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¨ U¸cKatlı ˙ Integraller ¨ U¸cKatlı ˙ Integraller Tıpkı tek de˘ gi¸ skenli fonksiyonlar i¸cin tek katlı, iki de˘ gi¸ skenli fonksiyonlar i¸cin ¸cift katlı integrali tanımladı˘ gımız gibi ¨ u¸cde˘ gi¸ skenli fonksiyonlar i¸ cin de ¨ u¸c katlı integrali tanımlayabiliriz. ¨ Once, f fonksiyonunun B = {(x, y, z )|a x b, c y d, r z s} dikd¨ ortgenler prizması ¨ uzerinde tanımlandı˘ gı basit durumu ele alalım.

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• Üç Katlı İntegraller

Üç Katlı İntegraller

Tıpkı tek değişkenli fonksiyonlar için tek katlı, iki değişkenli fonksiyonlariçin çift katlı integrali tanımladığımız gibi üç değişkenli fonksiyonlar içinde üç katlı integrali tanımlayabiliriz. Önce, f fonksiyonunun

B = {(x, y, z)|a ≤ x ≤ b, c ≤ y ≤ d, r ≤ z ≤ s}

dikdörtgenler prizması üzerinde tanımlandığı basit durumu ele alalım.

• Üç Katlı İntegraller

Triple Integrals � � � � � � � � � � � � � � � �

Just as we defined single integrals for functions of one variable and double integralsfor functions of two variables, so we can define triple integrals for functions of threevariables. Let’s first deal with the simplest case where is defined on a rectangular box:

The first step is to divide B into sub-boxes. We do this by dividing the interval into l subintervals of equal width , dividing into m subintervals ofwidth , and dividing into n subintervals of width . The planes through theendpoints of these subintervals parallel to the coordinate planes divide the box into

sub-boxes

which are shown in Figure 1. Each sub-box has volume .Then we form the triple Riemann sum

where the sample point is in . By analogy with the definition of adouble integral (12.1.5), we define the triple integral as the limit of the triple Riemannsums in (2).

Definition The triple integral of over the box is

if this limit exists.

Again, the triple integral always exists if is continuous. We can choose the samplepoint to be any point in the sub-box, but if we choose it to be the point weget a simpler-looking expression for the triple integral:

Just as for double integrals, the practical method for evaluating triple integrals is toexpress them as iterated integrals as follows.

Fubini’s Theorem for Triple Integrals If is continuous on the rectangular box, then

yyyB

f �x, y, z� dV � ys

r y

d

c y

b

a f �x, y, z� dx dy dz

B � �a, b� � �c, d � � �r, s�f4

yyyB

f �x, y, z� dV � lim l, m, n l �

�l

i�1 �

m

j�1 �

n

k�1 f �xi, yj, zk � �V

�xi, yj, zk �f

yyyB

f �x, y, z� dV � lim l, m, n l �

�l

i�1 �

m

j�1 �

n

k�1 f �xi jk* , yi jk* , zi jk* � �V

Bf3

Bi jk�xi jk* , yi jk* , zi jk* �

�l

i�1 �

m

j�1 �

n

k�1 f �xijk* , yijk* , zijk* � �V2

�V � �x �y �z

Bi jk � �xi�1, xi� � �yj�1, yj� � �zk�1, zk�

lmnB

�z�r, s��y�c, d ��x�xi�1, xi �

�a, b�

B � ��x, y, z� a � x � b, c � y � d, r � z � s1f

12.7

SECTION 12.7 TRIPLE INTEGRALS � 883

FIGURE 1

z

yx

B

z

yx

Bijk

ÎxÎy

Îz

İlk adım, B yi daha küçük kutularaayırmaktır. Bunu, [a, b] aralığını l tane eşit∆x uzunluğunda [xi−1, xi], [c, d] aralığınım tane eşit ∆y uzunluğunda, [r, s] aralığının tane eşit ∆z uzunluğunda altaralığabölerek yaparız. Bu aralıkların uç nokta-larından geçen, koordinat düzlemlerine par-alelolan düzlemler B kutusunu daha küçüklmn tane

Bijk = [xi−1, xi]× [yi−1, yi]× [zi−1, zi]

alt kutularına böler. Her bir küçük kutununhacmi ∆V = ∆x∆y∆z olur.

• Üç Katlı İntegraller

Daha sonra, (x∗ijk, y∗ijk, z

∗ijk) örnek noktaları Bijk nın içinde alınmak

üzere,l∑

i=1

m∑j=1

n∑k=1

f(x∗ijk, y∗ijk, z

∗ijk)∆V

üçlü Riemann toplamını oluştururuz. Çift katlı integral tanımına benzerbir biçimde, üç katlı integrali de yukarıdaki üçlü Riemann toplamlarınınlimiti olarak tanımlarız.

Tanım 1f nin B dikdörtgenler prizması üzerindeki üç katlı integrali, eğer varsa∫∫∫

Bf(x, y, z)dV = lim

l,m,n→∞

l∑i=1

m∑j=1

n∑k=1

f(x∗ijk, y∗ijk, z

∗ijk)∆V

limitidir.

• Üç Katlı İntegraller

Çift katlı integrallerde olduğu gibi, üç katlı integrallerin değerini bulmanınkolay yolu onları aşağıdaki gibi ardışık integraller olarak ifade etmektir:

Teorem 2 (Üç Katlı İntegraller için Fubini Teoremi)

B = [a, b]× [c, d]× [r, s] dikdörtgenler prizması üzerinde süreki olan bir ffonksiyonu için∫∫∫

Bf(x, y, z)dV =

∫ ba

∫ dc

∫ srf(x, y, z)dxdydz

dir.

• Üç Katlı İntegraller

Fubini Teoremi’nin sağ yanındaki ardışık integral önce (y ve z yi sabittutarak) x e, sonra (z yi sabit tutarak) y ye, en sonunda da z ye göreintegral almamız anlamına gelir. İntegrali almak için seçebileceğimiz diğerbeş sıralama da aynı sonucu verir. Örneğin, önce y, sonra z ve ensonunda x e göre integral alırsak∫∫∫

Bf(x, y, z)dV =

∫ dc

∫ sr

∫ baf(x, y, z)dydzdx

elde ederiz.

• Üç Katlı İntegraller

Örnek 3B = {(x, y, z)|0 ≤ x ≤ 0,−1 ≤ y ≤ 2, 0 ≤ z ≤ 3} dikdörtgenler prizmasıolmak üzere,

∫∫∫Bxyz2dV üç katlı integralini hesaplayınız.

Çözüm.

Altı integral sıralamasından herhangi birini seçebiliriz. Önce x, sonra y,en sonunda da z ye göre integral almayı seçersek,∫∫∫

Bxyz2dV =

∫ 30

∫ 2−1

∫ 10xyz2dxdydz =

∫ 30

∫ 2−1

[x2yz2

2

]x=1x=0

dydz

=

∫ 30

∫ 2−1

[yz2

2

]dydz =

∫ 30

[y2z2

4

]y=2y=−1

dz

=

∫ 30

[3z2

4

]dz =

[z3

4

]=

27

4elde ederiz.

• Üç Katlı İntegraller

Şimdi, çift katlı integraller için kullandığımız yönteme çok benzer birşekilde, üç boyutlu uzayda genel, sınırlı bir E bölgesi (bir cisim)üzerinde alınan üç katlı integrali tanımlıyoruz. E bölgesini içine alantipte bir B dikdörtgenler prizması belirliyoruz. Daha sonra, E üzerinde file aynı değerleri, B nin E dışında kalan noktalarında 0 değerini alan birF fonksiyonu tanımlıyoruz. f nin E üzerinde alınan üç katlı integrali∫∫∫

Ef(x, y, z)dV =

∫∫∫Bf(x, y, z)dV

olarak tanımlanır.

• Üç Katlı İntegraller

Şimdi f nin sürekli ve bölgenin basit tipte olduğu durumu ele alacağız.Bir cismin oluşturduğu E bölgesine, x ve y nin iki sürekli fonksiyonuarasında kalıyorsa, 1. Tipte bölge denir. Bunu, Şekilde gösterildiği gibiD bölgesi E nin xy-düzlemi üzerine izdüşümü olmak üzere,

E = {(x, y, z)|(x, y) ∈ D,u1(x, y) ≤ z ≤ u2(x, y)}

olarak ifade edebiliriz. E cisminin üst sınırının denklemi z = u2(x, y), altsınırının denklemi ise z = u1(x, y) olan yüzeylerden oluştuğuna dikkatediniz.

The iterated integral on the right side of Fubini’s Theorem means that we integratefirst with respect to (keeping and fixed), then we integrate with respect to (keep-ing fixed), and finally we integrate with respect to . There are five other possibleorders in which we can integrate, all of which give the same value. For instance, if weintegrate with respect to , then , and then , we have

EXAMPLE 1 Evaluate the triple integral , where is the rectangular boxgiven by

SOLUTION We could use any of the six possible orders of integration. If we choose tointegrate with respect to , then , and then , we obtain

Now we define the triple integral over a general bounded region E in three-dimensional space (a solid) by much the same procedure that we used for double integrals (12.3.2). We enclose in a box of the type given by Equation 1. Then wedefine a function so that it agrees with on but is 0 for points in that are out-side . By definition,

This integral exists if is continuous and the boundary of is “reasonably smooth.”The triple integral has essentially the same properties as the double integral (Proper-ties 6–9 in Section 12.3).

We restrict our attention to continuous functions and to certain simple types ofregions. A solid region is said to be of type 1 if it lies between the graphs of twocontinuous functions of and , that is,

where is the projection of onto the -plane as shown in Figure 2. Notice that theupper boundary of the solid is the surface with equation , while thelower boundary is the surface .

By the same sort of argument that led to (12.3.3), it can be shown that if is a type 1 region given by Equation 5, then

yyyE

f �x, y, z� dV � yyD

yu2�x, y�u1�x, y�

f �x, y, z� dz� dA6

Ez � u1�x, y�

z � u2�x, y�ExyED

E � ��x, y, z� �x, y� � D, u1�x, y� � z � u2�x, y�5yx

Ef

Ef

yyyE

f �x, y, z� dV � yyyB

F�x, y, z� dV

EBEfF

BE

� y3

0 3z2

4 dz �

z3

4 �03

�27

4

� y3

0 y

2

�1 yz2

2 dy dz � y

3

0

y 2z24 �y��1y�2

dz

yyyB

xyz2 dV � y3

0 y

2

�1 y

1

0 xyz2 dx dy dz � y

3

0

y2

�1

x 2yz22 �x�0x�1

dy dz

zyx

B � ��x, y, z� 0 � x � 1, �1 � y � 2, 0 � z � 3

BxxxB xyz2 dV

yyyB

f �x, y, z� dV � yb

a y

s

r y

d

c f �x, y, z� dy dz dx

xzy

zzyzyx

884 � CHAPTER 12 MULTIPLE INTEGRALS

FIGURE 2A type 1 solid region

z

0

xyD

E

z=u™ (x, y)

z=u¡ (x, y)

• Üç Katlı İntegraller

1. Tipte bir E bölgesi için∫∫∫Ef(x, y, z)dV =

∫∫D

[∫ u2(x,y)u1(x,y)

f(x, y, z)dz

]dA

olduğu gösterilebilir.

Denklemin sağ yanında içteki integralin anlamı, x ve y nin sabittutulduğu, dolayısıyla u1(x, y) ve u2(x, y) nin sabit olarak algılandığı,f(x, y, z) nin integralinin z ye göre alındığıdır.

• Üç Katlı İntegraller

The meaning of the inner integral on the right side of Equation 6 is that and areheld fixed, and therefore and are regarded as constants, while is integrated with respect to .

In particular, if the projection of onto the -plane is a type I plane region (asin Figure 3), then

and Equation 6 becomes

If, on the other hand, is a type II plane region (as in Figure 4), then

and Equation 6 becomes

EXAMPLE 2 Evaluate , where is the solid tetrahedron bounded by the fourplanes , , , and .

SOLUTION When we set up a triple integral it’s wise to draw two diagrams: one of the solid region (see Figure 5) and one of its projection on the -plane (see Figure 6). The lower boundary of the tetrahedron is the plane and the upper boundary is the plane (or ), so we use and in Formula 7. Notice that the planes and

intersect in the line (or ) in the -plane. So the projec-tion of is the triangular region shown in Figure 6, and we have

This description of as a type 1 region enables us to evaluate the integral as follows:

� 16 y1

0 �1 � x�3 dx �

1

6 � �1 � x�44 �0

1

�1

24

� 12 y1

0

� �1 � x � y�33 �y�0y�1�x

dx

� 12 y1

0 y

1�x

0 �1 � x � y�2 dy dx

� y1

0

y1�x

0

z22 �z�0z�1�x�y

dy dx yyyE

z dV � y1

0 y

1�x

0 y

1�x�y

0 z dz dy dx

E

E � ��x, y, z� 0 � x � 1, 0 � y � 1 � x, 0 � z � 1 � x � y9E

xyy � 1 � xx � y � 1z � 0x � y � z � 1u2�x, y� � 1 � x � y

u1�x, y� � 0z � 1 � x � yx � y � z � 1z � 0

xyDE

x � y � z � 1z � 0y � 0x � 0Exxx

E z dV

yyyE

f �x, y, z� dV � yd

c y

h2� y�

h1� y� y

u2�x, y�

u1�x, y� f �x, y, z� dz dx dy8

E � ��x, y, z� c � y � d, h1�y� � x � h2�y�, u1�x, y� � z � u2�x, y�

D

yyyE

f �x, y, z� dV � yb

a y

t2�x�

t1�x� y

u2�x, y�

u1�x, y� f �x, y, z� dz dy dx7

E � ��x, y, z� a � x � b, t1�x� � y � t2�x�, u1�x, y� � z � u2�x, y�

xyEDz

f �x, y, z�u2�x, y�u1�x, y�yx

SECTION 12.7 TRIPLE INTEGRALS � 885

FIGURE 3A type 1 solid region

z=u¡(x, y)

z=u™(x, y)

y=g™(x)y=g¡(x)

z

0

yx

a

D

E

b

FIGURE 4Another type 1 solid region

x

0

z

y

c d

z=u™(x, y)

x=h™(y)

x=h¡(y)

z=u¡(x, y)E

D

FIGURE 5

x

0

z

y(1, 0, 0)

(0, 1, 0)

(0, 0, 1)

E

z=1-x-y

z=0

0

y

1

x1y=0

y=1-x

D

FIGURE 6

Özel olarak, E bölgesinin xy-düzlemi üzerine izdüşümü olan D bölgesi I.Tipte ise

E = {(x, y, z)|a ≤ x ≤ b, g1(x) ≤ y ≤ g2(x), u1(x, y) ≤ z ≤ u2(x, y)}∫∫∫Ef(x, y, z)dV =

∫ ba

∫ g2(x)g1(x)

∫ u2(x,y)u1(x,y)

f(x, y, z)dzdydx

biçimini alır.

• Üç Katlı İntegraller

Örnek 4E bölgesi, x = 0, y = 0, z = 0 ve x+ y + z = 1 olarak verilen dört

düzlem tarafından sınırlanan düzgün dörtyüzlü olmak üzere

∫∫∫EzdV

integralini hesaplayınız.

Çözüm.

The meaning of the inner integral on the right side of Equation 6 is that and areheld fixed, and therefore and are regarded as constants, while is integrated with respect to .

In particular, if the projection of onto the -plane is a type I plane region (asin Figure 3), then

and Equation 6 becomes

If, on the other hand, is a type II plane region (as in Figure 4), then

and Equation 6 becomes

EXAMPLE 2 Evaluate , where is the solid tetrahedron bounded by the fourplanes , , , and .

SOLUTION When we set up a triple integral it’s wise to draw two diagrams: one of the solid region (see Figure 5) and one of its projection on the -plane (see Figure 6). The lower boundary of the tetrahedron is the plane and the upper boundary is the plane (or ), so we use and in Formula 7. Notice that the planes and

intersect in the line (or ) in the -plane. So the projec-tion of is the triangular region shown in Figure 6, and we have

This description of as a type 1 region enables us to evaluate the integral as follows:

� 16 y1

0 �1 � x�3 dx �

1

6 � �1 � x�44 �0

1

�1

24

� 12 y1

0

� �1 � x � y�33 �y�0y�1�x

dx

� 12 y1

0 y

1�x

0 �1 � x � y�2 dy dx

� y1

0

y1�x

0

z22 �z�0z�1�x�y

dy dx yyyE

z dV � y1

0 y

1�x

0 y

1�x�y

0 z dz dy dx

E

E � ��x, y, z� 0 � x � 1, 0 � y � 1 � x, 0 � z � 1 � x � y9E

xyy � 1 � xx � y � 1z � 0x � y � z � 1u2�x, y� � 1 � x � y

u1�x, y� � 0z � 1 � x � yx � y � z � 1z � 0

xyDE

x � y � z � 1z � 0y � 0x � 0Exxx

E z dV

yyyE

f �x, y, z� dV � yd

c y

h2� y�

h1� y� y

u2�x, y�

u1�x, y� f �x, y, z� dz dx dy8

E � ��x, y, z� c � y � d, h1�y� � x � h2�y�, u1�x, y� � z � u2�x, y�

D

yyyE

f �x, y, z� dV � yb

a y

t2�x�

t1�x� y

u2�x, y�

u1�x, y� f �x, y, z� dz dy dx7

E � ��x, y, z� a � x � b, t1�x� � y � t2�x�, u1�x, y� � z � u2�x, y�

xyEDz

f �x, y, z�u2�x, y�u1�x, y�yx

SECTION 12.7 TRIPLE INTEGRALS � 885

FIGURE 3A type 1 solid region

z=u¡(x, y)

z=u™(x, y)

y=g™(x)y=g¡(x)

z

0

yx

a

D

E

b

FIGURE 4Another type 1 solid region

x

0

z

y

c d

z=u™(x, y)

x=h™(y)

x=h¡(y)

z=u¡(x, y)E

D

FIGURE 5

x

0

z

y(1, 0, 0)

(0, 1, 0)

(0, 0, 1)

E

z=1-x-y

z=0

0

y

1

x1y=0

y=1-x

D

FIGURE 6

Üç katlı bir integrali oluştururken, biri Ecismi, diğeri E nin xy-düzlemi üzerine Dizdüşümü olan iki şekil çizmek yararlıdır.Düzgün dörtyüzlünün alt sınır z = 0düzlemi, üst sınırı x + y + z = 1 düzlemiolduğundan, u1(x, y) = 0 ve u2(x, y) =1− x− y alırız.

• Üç Katlı İntegraller

Çözüm.

x+ y + z = 1 ve z = 0 düzlemlerinin xy-düzlemindeki x+ y = 1doğrusunda kesiştiklerine dikkat ediniz. Dolayısıyla E nin izdüşümüŞekilde gösterilen üçgensel bölge olur ve

E = {(x, y, z)|0 ≤ x ≤ 1, 0 ≤ y ≤ 1− x, 0 ≤ z ≤ 1− x− y}

elde edilir.

The meaning of the inner integral on the right side of Equation 6 is that and areheld fixed, and therefore and are regarded as constants, while is integrated with respect to .

In particular, if the projection of onto the -plane is a type I plane region (asin Figure 3), then

and Equation 6 becomes

If, on the other hand, is a type II plane region (as in Figure 4), then

and Equation 6 becomes

EXAMPLE 2 Evaluate , where is the solid tetrahedron bounded by the fourplanes , , , and .

SOLUTION When we set up a triple integral it’s wise to draw two diagrams: one of the solid region (see Figure 5) and one of its projection on the -plane (see Figure 6). The lower boundary of the tetrahedron is the plane and the upper boundary is the plane (or ), so we use and in Formula 7. Notice that the planes and

intersect in the line (or ) in the -plane. So the projec-tion of is the triangular region shown in Figure 6, and we have

This description of as a type 1 region enables us to evaluate the integral as follows:

� 16 y1

0 �1 � x�3 dx �

1

6 � �1 � x�44 �0

1

�1

24

� 12 y1

0

� �1 � x � y�33 �y�0y�1�x

dx

� 12 y1

0 y

1�x

0 �1 � x � y�2 dy dx

� y1

0

y1�x

0

z22 �z�0z�1�x�y

dy dx yyyE

z dV � y1

0 y

1�x

0 y

1�x�y

0 z dz dy dx

E

E � ��x, y, z� 0 � x � 1, 0 � y � 1 � x, 0 � z � 1 � x � y9E

xyy � 1 � xx � y � 1z � 0x � y � z � 1u2�x, y� � 1 � x � y

u1�x, y� � 0z � 1 � x � yx � y � z � 1z � 0

xyDE

x � y � z � 1z � 0y � 0x � 0Exxx

E z dV

yyyE

f �x, y, z� dV � yd

c y

h2� y�

h1� y� y

u2�x, y�

u1�x, y� f �x, y, z� dz dx dy8

E � ��x, y, z� c � y � d, h1�y� � x � h2�y�, u1�x, y� � z � u2�x, y�

D

yyyE

f �x, y, z� dV � yb

a y

t2�x�

t1�x� y

u2�x, y�

u1�x, y� f �x, y, z� dz dy dx7

E � ��x, y, z� a � x � b, t1�x� � y � t2�x�, u1�x, y� � z � u2�x, y�

xyEDz

f �x, y, z�u2�x, y�u1�x, y�yx

SECTION 12.7 TRIPLE INTEGRALS � 885

FIGURE 3A type 1 solid region

z=u¡(x, y)

z=u™(x, y)

y=g™(x)y=g¡(x)

z

0

yx

a

D

E

b

FIGURE 4Another type 1 solid region

x

0

z

y

c d

z=u™(x, y)

x=h™(y)

x=h¡(y)

z=u¡(x, y)E

D

FIGURE 5

x

0

z

y(1, 0, 0)

(0, 1, 0)

(0, 0, 1)

E

z=1-x-y

z=0

0

y

1

x1y=0

y=1-x

D

FIGURE 6

E nin 1. Tipte bir bölge olarak gösterimi, integrali∫∫∫Ef(x, y, z)dV =

∫ 10

∫ 1−x0

∫ 1−x−y0

zdzdydx

=1

2

∫ 10

∫ 1−x0

(1− x− y)2dydx

=1

6

∫ 10

(1− x)3dx = 124

şeklinde bulmamızı sağlar.

• Üç Katlı İntegraller

E = {(x, y, z)|(y, z) ∈ D,u1(y, z) ≤ x ≤ u2(y, z)}şeklindeki E bölgesine 2. Tipte bölge denir. Bu kez, D bölgesi, E ninyz-düzlemi üzerine izdüşümüdür. Arka yüzey x = u1(y, z), ön yüzeyx = u2(y, z) olduğundan,∫∫∫

Ef(x, y, z)dV =

∫∫D

[∫ u2(y,z)u1(y,z)

f(x, y, z)dx

]dA

elde ederiz.A solid region is of type 2 if it is of the form

where, this time, is the projection of onto the -plane (see Figure 7). The backsurface is , the front surface is , and we have

Finally, a type 3 region is of the form

where is the projection of onto the -plane, is the left surface, andis the right surface (see Figure 8). For this type of region we have

In each of Equations 10 and 11 there may be two possible expressions for the inte-gral depending on whether is a type I or type II plane region (and corresponding toEquations 7 and 8).

EXAMPLE 3 Evaluate , where is the region bounded by the parab-oloid and the plane .

SOLUTION The solid is shown in Figure 9. If we regard it as a type 1 region, then weneed to consider its projection onto the -plane, which is the parabolic region inFigure 10. (The trace of in the plane is the parabola .)

From we obtain , so the lower boundary surface of is and the upper surface is . Therefore, the descriptionof as a type 1 region is

and so we obtain

yyyE

sx 2 � z 2 dV � y2

�2 y

4

x2 y

sy�x 2

�sy�x2 sx 2 � z 2 dz dy dx

E � {�x, y, z� �2 � x � 2, x 2 � y � 4, �sy � x 2 � z � sy � x 2}E

z � sy � x 2z � �sy � x 2Ez � �sy � x 2y � x 2 � z2

x0

y

y=4

y=≈

FIGURE 10Projection on xy-plane

FIGURE 9Region of integration

x

0

z

y4

E

y � x 2z � 0y � x 2 � z2xyD1

E

y � 4y � x 2 � z2Exxx

E sx 2 � z 2 dV

D

yyyE

f �x, y, z� dV � yyD

yu2�x, z�u1�x, z�

f �x, y, z� dy� dA11y � u2�x, z�

y � u1�x, z�xzED

E � ��x, y, z� �x, z� � D, u1�x, z� � y � u2�x, z�

yyyE

f �x, y, z� dV � yyD

yu2� y, z�u1� y, z�

f �x, y, z� dx� dA10x � u2�y, z�x � u1�y, z�

yzED

E � ��x, y, z� �y, z� � D, u1�y, z� � x � u2�y, z�

E

886 � CHAPTER 12 MULTIPLE INTEGRALS

0

z

yx E

D

x=u¡(y, z)

x=u™(y, z)

FIGURE 7A type 2 region

FIGURE 8A type 3 region

x

0

z

yy=u¡(x, z)

DE

y=u™(x, z)

• Üç Katlı İntegraller

E = {(x, y, z)|(x, z) ∈ D,u1(x, z) ≤ y ≤ u2(x, z)}şeklinde verilen bir E bölgesine 3. Tipte bölge denir. Burada, D bölgesiE nin xz-düzlemi üzerine izdüşümü, y = u1(x, z) sol, y = u2(x, z) sağyüzeydir. Bu tipteki bir bölge için∫∫∫

Ef(x, y, z)dV =

∫∫D

[∫ u2(x,z)u1(x,z)

f(x, y, z)dy

]dA

elde edilir.

A solid region is of type 2 if it is of the form

where, this time, is the projection of onto the -plane (see Figure 7). The backsurface is , the front surface is , and we have

Finally, a type 3 region is of the form

where is the projection of onto the -plane, is the left surface, andis the right surface (see Figure 8). For this type of region we have

In each of Equations 10 and 11 there may be two possible expressions for the inte-gral depending on whether is a type I or type II plane region (and corresponding toEquations 7 and 8).

EXAMPLE 3 Evaluate , where is the region bounded by the parab-oloid and the plane .

SOLUTION The solid is shown in Figure 9. If we regard it as a type 1 region, then weneed to consider its projection onto the -plane, which is the parabolic region inFigure 10. (The trace of in the plane is the parabola .)

From we obtain , so the lower boundary surface of is and the upper surface is . Therefore, the descriptionof as a type 1 region is

and so we obtain

yyyE

sx 2 � z 2 dV � y2

�2 y

4

x2 y

sy�x 2

�sy�x2 sx 2 � z 2 dz dy dx

E � {�x, y, z� �2 � x � 2, x 2 � y � 4, �sy � x 2 � z � sy � x 2}E

z � sy � x 2z � �sy � x 2Ez � �sy � x 2y � x 2 � z2

x0

y

y=4

y=≈

FIGURE 10Projection on xy-plane

FIGURE 9Region of integration

x

0

z

y4

E

y � x 2z � 0y � x 2 � z2xyD1

E

y � 4y � x 2 � z2Exxx

E sx 2 � z 2 dV

D

yyyE

f �x, y, z� dV � yyD

yu2�x, z�u1�x, z�

f �x, y, z� dy� dA11y � u2�x, z�

y � u1�x, z�xzED

E � ��x, y, z� �x, z� � D, u1�x, z� � y � u2�x, z�

yyyE

f �x, y, z� dV � yyD

yu2� y, z�u1� y, z�

f �x, y, z� dx� dA10x � u2�y, z�x � u1�y, z�

yzED

E � ��x, y, z� �y, z� � D, u1�y, z� � x � u2�y, z�

E

886 � CHAPTER 12 MULTIPLE INTEGRALS

0

z

yx E

D

x=u¡(y, z)

x=u™(y, z)

FIGURE 7A type 2 region

FIGURE 8A type 3 region

x

0

z

yy=u¡(x, z)

DE

y=u™(x, z)

• Üç Katlı İntegraller

Örnek 5E bölgesi, y = x2 + z2 paraboloidi ve y = 4 düzlemi tarafından sınırlanan

bölge olmak üzere,

∫∫∫E

√x2 + z2dV integralini hesaplayınız.

Çözüm.

Eğer E yi 1. Tipte bir bölge olarak düşünürsek, onun, Şekilde gösterilen,xy-düzlemi üzerindeki izdüşümü olan parabolik D1 bölgesini ele almamızgerekir.

A solid region is of type 2 if it is of the form

where, this time, is the projection of onto the -plane (see Figure 7). The backsurface is , the front surface is , and we have

Finally, a type 3 region is of the form

where is the projection of onto the -plane, is the left surface, andis the right surface (see Figure 8). For this type of region we have

In each of Equations 10 and 11 there may be two possible expressions for the inte-gral depending on whether is a type I or type II plane region (and corresponding toEquations 7 and 8).

EXAMPLE 3 Evaluate , where is the region bounded by the parab-oloid and the plane .

SOLUTION The solid is shown in Figure 9. If we regard it as a type 1 region, then weneed to consider its projection onto the -plane, which is the parabolic region inFigure 10. (The trace of in the plane is the parabola .)

From we obtain , so the lower boundary surface of is and the upper surface is . Therefore, the descriptionof as a type 1 region is

and so we obtain

yyyE

sx 2 � z 2 dV � y2

�2 y

4

x2 y

sy�x 2

�sy�x2 sx 2 � z 2 dz dy dx

E � {�x, y, z� �2 � x � 2, x 2 � y � 4, �sy � x 2 � z � sy � x 2}E

z � sy � x 2z � �sy � x 2Ez � �sy � x 2y � x 2 � z2

x0

y

y=4

y=≈

FIGURE 10Projection on xy-plane

FIGURE 9Region of integration

x

0

z

y4

E

y � x 2z � 0y � x 2 � z2xyD1

E

y � 4y � x 2 � z2ExxxE sx

2 � z 2 dV

D

yyyE

f �x, y, z� dV � yyD

yu2�x, z�u1�x, z�

f �x, y, z� dy� dA11y � u2�x, z�

y � u1�x, z�xzED

E � ��x, y, z� �x, z� � D, u1�x, z� � y � u2�x, z�

yyyE

f �x, y, z� dV � yyD

yu2� y, z�u1� y, z�

f �x, y, z� dx� dA10x � u2�y, z�x � u1�y, z�

yzED

E � ��x, y, z� �y, z� � D, u1�y, z� � x � u2�y, z�

E

886 � CHAPTER 12 MULTIPLE INTEGRALS

0

z

yx E

D

x=u¡(y, z)

x=u™(y, z)

FIGURE 7A type 2 region

FIGURE 8A type 3 region

x

0

z

yy=u¡(x, z)

DE

y=u™(x, z)

A solid region is of type 2 if it is of the form

where, this time, is the projection of onto the -plane (see Figure 7). The backsurface is , the front surface is , and we have

Finally, a type 3 region is of the form

where is the projection of onto the -plane, is the left surface, andis the right surface (see Figure 8). For this type of region we have

In each of Equations 10 and 11 there may be two possible expressions for the inte-gral depending on whether is a type I or type II plane region (and corresponding toEquations 7 and 8).

EXAMPLE 3 Evaluate , where is the region bounded by the parab-oloid and the plane .

SOLUTION The solid is shown in Figure 9. If we regard it as a type 1 region, then weneed to consider its projection onto the -plane, which is the parabolic region inFigure 10. (The trace of in the plane is the parabola .)

From we obtain , so the lower boundary surface of is and the upper surface is . Therefore, the descriptionof as a type 1 region is

and so we obtain

yyyE

sx 2 � z 2 dV � y2

�2 y

4

x2 y

sy�x 2

�sy�x2 sx 2 � z 2 dz dy dx

E � {�x, y, z� �2 � x � 2, x 2 � y � 4, �sy � x 2 � z � sy � x 2}E

z � sy � x 2z � �sy � x 2Ez � �sy � x 2y � x 2 � z2

x0

y

y=4

y=≈

FIGURE 10Projection on xy-plane

FIGURE 9Region of integration

x

0

z

y4

E

y � x 2z � 0y � x 2 � z2xyD1

E

y � 4y � x 2 � z2ExxxE sx

2 � z 2 dV

D

yyyE

f �x, y, z� dV � yyD

yu2�x, z�u1�x, z�

f �x, y, z� dy� dA11y � u2�x, z�

y � u1�x, z�xzED

E � ��x, y, z� �x, z� � D, u1�x, z� � y � u2�x, z�

yyyE

f �x, y, z� dV � yyD

yu2� y, z�u1� y, z�

f �x, y, z� dx� dA10x � u2�y, z�x � u1�y, z�

yzED

E � ��x, y, z� �y, z� � D, u1�y, z� � x � u2�y, z�

E

886 � CHAPTER 12 MULTIPLE INTEGRALS

0

z

yx E

D

x=u¡(y, z)

x=u™(y, z)

FIGURE 7A type 2 region

FIGURE 8A type 3 region

x

0

z

yy=u¡(x, z)

DE

y=u™(x, z)

• Üç Katlı İntegraller

Çözüm (devamı).

y = x2 + z2 den z = ∓√y − x2 elde edildiğinden E nin alt sınırı

z = −√y − x2 yüzeyi, üst sınırı z =

√y − x2 yüzeyidir. Dolayısıyla, E

nin 1. Tipte bir bölge olarak ifadesi

E ={

(x, y, z)| − 2 ≤ x ≤ 2, x2 ≤ y ≤ 4,−√y − x2 ≤ z ≤

√y − x2

}olur ve ∫∫∫

E

√x2 + z2dV =

∫ 2−2

∫ 4x2

∫ √y−x2−√y−x2

√x2 + z2dzdydx

elde ederiz. Bu, doğru olmakla birlikte hesaplanması son derece zor birifadedir.

• Üç Katlı İntegraller

Çözüm (devamı).Although this expression is correct, it is extremely difficult to evaluate. So let’s

instead consider as a type 3 region. As such, its projection onto the -plane isthe disk shown in Figure 11.

Then the left boundary of is the paraboloid and the right boundaryis the plane , so taking and in Equation 11, wehave

Although this integral could be written as

it’s easier to convert to polar coordinates in the -plane: , .This gives

Applications of Triple Integrals

Recall that if , then the single integral represents the area under thecurve from to , and if , then the double integral represents the volume under the surface and above . The correspondinginterpretation of a triple integral , where , is not veryuseful because it would be the “hypervolume” of a four-dimensional object and, ofcourse, that is very difficult to visualize. (Remember that is just the domain of thefunction ; the graph of lies in four-dimensional space.) Nonetheless, the triple inte-gral can be interpreted in different ways in different physical situa-tions, depending on the physical interpretations of , , and .

Let’s begin with the special case where for all points in . Then thetriple integral does represent the volume of :

For example, you can see this in the case of a type 1 region by putting in Formula 6:

yyyE

1 dV � yyD

yu2�x, y�u1�x, y�

dz� dA � yyD

�u2�x, y� � u1�x, y�� dA

f �x, y, z� � 1

V�E � � yyyE

dV12

EEf �x, y, z� � 1

f �x, y, z�zyxxxxE f �x, y, z� dV

ffE

f �x, y, z� � 0xxxE f �x, y, z� dVDz � f �x, y�

xxD f �x, y� dAf �x, y� � 0bay � f �x�x

ba f �x� dxf �x� � 0

� 2�4r 33 � r 55 �02

�128�

15

� y2�

0 y

2

0 �4 � r 2 �r r dr d� � y

2�

0 d� y

2

0 �4r 2 � r 4 � dr

yyyE

sx 2 � z 2 dV � yyD3

�4 � x 2 � z 2 �sx 2 � z 2 dA

z � r sin �x � r cos �xz

y2

�2 y

s4�x2

�s4�x2 �4 � x 2 � z2 �sx 2 � z 2 dz dx

� yyD3

�4 � x 2 � z 2 �sx 2 � z 2 dA

yyyE

sx 2 � z 2 dV � yyD3

y4x2�z2

sx 2 � z 2 dy� dAu2�x, z� � 4u1�x, z� � x 2 � z2y � 4

y � x 2 � z2Ex 2 � z2 � 4

xzD3E

SECTION 12.7 TRIPLE INTEGRALS � 887

FIGURE 11Projection on xz-plane

x0

z

_2 2

| The most difficult step in evaluating atriple integral is setting up an expressionfor the region of integration (such asEquation 9 in Example 2). Rememberthat the limits of integration in the innerintegral contain at most two variables,the limits of integration in the middle integral contain at most one variable,and the limits of integration in the outerintegral must be constants.

E yi 3. Tipte bir bölge olarak düşünelim. Bu du-rumda xz-düzlemi üzerindeki D3 izdüşümü Şekildegösterilen x2 + z2 ≤ 4 dairesi olur. E nin sol sınırıy = x2 + z2 paraboloidi, sağ sınırı y = 4 düzlemiolduğundan, u1(x, z) = x

2 + z2 ve u2(x, z) = 4alarak∫∫∫

E

√x2 + z2dV =

∫∫D3

[∫ 4x2+z2

√x2 + z2dy

]dA

=

∫∫D3

(4− x2 − z2)√x2 + z2dA

elde ederiz. Bu integral∫ 2−2

∫ 4−x2−√4−x2

(4− x2 − z2)√x2 + z2dzdx

• Üç Katlı İntegraller

Çözüm (devamı).

şeklinde yazılabilirse de, integrali kutupsal koordinatlara çevirmek dahakolaydır. xz-düzleminde x = r cos θ, z = r sin θ alalım. Bu bize∫∫∫

E

√x2 + z2dV =

∫∫D3

(4− x2 − z2)√x2 + z2dA

=

∫ 2π0

∫ 20

(4− r2)rrdrdθ

=

∫ 2π0

∫ 20

(4r2 − r4)drdθ

=128π

15

verir.

• Üç Katlı İntegraller Üç Katıl integrallerin Uygulamaları

Üç Katıl integrallerin Uygulamaları

f(x) ≥ 0 olduğunda tek katlı∫ ba f(x)dx integralinin a dan b ye kadar

y = f(x) eğrisinin altında kalan alanı, f(x, y) ≥ 0 olduğunda çift katlı∫∫D f(x, y)dA integralinin z = f(x, y) yüzeyinin altında ve D nin

üstünde kalan hacmi verdiğini anımsayalım.

f(x, y, z) ≥ 0 olduğunda∫∫∫

E f(x, y, z)dV integralinin benzer şekildeyorumlanması bize, gözümüzde canlandırması çok zor olan, dört-boyutlubir cismin “hiper-hacmi” ni vereceğinden pek yararlı değildir. (E ninyalnızca f fonksiyonunun tanım kümesi olduğunu ve f grafiğinindört-boyutlu uzayda olduğunu anımsayınız.)

• Üç Katlı İntegraller Üç Katıl integrallerin Uygulamaları

E deki her nokta için f(x, y, z) = 1 olan özel durumla başlayalım. Budurumda üç katlı integral gerçekten de E nin hacmini verir:∫∫∫

EdV.

Bunu, örneğin, 1. Tipte bir bölge için üç katlı integralde f(x, y, z) = 1alarak görebilirsiniz:∫∫∫

E1dV =

∫∫D

[∫ u2(x,y)u1(x,y)

dz

]dA =

∫∫D

[u2(x, y)− u1(x, y)]dA.

Bu ifadenin z = u1(x, y) ve z = u2(x, y) yüzeyleri arasında kalanbölgenin hacmini verdiğini biliyoruz.

• Üç Katlı İntegraller Üç Katıl integrallerin Uygulamaları

Örnek 6x+ 2y + z = 2, x = 2y, x = 0 ve z = 0 düzlemleri tarafından sınırlananT düzgün dörtyüzlüsünün hacmini bir üç katlı integral kullanarak bulunuz.

Çözüm.

and from Section 12.3 we know this represents the volume that lies between the sur-faces and .

EXAMPLE 4 Use a triple integral to find the volume of the tetrahedron bounded bythe planes , , , and .

SOLUTION The tetrahedron and its projection on the -plane are shown in Figures 12 and 13. The lower boundary of is the plane and the upper boundary is the plane , that is, . Therefore, we have

by the same calculation as in Example 4 in Section 12.3.

(Notice that it is not necessary to use triple integrals to compute volumes. Theysimply give an alternative method for setting up the calculation.)

All the applications of double integrals in Section 12.5 can be immediately ex-tended to triple integrals. For example, if the density function of a solid object thatoccupies the region is , in units of mass per unit volume, at any given point

, then its mass is

and its moments about the three coordinate planes are

Mxy � yyyE

z��x, y, z� dV

Mxz � yyyE

y��x, y, z� dVMyz � yyyE

x��x, y, z� dV14

m � yyyE

��x, y, z� dV13

�x, y, z���x, y, z�E

FIGURE 12 FIGURE 13

y= x2

”1,   ’12

x+2y=2 ”or y=1-    ’x2

D

y

0

1

x1

(0, 1, 0)

(0, 0, 2)

y

x

0

z

x+2y+z=2x=2y

”1,  , 0’12

T

� y1

0 y

1�x�2

x�2 �2 � x � 2y� dy dx � 13

V�T� � yyyT

dV � y1

0 y

1�x�2

x�2 y

2�x�2y

0 dz dy dx

z � 2 � x � 2yx � 2y � z � 2z � 0T

xyDT

z � 0x � 0x � 2yx � 2y � z � 2T

z � u2�x, y�z � u1�x, y�

888 � CHAPTER 12 MULTIPLE INTEGRALS

and from Section 12.3 we know this represents the volume that lies between the sur-faces and .

EXAMPLE 4 Use a triple integral to find the volume of the tetrahedron bounded bythe planes , , , and .

SOLUTION The tetrahedron and its projection on the -plane are shown in Figures 12 and 13. The lower boundary of is the plane and the upper boundary is the plane , that is, . Therefore, we have

by the same calculation as in Example 4 in Section 12.3.

(Notice that it is not necessary to use triple integrals to compute volumes. Theysimply give an alternative method for setting up the calculation.)

All the applications of double integrals in Section 12.5 can be immediately ex-tended to triple integrals. For example, if the density function of a solid object thatoccupies the region is , in units of mass per unit volume, at any given point

, then its mass is

and its moments about the three coordinate planes are

Mxy � yyyE

z��x, y, z� dV

Mxz � yyyE

y��x, y, z� dVMyz � yyyE

x��x, y, z� dV14

m � yyyE

��x, y, z� dV13

�x, y, z���x, y, z�E

FIGURE 12 FIGURE 13

y= x2

”1,   ’12

x+2y=2 ”or y=1-    ’x2

D

y

0

1

x1

(0, 1, 0)

(0, 0, 2)

y

x

0

z

x+2y+z=2x=2y

”1,  , 0’12

T

� y1

0 y

1�x�2

x�2 �2 � x � 2y� dy dx � 13

V�T� � yyyT

dV � y1

0 y

1�x�2

x�2 y

2�x�2y

0 dz dy dx

z � 2 � x � 2yx � 2y � z � 2z � 0T

xyDT

z � 0x � 0x � 2yx � 2y � z � 2T

z � u2�x, y�z � u1�x, y�

888 � CHAPTER 12 MULTIPLE INTEGRALS

• Üç Katlı İntegraller Üç Katıl integrallerin Uygulamaları

Çözüm (devamı).

T nin alt sınırı z = 0 düzlemi ve üst sınırı x+ 2y + z = 2 ya daz = 2− x− 2y düzlemi olduğundan, yapılan hesaplamayla

V (T ) =

∫∫∫TdV =

∫ 10

∫ 1−x/2x/2

∫ 2−x−2y0

dzdydx

=

∫ 10

∫ 1−x/2x/2

(2− x− 2y)dydx = 13

bulunur.

Hacim hesabı için üç katlı integralleri kullanmanın mutlaka gerekliolmadığına bunun yalnızca farklı bir hesaplama olanağı sağladığına dikkatediniz.

• Silindirik ve Küresel Koordinatlarda Üç Katlı integraller Silindirik Koordinatlar

Silindirik ve Küresel Koordinatlarda Üç Katlı integrallerSilindirik Koordinatlar

Silindirik koordinat sisteminde, üç boyutlu uzaydaki bir P noktası sıralı(r, θ, z) üçlüsünü temsil eder. Burada r ve θ değişkenleri P ninxy-düzlemine izdüşümünün kutupsal koordinatları, z ise P ninxy-düzlemine olan yönlü uzaklığıdır.

Triple Integrals in Cylindrical and Spherical Coordinates � � � �

We saw in Section 12.4 that some double integrals are easier to evaluate using polarcoordinates. In this section we see that some triple integrals are easier to evaluateusing cylindrical or spherical coordinates.

Cylindrical Coordinates

Recall from Section 9.7 that the cylindrical coordinates of a point are , where, , and are shown in Figure 1. Suppose that is a type 1 region whose projection

on the -plane is conveniently described in polar coordinates (see Figure 2). In par-ticular, suppose that is continuous and

where is given in polar coordinates by

FIGURE 2

0

z

x

yD

r=h¡(¨)

r=h™(¨)

z=u™(x, y)

z=u¡(x, y)

D � ��r, �� � � � � , h1��� � r � h2���

D

E � ��x, y, z� �x, y� � D, u1�x, y� � z � u2�x, y�

f

xyDEz�r

�r, �, z�P

12.8

SECTION 12.8 TRIPLE INTEGRALS IN CYLINDRICAL AND SPHERICAL COORDINATES � 893

Volumes of Hyperspheres

In this project we find formulas for the volume enclosed by a hypersphere in -dimensionalspace.

1. Use a double integral and the trigonometric substitution , together with Formula 64 in the Table of Integrals, to find the area of a circle with radius .

2. Use a triple integral and trigonometric substitution to find the volume of a sphere with radius .

3. Use a quadruple integral to find the hypervolume enclosed by the hyperspherein . (Use only trigonometric substitution and the reduction

formulas for or .)

4. Use an -tuple integral to find the volume enclosed by a hypersphere of radius in -dimensional space . [Hint: The formulas are different for even and odd.]nn�nn

rn

x cosnx dxx sinnx dx�4x 2 � y 2 � z 2 � w 2 � r 2

r

ry � r sin �

n

DiscoveryProject

FIGURE 1

z

0

x

y

P(r, ̈ , z)

r

Silindirik koordinatlardan Kartezyen koordi-natlara geçmek için aşağıdaki denklemleri kul-lanırız:

x = r cos θ y = r sin θ z = z.

Kartezyen koordinatlardan silindirik koordinat-lara geçmek için ise

r2 = x2 + y2 tan θ =y

xz = z

denklemlerini kullanabiliriz.

• Silindirik ve Küresel Koordinatlarda Üç Katlı integraller Silindirik Koordinatlar

E bölgesi, xy-düzlemi üzerine D izdüşümü kutupsal koordinatlarlabetimlenmeye uygun 1. Tipte bir bölge olsun. Özel olarak, f nin süreklive D bölgesi kutupsal koordinatlarda

D = {(x, y)|α ≤ θ ≤ β, h1(θ) ≤ r ≤ h2(θ)}

olarak verilmek üzere,

E = {(x, y, z)|(x, y) ∈ D,u1(x, y) ≤ z ≤ u2(x, y)}

olduğunu varsayalım.

Triple Integrals in Cylindrical and Spherical Coordinates � � � �

We saw in Section 12.4 that some double integrals are easier to evaluate using polarcoordinates. In this section we see that some triple integrals are easier to evaluateusing cylindrical or spherical coordinates.

Cylindrical Coordinates

Recall from Section 9.7 that the cylindrical coordinates of a point are , where, , and are shown in Figure 1. Suppose that is a type 1 region whose projection

on the -plane is conveniently described in polar coordinates (see Figure 2). In par-ticular, suppose that is continuous and

where is given in polar coordinates by

FIGURE 2

0

z

x

yD

r=h¡(¨)

r=h™(¨)

z=u™(x, y)

z=u¡(x, y)

D � ��r, �� � � � � , h1��� � r � h2���

D

E � ��x, y, z� �x, y� � D, u1�x, y� � z � u2�x, y�

f

xyDEz�r

�r, �, z�P

12.8

SECTION 12.8 TRIPLE INTEGRALS IN CYLINDRICAL AND SPHERICAL COORDINATES � 893

Volumes of Hyperspheres

In this project we find formulas for the volume enclosed by a hypersphere in -dimensionalspace.

1. Use a double integral and the trigonometric substitution , together with Formula 64 in the Table of Integrals, to find the area of a circle with radius .

2. Use a triple integral and trigonometric substitution to find the volume of a sphere with radius .

3. Use a quadruple integral to find the hypervolume enclosed by the hyperspherein . (Use only trigonometric substitution and the reduction

formulas for or .)

4. Use an -tuple integral to find the volume enclosed by a hypersphere of radius in -dimensional space . [Hint: The formulas are different for even and odd.]nn�nn

rn

x cosnx dxx sinnx dx�4x 2 � y 2 � z 2 � w 2 � r 2

r

ry � r sin �

n

DiscoveryProject

FIGURE 1

z

0

x

y

P(r, ̈ , z)

r

• Silindirik ve Küresel Koordinatlarda Üç Katlı integraller Silindirik Koordinatlar

Ef(x, y, z)dV =

∫ βα

∫ h2(θ)h1(θ)

∫ u2(r sin θ)u1(r cos θ)

f(r cos θ, r sin θ, z)rdzdrdθ.

Bu formülü kullanmak, E bölgesi silindirik koordinatlarla kolaycabetimlemeye uygun olduğu ve özellikle f(x, y, z) de x2 + y2 ifadesigeçtiği zaman yararlıdır.

• Silindirik ve Küresel Koordinatlarda Üç Katlı integraller Silindirik Koordinatlar

Örnek 7E cismi, x2 + y2 = 1 silindirinin içinde, z = 4 düzleminin altında vez = 1− x2 − y2 paraboloidinin üstündedir. Bu bölge üzerinde,∫∫∫

E

√x2 + y2dV integralini hesaplayınız.

Çözüm.

We know from Equation 12.7.6 that

But we also know how to evaluate double integrals in polar coordinates. In fact, com-bining Equation 1 with Equation 12.4.3, we obtain

Formula 2 is the formula for triple integration in cylindrical coordinates. It saysthat we convert a triple integral from rectangular to cylindrical coordinates by writing

, , leaving as it is, using the appropriate limits of integrationfor , , and , and replacing by . (Figure 3 shows how to remember this.)It is worthwhile to use this formula when is a solid region easily described in cylin-drical coordinates, and especially when the function involves the expression

.

EXAMPLE 1 A solid lies within the cylinder , below the plane ,and above the paraboloid . (See Figure 4.) The density at any pointis proportional to its distance from the axis of the cylinder. Find the mass of .

SOLUTION In cylindrical coordinates the cylinder is and the paraboloid is, so we can write

Since the density at is proportional to the distance from the -axis, the den-sity function is

where is the proportionality constant. Therefore, from Formula 12.7.13, the massof is

EXAMPLE 2 Evaluate .

SOLUTION This iterated integral is a triple integral over the solid region

E � {�x, y, z� �2 � x � 2, �s4 � x 2 � y � s4 � x 2, sx 2 � y 2 � z � 2}

y2

�2 y

s4�x2

�s4�x2 y

2

sx2�y2 �x 2 � y 2 � dz dy dx

� 2�Kr 3 � r 55 �01

�12�K

5

� y2�

0 y

1

0 Kr 2 �4 � �1 � r 2 �� dr d� � K y

2�

0 d� y

1

0 �3r 2 � r 4 � dr

m � yyyE

Ksx 2 � y 2 dV � y2�

0 y

1

0 y

4

1�r2 �Kr� r dz dr d�

EK

f �x, y, z� � Ksx 2 � y 2 � Kr

z�x, y, z�

E � ��r, �, z� 0 � � � 2�, 0 � r � 1, 1 � r 2 � z � 4

z � 1 � r 2

r � 1

Ez � 1 � x 2 � y 2

z � 4x 2 � y 2 � 1E

x 2 � y2f �x, y, z�

Er dz dr d�dV�rz

zy � r sin �x � r cos �

yyyE

f �x, y, z� dV � y

� y

h2���

h1��� y

u2�r cos �, r sin ��

u1�r cos �, r sin �� f �r cos �, r sin �, z� r dz dr d�2

yyyE

f �x, y, z� dV � yyD

yu2�x, y�u1�x, y�

f �x, y, z� dz� dA1

894 � CHAPTER 12 MULTIPLE INTEGRALS

z

dz

drr d¨

r

z

0

(1, 0, 0)

(0, 0, 1)

(0, 0, 4)

z=4

x

y

[email protected]

FIGURE 3Volume element in cylindricalcoordinates: dV=r dz dr d¨

FIGURE 4

Silindirik koordinatlarda, silindirin ifadesi r = 1 veparaboloidin ifadesi z = 1− r2 olduğundan

E = {(x, y, z)|0 ≤ θ ≤ 2π, 0 ≤ r ≤ 1, 1−r2 ≤ z ≤ 4}

yazabiliriz. Dolayısıyla,∫∫∫E

√x2 + y2dV =

∫ 2π0

∫ 10

∫ 41−r2

rrdzdrdθ

=12π

5olur.

• Silindirik ve Küresel Koordinatlarda Üç Katlı integraller Silindirik Koordinatlar

Örnek 8∫ 2−2

∫ √4−x2−√4−x2

∫ 2√x2+y2

(x2 + y2)dzdydx integralini hesaplayınız.

Çözüm.Bu integral

E ={

(x, y, z)| − 2 ≤ x ≤ 2,−√

4− x2 ≤ y ≤√

4− x2,√x2 + y2 ≤ z ≤ 2

}bölgesi üzerinde bir integraldir ve E nin xy-düzlemi üzerine izdüşümüx2 + y2 ≤ 4 dairesidir. E nin alt yüzeyi z =

√x2 + y2 konisi, üst yüzeyi

de z = 2 düzlemidir.

• Silindirik ve Küresel Koordinatlarda Üç Katlı integraller Silindirik Koordinatlar

Çözüm (devamı).

Bu bölgenin silindirik koordinatlardaki ifadesi çok daha basittir:

E = {(r, θ, z)|0 ≤ θ ≤ 2π, 0 ≤ r ≤ 2, r ≤ z ≤ 2} .

and the projection of onto the -plane is the disk . The lower sur-face of is the cone and its upper surface is the plane . (SeeFigure 5.) This region has a much simpler description in cylindrical coordinates:

Therefore, we have

Spherical Coordinates

In Section 9.7 we defined the spherical coordinates of a point (see Figure 6)and we demonstrated the following relationships between rectangular coordinates andspherical coordinates:

In this coordinate system the counterpart of a rectangular box is a spherical wedge

where , , and . Although we defined triple integrals bydividing solids into small boxes, it can be shown that dividing a solid into small spher-ical wedges always gives the same result. So we divide into smaller spherical wedges

by means of equally spaced spheres , half-planes , and half-cones. Figure 7 shows that is approximately a rectangular box with dimensions

, (arc of a circle with radius angle ), and (arc of a circlewith radius angle ). So an approximation to the volume of is given by

FIGURE 7

z

0

x

yri=∏i sin ˙k

ri Î¨=∏i sin ˙k Î¨

∏i Î˙

∏i sin ˙k Î¨Î∏

Î˙˙k

Î¨

���� � �� i ��� � �� i sin �k ��� � � i2 sin �k �� �� ��

Eijk��� i sin �k,� i sin �k ����� i,� i ����

Eijk� � �k� � � j� � � iEijk

E

d � c � � � � � 2�a � 0

E � ���, �, �� a � � � b, � � � � , c � � � d

z � � cos �y � � sin � sin �x � � sin � cos �3

��, �, ��

� 2� [ 12 r 4 � 15 r 5 ]02

�16�

5

� y2�

0 d� y

2

0 r 3�2 � r� dr

� y2�

0 y

2

0 y

2

r r 2 r dz dr d�

y2

�2 y

s4�x2

�s4�x2 y

2

sx2�y2 �x 2 � y 2 � dz dy dx � yyy

E

�x 2 � y 2 � dV

E � ��r, �, z� 0 � � � 2�, 0 � r � 2, r � z � 2

z � 2z � sx 2 � y 2Ex 2 � y 2 � 4xyE

SECTION 12.8 TRIPLE INTEGRALS IN CYLINDRICAL AND SPHERICAL COORDINATES � 895

FIGURE 5

z=œ„„„„„≈+¥

z

x 2y2

z=2

2

z

0

x

y

P(∏, ̈ , ̇ )∏

˙

¨

FIGURE 6Spherical coordinates of P

Dolayısıyla∫ 2−2

∫ √4−x2−√4−x2

∫ 2√x2+y2

(x2 + y2)dzdydx

=

∫ 2π0

∫ 20

∫ 2rr2rdzdrdθ

=16π

5

elde edilir.

• Silindirik ve Küresel Koordinatlarda Üç Katlı integraller Küresel Koordinatlar

Küresel Koordinatlar

Uzaydaki bir P noktasının (ρ, θ, φ) küresel koordinatları Şekildegösterilmiştir. Burada, ρ = |OP |, P den başlangıç noktasına olanuzaklığı, θ silindirik koordinatlardaki aynı açıyı, φ ise pozitif z-ekseni ileOP arasındaki açıyı göstermektedir. ρ ≥ 0 ve 0 ≤ φ ≤ π olduğuna dikkatediniz.

Spherical Coordinates

The spherical coordinates of a point in space are shown in Figure 5,where is the distance from the origin to , is the same angle as in cylin-drical coordinates, and is the angle between the positive -axis and the line segment

. Note that

The spherical coordinate system is especially useful in problems where there is sym-metry about a point, and the origin is placed at this point. For example, the sphere withcenter the origin and radius has the simple equation (see Figure 6); this is thereason for the name “spherical” coordinates. The graph of the equation is a ver-tical half-plane (see Figure 7), and the equation represents a half-cone with the-axis as its axis (see Figure 8).

The relationship between rectangular and spherical coordinates can be seen fromFigure 9. From triangles and we have

But and , so to convert from spherical to rectangular coordi-nates, we use the equations

Also, the distance formula shows that

We use this equation in converting from rectangular to spherical coordinates.

�2 � x 2 � y 2 � z24

z � � cos �y � � sin � sin x � � sin � cos 3

y � r sin x � r cos

r � � sin �z � � cos �

OPP�OPQ

z� � c

� c� � cc

0 � � � �� � 0

OPz�

P� � � OP �P��, , ��

696 � CHAPTER 9 VECTORS AND THE GEOMETRY OF SPACE

0

z

x

y

FIGURE 6 ∏=c, a sphere FIGURE 8 ˙=c, a half-cone

0

z

c

π/2

• Silindirik ve Küresel Koordinatlarda Üç Katlı integraller Küresel Koordinatlar

Buna göre, küresel koordinatlardan Kartezyen koordinatlara geçiş

x = ρ sinφ cos θ y = ρ sinφ sin θ z = ρ cosφ.

denklemleri ile verilir. Uzaklık formülünden,

ρ2 = x2 + y2 + z2

dir. Bu denklemi Kartezyen koordinatlardan küresel koordinatlarageçerken kullanırız.

• Silindirik ve Küresel Koordinatlarda Üç Katlı integraller Küresel Koordinatlar

Bu koordinat sisteminde bir dikdörtgenler prizmasına,

E = {(ρ, θ, φ)|a ≤ ρ ≤ b, α ≤ θ ≤ β, c ≤ φ ≤ d}

küresel yarığı karşı gelir. Burada a ≥ 0, β − α ≤ 2π ve d− c ≤ π dir.

Ef(x, y, z)dV

=

∫ dc

∫ βα

∫ baf(ρ sinφ cos θ, ρ sinφ sin θ, ρ cosφ)ρ2 sinφdρdθdφ.

• Silindirik ve Küresel Koordinatlarda Üç Katlı integraller Küresel Koordinatlar

Örnek 9B bölgesi B = {(x, y, z)|x2 + y2 + z2 ≤ 1} olmak üzere∫∫∫

Be(x

2+y2+z2)3/2dV integralini hesaplayınız.

Çözüm.

B nin sınırı bir küre olduğu için, küresel koordinatları kullanırız:

B = {(ρ, θ, φ)|0 ≤ ρ ≤ 1, 0 ≤ θ ≤ 2π, 0 ≤ φ ≤ π}.

Ayrıca x2 + y2 + z2 = ρ2 olduğundan, küresel koordinatları kullanmakuygun olur. Buradan∫∫∫

Be(x

2+y2+z2)3/2dV =

∫ π0

∫ 2π0

∫ 10e(ρ

2)3/2ρ2 sinφdρdθdφ

=4π

3(e− 1) bulunur.

• Silindirik ve Küresel Koordinatlarda Üç Katlı integraller Küresel Koordinatlar

NotÖnceki Örnekteki integrali küresel koordinatları kullanmadan hesaplamakson derece sıkıntılı olurdu. Mesela Kartezyen koordinatlarda ardışıkintegral ∫ 1

−1

∫ √1−x2−√1−x2

∫ √1−x2−y2−√

1−x2−y2e(x

2+y2+z2)3/2dzdydx

olurdu.

• Silindirik ve Küresel Koordinatlarda Üç Katlı integraller Küresel Koordinatlar

Örnek 10z =

√x2 + y2 konisinin üstünde ve x2 + y2 + z2 = z küresinin altında

kalan cismin hacmini küresel koordinatları kullanarak bulunuz.

Çözüm.

SOLUTION Since the boundary of is a sphere, we use spherical coordinates:

In addition, spherical coordinates are appropriate because

Thus, (4) gives

NOTE � It would have been extremely awkward to evaluate the integral in Example 3without spherical coordinates. In rectangular coordinates the iterated integral wouldhave been

EXAMPLE 4 Use spherical coordinates to find the volume of the solid that lies abovethe cone and below the sphere . (See Figure 9.)

SOLUTION Notice that the sphere passes through the origin and has center . Wewrite the equation of the sphere in spherical coordinates as

The equation of the cone can be written as

This gives , or . Therefore, the description of the solid inspherical coordinates is

E � ���, �, �� 0 � � � 2�, 0 � � � ��4, 0 � � � cos �

E� � ��4sin � � cos �

� cos � � s�2 sin2� cos2� � �2 sin 2� sin2� � � sin �

� � cos �or�2 � � cos �

(0, 0, 12 )

yx

z(0, 0, 1)

≈+¥[email protected]=z

z=œ„„„„„≈+¥π4

FIGURE 9

x 2 � y 2 � z2 � zz � sx 2 � y 2

y1

�1 y

s1�x2

�s1�x2 y

s1�x2�y2

�s1�x2�y2 e �x

2�y2�z2�3�2 dz dy dx

� [�cos �]0� �2�� [ 13e �3 ]01 �4�

3 �e � 1�

� y�

0 sin � d� y

2�

0 d� y

1

0 �2e �

3 d�

yyyB

e �x2�y2�z2�3�2 dV � y

0 y

2�

0 y

1

0 e��

2�3�2�2 sin � d� d� d�

x 2 � y 2 � z2 � �2

B � ���, �, �� 0 � � � 1, 0 � � � 2�, 0 � � � �

B

SECTION 12.8 TRIPLE INTEGRALS IN CYLINDRICAL AND SPHERICAL COORDINATES � 897

FIGURE 10

� Figure 10 gives another look (this time drawn by Maple) at the solid ofExample 4.

Kürenin başlangıç noktasından geçtiğine vemerkezinin (0, 0, 12) noktası olduğuna dikkatediniz. Küre denklemini küresel koordinat-larda

ρ2 = ρ cosφ ya da ρ = cosφ.

biçiminde yazarız. Koninin denklemi de

ρ cosφ =

√ρ2 sin2 φ cos2 θ + ρ2 sin2 φ sin2 θ = ρ sinφ

olur.

• Silindirik ve Küresel Koordinatlarda Üç Katlı integraller Küresel Koordinatlar

Çözüm (devamı).

Buradan, sinφ = cosφ, φ = π/4 Dolayısıyla, E cisminin küreselkoordinatlardaki ifadesi

E = {(ρ, θ, φ)|0 ≤ θ ≤ 2π, 0 ≤ φ ≤ π/4, 0 ≤ ρ ≤ cosφ}

olur.

6.

� � � � � � � � � � � � �

7–14 � Use cylindrical coordinates.

7. Evaluate , where is the region that liesinside the cylinder and between the planes

and .

8. Evaluate , where is the solid in the firstoctant that lies beneath the paraboloid .

9. Evaluate , where is the solid that lies between the cylinders and , above the

-plane, and below the plane .z � x � 2xyx 2 � y 2 � 4x 2 � y 2 � 1

ExxxE y dV

z � 1 � x 2 � y 2ExxxE �x

3 � xy 2 � dV

z � 4z � �5x 2 � y 2 � 16

ExxxE sx2 � y 2 dV

z

x y21

1–4 � Sketch the solid whose volume is given by the integraland evaluate the integral.

1. 2.

3.

4.

� � � � � � � � � � � � �

5–6 � Set up the triple integral of an arbitrary continuous func-tion in cylindrical or spherical coordinates over thesolid shown.

5. z

xy

3

2

f �x, y, z�

y2�

0 y

��2 y

2

1 �2 sin � d� d� d�

y��6

0 y

��2

0 y

3

0 �2 sin � d� d� d�

y��2

0 y

2

0 y

9�r 2

0 r dz dr d�y

4

0 y

2�

0 y

4

r r dz d� dr

898 � CHAPTER 12 MULTIPLE INTEGRALS

Exercises � � � � � � � � � � � � � � � � � � � � � � � � � �12.8

Figure 11 shows how E is swept out if we integrate first with respect to , then ,and then . The volume of E is

FIGURE 11¨ varies from 0 to 2π.∏ varies from 0 to cos ˙ while

˙ and ¨ are constant.˙ varies from 0 to π/4 while¨ is constant.

z

yx

z

yx

z

yx

�2�

3 y

��4

0 sin � cos3� d� �

2�

3 � cos4�4 �0

��4

��

8

� y2�

0 d� y

��4

0 sin � �33 �

��0

��cos �

d�

V�E� � yyyE

dV � y2�

0 y

��4

0 y

cos �

0 �2 sin � d� d� d�

���

6.

� � � � � � � � � � � � �

7–14 � Use cylindrical coordinates.

7. Evaluate , where is the region that liesinside the cylinder and between the planes

and .

8. Evaluate , where is the solid in the firstoctant that lies beneath the paraboloid .

9. Evaluate , where is the solid that lies between the cylinders and , above the

-plane, and below the plane .z � x � 2xyx 2 � y 2 � 4x 2 � y 2 � 1

ExxxE y dV

z � 1 � x 2 � y 2ExxxE �x

3 � xy 2 � dV

z � 4z � �5x 2 � y 2 � 16

ExxxE sx2 � y 2 dV

z

x y21

1–4 � Sketch the solid whose volume is given by the integraland evaluate the integral.

1. 2.

3.

4.

� � � � � � � � � � � � �

5–6 � Set up the triple integral of an arbitrary continuous func-tion in cylindrical or spherical coordinates over thesolid shown.

5. z

xy

3

2

f �x, y, z�

y2�

0 y

��2 y

2

1 �2 sin � d� d� d�

y��6

0 y

��2

0 y

3

0 �2 sin � d� d� d�

y��2

0 y

2

0 y

9�r 2

0 r dz dr d�y

4

0 y

2�

0 y

4

r r dz d� dr

898 � CHAPTER 12 MULTIPLE INTEGRALS

Exercises � � � � � � � � � � � � � � � � � � � � � � � � � �12.8

Figure 11 shows how E is swept out if we integrate first with respect to , then ,and then . The volume of E is

FIGURE 11¨ varies from 0 to 2π.∏ varies from 0 to cos ˙ while

˙ and ¨ are constant.˙ varies from 0 to π/4 while¨ is constant.

z

yx

z

yx

z

yx

�2�

3 y

��4

0 sin � cos3� d� �

2�

3 � cos4�4 �0

��4

��

8

� y2�

0 d� y

��4

0 sin � �33 �

��0

��cos �

d�

V�E� � yyyE

dV � y2�

0 y

��4

0 y

cos �

0 �2 sin � d� d� d�

���

6.

� � � � � � � � � � � � �

7–14 � Use cylindrical coordinates.

7. Evaluate , where is the region that liesinside the cylinder and between the planes

and .

8. Evaluate , where is the solid in the firstoctant that lies beneath the paraboloid .

9. Evaluate , where is the solid that lies between the cylinders and , above the

-plane, and below the plane .z � x � 2xyx 2 � y 2 � 4x 2 � y 2 � 1

ExxxE y dV

z � 1 � x 2 � y 2ExxxE �x

3 � xy 2 � dV

z � 4z � �5x 2 � y 2 � 16

ExxxE sx2 � y 2 dV

z

x y21

1–4 � Sketch the solid whose volume is given by the integraland evaluate the integral.

1. 2.

3.

4.

� � � � � � � � � � � � �

5–6 � Set up the triple integral of an arbitrary continuous func-tion in cylindrical or spherical coordinates over thesolid shown.

5. z

xy

3

2

f �x, y, z�

y2�

0 y

��2 y

2

1 �2 sin � d� d� d�

y��6

0 y

��2

0 y

3

0 �2 sin � d� d� d�

y��2

0 y

2

0 y

9�r 2

0 r dz dr d�y

4

0 y

2�

0 y

4

r r dz d� dr

898 � CHAPTER 12 MULTIPLE INTEGRALS

Exercises � � � � � � � � � � � � � � � � � � � � � � � � � �12.8

Figure 11 shows how E is swept out if we integrate first with respect to , then ,and then . The volume of E is

FIGURE 11¨ varies from 0 to 2π.∏ varies from 0 to cos ˙ while

˙ and ¨ are constant.˙ varies from 0 to π/4 while¨ is constant.

z

yx

z

yx

z

yx

�2�

3 y

��4

0 sin � cos3� d� �

2�

3 � cos4�4 �0

��4

��

8

� y2�

0 d� y

��4

0 sin � �33 �

��0

��cos �

d�

V�E� � yyyE

dV � y2�

0 y

��4

0 y

cos �

0 �2 sin � d� d� d�

���

ρ, 0 dan cosφ ye değişir φ, 0 dan π/4 e değişir θ, 0 dan 2π ye değişir

• Silindirik ve Küresel Koordinatlarda Üç Katlı integraller Küresel Koordinatlar

Çözüm (devamı).

E nin hacmi

V (E) =

∫∫∫EdV =

∫ 2π0

∫ π/40

∫ cosφ0

ρ2 sinφdρdφdθ

=

∫ 2π0

∫ π/40

sinφ

[ρ3

3

]ρ=cosφρ=0

=2π

3

∫ π/40

sinφ cos3 φdφ

8olur.

Üç Katlı IntegrallerSilindirik ve Küresel Koordinatlarda Üç Katlı integraller