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15. MULTIPLE INTEGRALS. MULTIPLE INTEGRALS. 15.6 Triple Integrals. In this section, we will learn about: Triple integrals and their applications. TRIPLE INTEGRALS.

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15

MULTIPLE INTEGRALS

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MULTIPLE INTEGRALS

15.6

Triple Integrals

  • In this section, we will learn about:
  • Triple integrals and their applications.
triple integrals
TRIPLE INTEGRALS
  • Just as we defined single integrals for functions of one variable and double integrals for functions of two variables, so we can define triple integrals for functions of three variables.
triple integrals4
TRIPLE INTEGRALS

Equation 1

  • Let’s first deal with the simplest case where f is defined on a rectangular box:
triple integrals5
TRIPLE INTEGRALS
  • The first step is to divide B into sub-boxes—by dividing:
    • The interval [a, b] into lsubintervals [xi-1, xi] of equal width Δx.
    • [c, d] into m subintervals of width Δy.
    • [r, s] into n subintervals of width Δz.
triple integrals6
TRIPLE INTEGRALS
  • The planes through the endpoints of these subintervals parallel to the coordinate planes divide the box B into lmn sub-boxes
    • Each sub-box has volume ΔV = ΔxΔyΔz
triple integrals7
TRIPLE INTEGRALS

Equation 2

  • Then, we form the triple Riemann sum
  • where the sample point is in Bijk.
triple integrals8
TRIPLE INTEGRALS
  • By analogy with the definition of a double integral (Definition 5 in Section 15.1), we define the triple integral as the limit of the triple Riemann sums in Equation 2.
triple integral
TRIPLE INTEGRAL

Definition 3

  • The triple integralof f over the box B is:
  • if this limit exists.
    • Again, the triple integral always exists if fis continuous.
triple integrals10
TRIPLE INTEGRALS
  • We can choose the sample point to be any point in the sub-box.
  • However, if we choose it to be the point (xi, yj, zk) we get a simpler-looking expression:
triple integrals11
TRIPLE INTEGRALS
  • Just as for double integrals, the practical method for evaluating triple integrals is to express them as iterated integrals, as follows.
fubini s th triple integrals
FUBINI’S TH. (TRIPLE INTEGRALS)

Theorem 4

  • If f is continuous on the rectangular box B = [a, b] x [c, d] x [r, s], then
fubini s th triple integrals13
FUBINI’S TH. (TRIPLE INTEGRALS)
  • The iterated integral on the right side of Fubini’s Theorem means that we integrate in the following order:
    • With respect to x (keeping y and z fixed)
    • With respect to y (keeping z fixed)
    • With respect to z
fubini s th triple integrals14
FUBINI’S TH. (TRIPLE INTEGRALS)
  • There are five other possible orders in which we can integrate, all of which give the same value.
    • For instance, if we integrate with respect to y, then z, and then x, we have:
fubini s th triple integrals15
FUBINI’S TH. (TRIPLE INTEGRALS)

Example 1

  • Evaluate the triple integral where B is the rectangular box
fubini s th triple integrals16
FUBINI’S TH. (TRIPLE INTEGRALS)

Example 1

  • We could use any of the six possible orders of integration.
  • If we choose to integrate with respect to x, then y, and then z, we obtain the following result.
integral over bounded region
INTEGRAL OVER BOUNDED REGION
  • Now, we define the triple integral over a general bounded region Ein three-dimensional space (a solid) by much the same procedure that we used for double integrals.
    • See Definition 2 in Section 15.3
integral over bounded region19
INTEGRAL OVER BOUNDED REGION
  • We enclose E in a box B of the type given by Equation 1.
  • Then, we define a function F so that it agrees with f on E but is 0 for points in B that are outside E.
integral over bounded region20
INTEGRAL OVER BOUNDED REGION
  • By definition,
    • This integral exists if f is continuous and the boundary of E is “reasonably smooth.”
    • The triple integral has essentially the same properties as the double integral (Properties 6–9 in Section 15.3).
integral over bounded region21
INTEGRAL OVER BOUNDED REGION
  • We restrict our attention to:
    • Continuous functions f
    • Certain simple types of regions
type 1 region
TYPE 1 REGION
  • A solid region is said to be of type 1if it lies between the graphs of two continuous functions of x and y.
type 1 region23
TYPE 1 REGION

Equation 5

  • That is, where D is the projection of Eonto the xy-plane.
type 1 regions
TYPE 1 REGIONS
  • Notice that:
    • The upper boundary of the solid E is the surface with equation z = u2(x, y).
    • The lower boundary is the surface z = u1(x, y).
type 1 regions25
TYPE 1 REGIONS

Equation/Formula 6

  • By the same sort of argument that led to Formula 3 in Section 15.3, it can be shown that, if E is a type 1 region given by Equation 5, then
type 1 regions26
TYPE 1 REGIONS
  • The meaning of the inner integral on the right side of Equation 6 is that x and yare held fixed.
  • Therefore,
    • u1(x, y) and u2(x, y) are regarded as constants.
    • f(x, y, z) is integrated with respect to z.
type 1 regions27
TYPE 1 REGIONS
  • In particular, if the projection D of E onto the xy-plane is a type I plane region, then
type 1 regions28
TYPE 1 REGIONS

Equation 7

  • Thus, Equation 6 becomes:
type 1 regions29
TYPE 1 REGIONS
  • If, instead, D is a type II plane region, then
type 1 regions30
TYPE 1 REGIONS

Equation 8

  • Then, Equation 6 becomes:
type 1 regions31
TYPE 1 REGIONS

Example 2

  • Evaluate where E is the solid tetrahedron bounded by the four planesx = 0, y = 0, z = 0, x + y + z = 1
type 1 regions32
TYPE 1 REGIONS

Example 2

  • When we set up a triple integral, it’s wise to draw twodiagrams:
    • The solid region E
    • Its projection D on the xy-plane
type 1 regions33
TYPE 1 REGIONS

Example 2

  • The lower boundary of the tetrahedron is the plane z = 0 and the upper boundary is the plane x + y + z = 1 (or z = 1 – x – y).
    • So, we use u1(x, y) = 0 and u2(x, y) = 1 – x – yin Formula 7.
type 1 regions34
TYPE 1 REGIONS

Example 2

  • Notice that the planes x + y + z = 1 and z = 0 intersect in the line x + y = 1 (or y = 1 – x) in the xy-plane.
    • So, the projection of Eis the triangular region shown here, and we have the following equation.
type 1 regions35
TYPE 1 REGIONS

E. g. 2—Equation 9

  • This description of E as a type 1 region enables us to evaluate the integral as follows.
type 1 regions36
TYPE 1 REGIONS

Example 2

type 2 region
TYPE 2 REGION
  • A solid region E is of type 2if it is of the form where D is the projection of Eonto the yz-plane.
type 2 region38
TYPE 2 REGION
  • The back surface is x = u1(y, z).
  • The front surface is x = u2(y, z).
type 2 region39
TYPE 2 REGION

Equation 10

  • Thus, we have:
type 3 region
TYPE 3 REGION
  • Finally, a type 3region is of the form
  • where:
    • D is the projection of Eonto the xz-plane.
    • y = u1(x, z) is the left surface.
    • y = u2(x, z) is the right surface.
type 3 region41
TYPE 3 REGION

Equation 11

  • For this type of region, we have:
type 2 3 regions
TYPE 2 & 3 REGIONS
  • In each of Equations 10 and 11, there may be two possible expressions for the integral depending on:
    • Whether D is a type I or type II plane region (and corresponding to Equations 7 and 8).
bounded regions
BOUNDED REGIONS

Example 3

  • Evaluate where E is the region bounded by the paraboloid y = x2 + z2 and the plane y = 4.
type 1 regions44
TYPE 1 REGIONS

Example 3

  • The solid E is shown here.
  • If we regard it as a type 1 region, then we need to consider its projection D1onto the xy-plane.
type 1 regions45
TYPE 1 REGIONS

Example 3

  • That is the parabolic region shown here.
    • The trace of y = x2 + z2in the plane z = 0 is the parabola y = x2
type 1 regions46
TYPE 1 REGIONS

Example 3

  • From y = x2 + z2, we obtain:
    • So, the lower boundary surface of E is:
    • The upper surface is:
type 1 regions47
TYPE 1 REGIONS

Example 3

  • Therefore, the description of E as a type 1 region is:
type 1 regions48
TYPE 1 REGIONS

Example 3

  • Thus, we obtain:
    • Though this expression is correct, it is extremely difficult to evaluate.
type 3 regions
TYPE 3 REGIONS

Example 3

  • So, let’s instead consider E as a type 3 region.
    • As such, its projection D3onto the xz-plane is the disk x2 + z2≤ 4.
type 3 regions50
TYPE 3 REGIONS

Example 3

  • Then, the left boundary of E is the paraboloid y = x2 + z2.
  • The right boundary is the plane y = 4.
type 3 regions51
TYPE 3 REGIONS

Example 3

  • So, taking u1(x, z) = x2 + z2 and u2(x, z) = 4 in Equation 11, we have:
type 3 regions52
TYPE 3 REGIONS

Example 3

  • This integral could be written as:
  • However, it’s easier to convert to polar coordinates in the xz-plane: x = r cos θ, z = r sin θ
type 3 regions53
TYPE 3 REGIONS

Example 3

  • That gives:
applications of triple integrals
APPLICATIONS OF TRIPLE INTEGRALS
  • Recall that:
    • If f(x) ≥ 0, then the single integral represents the area under the curve y = f(x) from a to b.
    • If f(x, y) ≥ 0, then the double integral represents the volume under the surface z = f(x, y) and above D.
applications of triple integrals55
APPLICATIONS OF TRIPLE INTEGRALS
  • The corresponding interpretation of a triple integral , where f(x, y, z) ≥ 0, is not very useful.
    • It would be the “hypervolume” of a four-dimensional (4-D) object.
    • Of course, that is very difficult to visualize.
applications of triple integrals56
APPLICATIONS OF TRIPLE INTEGRALS
  • Remember that E is just the domainof the function f.
    • The graph of f lies in 4-D space.
applications of triple integrals57
APPLICATIONS OF TRIPLE INTEGRALS
  • Nonetheless, the triple integralcan be interpreted in different ways in different physical situations.
    • This depends on the physical interpretations of x, y, z and f(x, y, z).
    • Let’s begin with the special case where f(x, y, z) = 1 for all points in E.
applns of triple integrals
APPLNS. OF TRIPLE INTEGRALS

Equation 12

  • Then, the triple integral does represent the volume of E:
applns of triple integrals59
APPLNS. OF TRIPLE INTEGRALS
  • For example, you can see this in the case of a type 1 region by putting f(x, y, z) = 1 in Formula 6:
applns of triple integrals60
APPLNS. OF TRIPLE INTEGRALS
  • From Section 15.3, we know this represents the volume that lies between the surfaces z = u1(x, y) and z = u2(x, y)
applications
APPLICATIONS

Example 4

  • Use a triple integral to find the volume of the tetrahedron T bounded by the planes
  • x + 2y +z = 2
  • x = 2y
  • x = 0
  • z = 0
applications62
APPLICATIONS

Example 4

  • The tetrahedron T and its projection Don the xy-plane are shown.
applications63
APPLICATIONS

Example 4

  • The lower boundary of T is the plane z = 0.
  • The upper boundary is the plane x + 2y + z = 2, that is, z = 2 – x – 2y
applications64
APPLICATIONS

Example 4

  • So, we have:
    • This is obtained by the same calculation as in Example 4 in Section 15.3
applications65
APPLICATIONS

Example 4

  • Notice that it is not necessary to use triple integrals to compute volumes.
    • They simply give an alternative method for setting up the calculation.
applications66
APPLICATIONS
  • All the applications of double integrals in Section 15.5 can be immediately extended to triple integrals.
applications67
APPLICATIONS
  • For example, suppose the density function of a solid object that occupies the region Eis: ρ(x, y, z) in units of mass per unit volume, at any given point (x, y, z).
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MASS

Equation 13

  • Then, its massis:
moments
MOMENTS

Equations 14

  • Its momentsabout the three coordinate planes are:
center of mass
CENTER OF MASS

Equations 15

  • The center of massis located at the point , where:
    • If the density is constant, the center of mass of the solid is called the centroidof E.
moments of inertia
MOMENTS OF INERTIA

Equations 16

  • The moments of inertiaabout the three coordinate axes are:
total electric charge
TOTAL ELECTRIC CHARGE
  • As in Section 15.5, the total electric chargeon a solid object occupying a region Eand having charge density σ(x, y, z) is:
joint density function
JOINT DENSITY FUNCTION
  • If we have three continuous random variables X, Y, and Z, their joint density function is a function of three variables such that the probability that (X, Y, Z) lies in E is:
joint density function74
JOINT DENSITY FUNCTION
  • In particular,
  • The joint density function satisfies: f(x, y, z) ≥ 0
applications75
APPLICATIONS

Example 5

  • Find the center of mass of a solid of constant density that is bounded by the parabolic cylinder x = y2 and the planes x = z, z = 0, and x = 1.
applications76
APPLICATIONS

Example 5

  • The solid E and its projection onto the xy-plane are shown.
applications77
APPLICATIONS

Example 5

  • The lower and upper surfaces of E are the planes z = 0 and z = x.
    • So,we describe Eas a type 1 region:
applications78
APPLICATIONS

Example 5

  • Then, if the density is ρ(x, y, z), the mass is:
applications79
APPLICATIONS

Example 5

applications80
APPLICATIONS

Example 5

  • Due to the symmetry of E and ρabout the xz-plane, we can immediately say that Mxz = 0, and therefore .
  • The other moments are calculated as follows.
applications81
APPLICATIONS

Example 5

applications82
APPLICATIONS

Example 5

applications83
APPLICATIONS

Example 5

applications84
APPLICATIONS

Example 5

  • Therefore, the center of mass is: