MULTIPLE INTEGRALS

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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

MULTIPLE INTEGRALS

15.6

Triple Integrals

• In this section, we will learn about:
• Triple integrals and their applications.
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 INTEGRALS

Equation 1

• Let’s first deal with the simplest case where f is defined on a rectangular box:
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 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 INTEGRALS

Equation 2

• Then, we form the triple Riemann sum
• where the sample point is in Bijk.
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

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 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 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)

Theorem 4

• If f is continuous on the rectangular box B = [a, b] x [c, d] x [r, s], then
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 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 INTEGRALS)

Example 1

• Evaluate the triple integral where B is the rectangular box
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
• 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 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 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 REGION
• We restrict our attention to:
• Continuous functions f
• Certain simple types of regions
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 REGION

Equation 5

• That is, where D is the projection of Eonto the xy-plane.
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 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 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 REGIONS
• In particular, if the projection D of E onto the xy-plane is a type I plane region, then
TYPE 1 REGIONS

Equation 7

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

Equation 8

• Then, Equation 6 becomes:
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 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 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 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 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 REGIONS

Example 2

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 REGION
• The back surface is x = u1(y, z).
• The front surface is x = u2(y, z).
TYPE 2 REGION

Equation 10

• Thus, we have:
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 REGION

Equation 11

• For this type of region, we have:
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

Example 3

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

Example 3

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

Example 3

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

Example 3

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

Example 3

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

Example 3

• That gives:
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 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 INTEGRALS
• Remember that E is just the domainof the function f.
• The graph of f lies in 4-D space.
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

Equation 12

• Then, the triple integral does represent the volume of E:
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 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

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
APPLICATIONS

Example 4

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

Example 4

• So, we have:
• This is obtained by the same calculation as in Example 4 in Section 15.3
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.
APPLICATIONS
• All the applications of double integrals in Section 15.5 can be immediately extended to triple integrals.
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).
MASS

Equation 13

• Then, its massis:
MOMENTS

Equations 14

• Its momentsabout the three coordinate planes are:
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

Equations 16

• The moments of inertiaabout the three coordinate axes are:
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
• 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 FUNCTION
• In particular,
• The joint density function satisfies: f(x, y, z) ≥ 0
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.
APPLICATIONS

Example 5

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

Example 5

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

Example 5

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.
APPLICATIONS

Example 5

APPLICATIONS

Example 5

APPLICATIONS

Example 5

APPLICATIONS

Example 5

• Therefore, the center of mass is: