Triple Int

8/20/2019 Triple Int

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Calculus 3

Lia Vas

Triple Integrals

Let f (x,y,z ) be a function of three variables

defined on a solid region E consisting of points(x,y,z ) such that

a ≤ x ≤ b,

c(x) ≤ y ≤ d(x),

g(x, y) ≤ z ≤ h(x, y)

The triple integral of f over E is E

f (x,y,z ) dx dy dz

can be computed as follows. E

f (x,y,z ) dx dy dz = ba

d(x)c(x)

h(x,y)g(x,y)

f (x,y,z ) dz

dy

dx

In case the region is given with a bit differentdependence between the bounds, for example as

c ≤ y ≤ d,

a(y) ≤ x ≤ b(y),

g(x, y) ≤ z ≤ h(x, y)

the order of integration has to be different and,in this case, as follows.

E f (x,y,z ) dx dy dz =

dc

b(y)a(y)

h(x,y)g(x,y)

f (x,y,z ) dz

dx

dy

Yet another possible scenario is given in thenext figure. In this case, the region is given bythe bounds as follows.

a ≤ x ≤ b,

g(x) ≤ z ≤ h(x),c(x, z ) ≤ y ≤ d(x, z )

The order of integration in this case is as follows. E

f (x,y,z ) dx dy dz =

ba

h(x)g(x)

d(x,z)c(x,z)

f (x,y,z ) dy

dz

dx

1



Applications of triple integral

We exhibit the use of triple integrals for computing (1) volumes of solid regions in space (2) theaverage value of functions f (x,y,z ), and (3) the mass and the center of mass of a solid region withdensity ρ = ρ(x,y,z ).

Volume. Recall that the area of a region D in xy-plane can be obtained by integrating the areaelement dA = dxdy over D,

A =

Ddxdy.

Analogously, the volume of a solid region E in xyz -space can be obtained by integrating the volumeelement dV = dxdydz over E

V =

dx dy dz.

In particular, if E is the solid region be-tween surfaces z = h(x, y) and z = g(x, y) with

h(x, y) ≥ g(x, y) and D is its projection on xy-plane, then the volume of E can be computed as

V (E ) =

E dx dy dz =

D

z

h(x,y)g(x,y)

dx dy =

D(h(x, y)−g(x, y)) dx dy.

The average value of function. Recall that

• the average value of y = f (x) over interval [a, b] is found as f ave = 1b−a

ba f (x) dx and

• the average value of z = f (x, y) over region D is found as f ave = 1A(D)

D f (x, y) dx dy.

In complete analogy, the average value of u = f (x,y,z ) over a region E in xyz -space is definedas the value f ave such that the triple integral of f (x,y,z ) is equal to the product of the volume of f ave and the volume V (E ) of E . Thus,

f ave = 1V (E )

E f (x,y,z ) dx dy dz

The mass and the center of mass. If E is a solid in space, its mass m, volume V and densityρ

are related by ρ

=

m

V if the density is a constant function. If the density at a point (x,y,z

) is notconstant throughout the region E and it is given by a continuous function ρ(x,y,z ), then ρ is thequotient of differentials dm

dV . Thus, the total mass can be found by integration.

m =

E dm =

E

ρ(x,y,z ) dV ⇒

m =

E ρ(x,y,z ) dx dy dz.

2



The formulas for the coordinates (x,y,z ) of the center of mass of E can be obtained bysimilar arguments using the formulas for the mo-ments about the three axis and are given by

x = 1

m

E

x ρ(x,y,z ) dx dy dz

y = 1

m

E

y ρ(x,y,z ) dx dy dz

z = 1

m

E

z ρ(x,y,z ) dx dy dz

Practice problems.

1. Evaluate the triple integral

a) E

x3y2z dx dy dz

where E = { (x,y,z ) | 1 ≤ x ≤ 2, 0 ≤ y ≤ x, 0 ≤ z ≤ y2 }b)

E 2x dx dy dz

where E = { (x,y,z ) | 0 ≤ y ≤ 2, 0 ≤ x ≤ √ 4 − y2, 0 ≤ z ≤ y }

c)

E 6xy dx dy dz

where E lies under the plane z = x + y + 1 and above the region in the xy-plane boundedby the curves y =

√ x, y = 0 and x = 1.

d) E

xy dx dy dz

where E is the solid tetrahedron with vertices (0,0,0), (1, 0, 0), (0, 2, 0) and (0, 0, 3).

2. Find the volume of the tetrahedron bounded by the coordinate planes and the plane 2x + 3y +6z = 12.

3. Find the average value of the function f (x,y,z ) = xyz over the cube with side length 4 thatlies in the first octant with one vertex in the origin and edges parallel to the coordinate axes.

