of the outer disk:

The next example gives a further illustration of the procedure.

FIGURE 10

rinrout

A � � �outer radius�2 � � �inner radius�2

routrin

A � ��radius�2

A�y�A�x�

V � yb

a A�x� dx or V � y

d

c A�y� dy

FIGURE 9

4

2

x0 1x

y=≈y=x

y=2

y

xx

y=2

2-≈

≈

2-x

x


Module 6.2 illustrates the forma-tion and computation of volumes

using disks, washers, and shells.

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