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UIUC MATH 241 - 15_07

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Copyright © Cengage Learning. All rights reserved. 15 Multiple IntegralsCopyright © Cengage Learning. All rights reserved. 15.7 Triple Integrals3 3 Triple Integrals Learning objectives: § evaluate triple integrals over boxes by expressing them as iterated integrals § evaluate triple integrals over more general regions (of type 1, type 2, and/or type 3) by expressing them as iterated integrals § set up and evaluate triple integrals in applications (volume, mass, moments, center of mass, centroid, moments of inertia, electric charge, probability)4 4 Triple Integrals We have defined single integrals for functions of one variable and double integrals for functions of two variables. We can similarly define triple integrals for functions of three variables. Let’s first deal with the simplest case where f is defined on a rectangular box: B = {(x, y, z) | a ≤ x ≤ b, c ≤ y ≤ d, r ≤ z ≤ s} The first step is to divide B into sub-boxes. We do this by dividing the interval [a, b] into l subintervals [xi – 1, xi] of equal width Δx, dividing [c, d ] into m subintervals of width Δy, and dividing [r, s] into n subintervals of width Δz.5 5 Triple Integrals The planes through the endpoints of these subintervals parallel to the coordinate planes divide the box B into l • m • n sub-boxes Bijk = [xi – 1, xi] × [yj – 1, yj] × [zk – 1, zk] which are shown in Figure 1. Each sub-box has volume ΔV = Δx Δy Δz. Figure 16 6 Triple Integrals Then we form the triple Riemann sum where the sample point is in Bijk. By analogy with the definition of a double integral, we define the triple integral as the limit of the triple Riemann sums in .7 7 Triple Integrals Again, the triple integral always exists if f is continuous. We can choose the sample point to be any point in the sub-box, but if we choose it to be the point (xi, yj, zk) we get a simpler-looking expression for the triple integral:8 8 Triple Integrals Just as for double integrals, the practical method for evaluating triple integrals is to express them as iterated integrals as follows. The iterated integral on the right side of Fubini’s Theorem means that we integrate first with respect to x (keeping y and z fixed), then we integrate with respect to y (keeping z fixed), and finally we integrate with respect to z.9 9 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 have10 10 Triple Integrals 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. 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. We restrict our attention to continuous functions f and to certain simple types of regions.11 11 Triple Integrals A solid region E is said to be of type 1 if it lies between the graphs of two continuous functions of x and y, that is, E = {(x, y, z) | (x, y) ∈ D, u1(x, y) ≤ z ≤ u2(x, y)} where D is the projection of E onto the xy-plane as shown in Figure 2. Figure 2 A type 1 solid region12 12 Triple Integrals Notice that the upper boundary of the solid E is the surface with equation z = u2(x, y), while the lower boundary is the surface z = u1(x, y). By the same sort of argument, it can be shown that if E is a type 1 region given by Equation 5, then The meaning of the inner integral on the right side of Equation 6 is that x and y are held fixed, and therefore u1(x, y) and u2(x, y) are regarded as constants, while f (x, y, z) is integrated with respect to z.13 13 Triple Integrals In particular, if the projection D of E onto the xy-plane is a type I plane region (as in Figure 3), then E = {(x, y, z) | a ≤ x ≤ b, g1(x) ≤ y ≤ g2(x), u1(x, y) ≤ z ≤ u2(x,y)} and Equation 6 becomes Figure 3 A type 1 solid region where the projection D is a type I plane region14 14 Triple Integrals If, on the other hand, D is a type II plane region (as in Figure 4), then E = {(x, y, z) | c ≤ y ≤ d, h1(y) ≤ x ≤ h2(y), u1(x, y) ≤ z ≤ u2(x, y)} and Equation 6 becomes Figure 4 A type 1 solid region with a type II projection15 15 A solid region E is of type 2 if it is of the form E = {(x, y, z) | (y, z) ∈ D, u1(y, z) ≤ x ≤ u2(y, z)} where, this time, D is the projection of E onto the yz-plane (see Figure 7). The back surface is x = u1(y, z), the front surface is x = u2(y, z), and we have Figure 7 A type 2 region Triple Integrals16 16 Triple Integrals Finally, a type 3 region is of the form E = {(x, y, z) | (x, z) ∈ D, u1(x, z) ≤ y ≤ u2(x, z)} where D is the projection of E onto the xz-plane, y = u1(x, z) is the left surface, and y = u2(x, z) is the right surface (see Figure 8). Figure 8 A type 3 region17 17 Triple Integrals For this type of region we have 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).18 18 Applications of Triple Integrals19 19 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, and if f (x, y) ≥ 0, then the double integral ∫∫D f (x, y) dA represents the volume under the surface z = f (x, y) and above D. The corresponding interpretation of a triple integral ∫∫∫E f (x, y, z) dV, where f (x, y, z) ≥ 0, is not very useful because it would be the “hypervolume” of a four-dimensional object and, of course, that is very difficult to visualize. (Remember that E is just the domain of the function f ; the graph of f lies in four-dimensional space.)20 20 Applications of Triple Integrals Nonetheless, the triple integral ∫∫∫E f (x, y, z) dV can be interpreted in different ways in different physical situations, depending 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. Then the triple integral does represent the volume of E:21 21 Applications of Triple Integrals For …


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UIUC MATH 241 - 15_07

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