4. Find the mass and the center of mass of the solid E given in problem 1c) and that has thedensity function ρ(x,y,z ) = 2.

3



Solutions.

1. a) 2

1

x0

y20 x3y2z dx dy dz =

21 x3dx

x0 y2dy

y20 z dz =

21 x3dx

x0 y2dy z2

2|y20 =

21 x3dx

x0

y6

2 dy =

21 x3dx y7

14|x0 =

21

x10

14 dx = x11

14(11)|21 = 13.29.

b) Here you have to evaluate the integral with respect to y last since all the other variables

have y in the bounds. 20 dy √

4−y20 2x dx y0 dz = 20 dy

√ 4−y2

0 2x dx y = 20 y dy x2

|

√ 4−y2

0 = 20 y(4− y2) dy = (2y2 − y4

4 )|20 = 8 − 4 = 4.

c) Sketch the region in xy-plane first. The x-bounds are 0 ≤ x ≤ 1. The y-bounds are0 ≤ y ≤ √ x. The z -bounds are determined by the plane z = x + y + 1 and the xy-plane

and so 0 ≤ z ≤ x + y + 1. So 1

0

√ x0

x+y+10 6xy dx dy dz =

10 6xdx

√ x0 y dy

x+y+10 dz = 1

0 6xdx √ x

0 y dy(x + y + 1) = 10 6xdx (xy2

2 + y3

3 + y2

2 )|

√ x

0 = 10 6xdx (xx

2 + x3/2

3 + x

2) = 65

28.

d) First, find the equation of the plane determined by P = (1, 0, 0), Q = (0, 2, 0) and R =

(0, 0, 3). Vectors−→P Q = (−1, 2, 0) and

−→P R = (−1, 0, 3) are in the plane so the vector perpendic-

ular to the plane can be taken to be the cross product−→P Q×−→P R =

i j k

−1 2 0

−1 0 3

= (6, 3, 2). The

equation of plane, using point P for example, is 6(x− 1 ) + 3y + 2z = 0 ⇒ 6x + 3y + 2z = 6 ⇒z = 3 − 3x− 3

2y.

Thus, the upper bound for z is 3 − 3x − 32

y. The lower bound for z is 0. In xy-plane we havea triangle with vertices (0,0), (1,0) and (0,2). So, the bounds for x are 0 ≤ x ≤ 1. The lowerbound for y is 0 and the upper bound is the line passing (0,2) and (1,0). The equation of thisline is y = −2x + 2. Thus,

xy dx dy dz = 1

0 xdx −2x+2

0 ydy 3−3x−3

2y

0 dz = 1

0 xdx −2x+2

0 ydy (3 − 3x − 32

y) = 10 xdx (3y2

2 − 3xy2

2 − y3

2 )|−2x+20 =

10 xdx (3 (−2x+2)2

2 − 3x (−2x+2)2

2 − (−2x+2)3

2 ) = use calculator= 1

10.

2. The upper bound for z can be obtained by solving 2x + 3y + 6z = 12 for z. So, 0 ≤ z ≤2 − 1

3x − 1

2y. The projection of the relevant region in xy-plane is a triangle determined by the

coordinate axes and by the line 2x + 3y + 6(0) = 12 ⇒ y = 4 − 23

x (alternatively, find theequation of the line passing (6,0) and (0,4)). The bounds for x are 0 ≤ x ≤ 6. The volume is

V = 60 dx

4−2

3x

0 dy 2−1

3x−1

2y

0 dz = 60 dx

4−2

3x

0 dy (2− 13

x− 12

y) = 60 dx (2y− 1

3xy− y2

4 )|4−

2

3x

0 = 60 dx (2(4 − 2

3x) − 1

3x(4− 2

3x) − (4− 2

3x)2

4 ) = use calculator = 8.

3. When integrating over the cube, the bounds for all three variables are 0 and 4. The volume of the cube of side 4 is 43 = 64 (which agrees with the fact that V =

40

40

40 dxdydz = x|40y|40z |40 =

64).

The average value is f ave = 164

40

40

40 xyz dxdydz = 1

64x2

2|40 y

2

2|40 z

2

2|40 = 1

6483 = 8.

4. The bounds are the same as in problem 1c. The mass is m = 1

0

√ x0

x+y+10 2 dx dy dz = 1

0 2 dx √ x

0 (x + y + 1)dy = 10 2(x

√ x + x

2 +√

x) dx = 2.633.

The x-coordinate is x = 12.633

10

√ x0

x+y+10 2xdx dy dz = 1

2.633

10 2x(x

√ x + x

2 +

√ x) dx =

1.7052.633

= 0.647. Similarly you find that y = 0.418, and z = 1.032.

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Triple Int