12.5 Triple Integrals

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1 1.5 Triple Integrals Arkansas Tech University MATH 94: Calculus III r. Marcel B Finan In Sections , we showed how a function of two variables can be integrated over a region in -space and how integration over a region is equivalent to an iterated or double integral over two intervals. This concept can be extended to integration over a solid region in -space using triple integrals. Let f(x, y, z) be a continuous function on : a x b, c y d, e z f. Partition the interval a x b into n equal subintervals using the mesh points a x < x < x < < x n < x n b with x b a n denoting the length of each subinterval. Similarly, partition c y d into m subintervals using the mesh points c y < y 1 < y < < y m < y m d with y d c m denoting the length of each subinterval. Finally, partition e z f into l subintervals using the mesh points e z < z 1 < z < < z l < z l f with z f e l. This way, the box is partitioned into mnl smaller boxes each of volume equals to x y z as shown in Figure Figure Let ijk be a typical box. Pick a point (x i, y j, z k ) in this box. We define the triple integral of f(x, y, z) over as the limit of a iemann sum lim l m l,m,n k1 j1 i1 n f(x i, yj, zk) x y z provided that the limit exists. We can show that the integral over the solid region is equivalent to a triple integral over three intervals. The argument is similar to the argument we used for the double integral and is therefore not repeated here. We state formally that: f d b f(x, y, z)dxdydz. xample Integrate f(x, y, z) 1 + xyz over the cube of length side 4. e c a 1

2 We have (1 + xyz)dxdydz (4 + 8yz)dydz y + 4y z 4 dz ( z)dz 16z + z x + x yz 4 dydz Next, we define the triple integral of f(x, y, z) over a general bounded region in three-dimensional space (a solid) by much the same procedure that we used for double integrals. We enclose in a box of the type given above. Then we define a function F (x, y, z) so that it agrees with f(x, y, z) on but is for points in that are outside. By definition, F (x, y, z)dv. A solid region is said to be of Type I if {(x, y, z) : (x, y) in, u 1 (x, y) z u (x, y)} where is the projection of onto the xy plane. In this case, we define [ ] u(x,y) f(x, y, z)dz da. u 1(x,y) If is the -dimensional region of Type I (See Section 1.), then we have {(x, y) : a x b, g 1 (x) y g (x)} F (x, y, z)dv If is the -dimensional region of Type II then we have b g(x) u(x,y) a g 1(x) {(x, y) : f 1 (y) x f (y), c y d} F (x, y, z)dv u 1(x,y) d f(y) u(x,y) c f 1(y) u 1(x,y) f(x, y, z)dzdydx. f(x, y, z)dzdxdy. The triple integral has essentially the same properties as the double integral (See Section 1.). Also, note that for f(x, y, z) 1, the triple integral dv gives the volume of.

3 xample 1.5. Find the volume under the plane z 1 x 4 y 8 x 8, y 16. and above the rectangle The region is shown in Figure Thus, we see that as x goes from to 8 and y from to 16, z ranges from to the plane z 1 x 4 y 8. Thus the limits of integration are a, b 8, c, d 16, e, and f 1 x 4 y 8. The volume is V x 4 y 8 1y xy 4 y 16 Figure 1.5. dzdydx 16 dx 8 6 (176 4x)dx 176x x 8 18 (1 x 4 y 8 )dydz xample 1.5. Set up the integral representing the volume of the solid ice cream cone bounded by the cone z x + y and the sphere z 1 x y. The solid is shown in Figure 1.5. Figure 1.5.

4 The ice cream cone is between these two surfaces: x + y z 1 x y This gives the range of z as a function of x and y. Now we need to find the maximum range of x and y. Inside the ice cream cone, the maximum range of x and y occurs where the two surfaces meet, i.e., where the ice cream (the sphere) meets the cone. From the figure, you can see that the surfaces meet in a circle, and the range of x and y is the disk that is the interior of that circle. The surfaces meet when x + y 1 x y, which means x + y 1 x y or x + y 1. We have shown that in the ice cream cone x + y 1. This gives that and 1 x y 1 x 1. 1 x Hence, the volume of the ice cream cone is 1 1 x 1 x y dzdydx 1 1 x x +y A solid region is said to be of Type II if {(x, y, z) : (y, z) in, v 1 (y, z) x v (y, z)} where is the projection of onto the yz plane. In this case, we have [ ] v(y,z) f(x, y, z)dx da v 1(y,z) and as in the case of Type I, the limits of are determined according to whether is a Type I or II plane region. A solid region is said to be of Type III if then we have {(x, y, z) : (x, z) in, w 1 (x, z) y w (x, z)} [ ] w(x,z) f(x, y, z)dy da. w 1(x,z) In each of the Types II and III there may be two possible expressions for the integral depending on whether is a Type I or Type II plane region. 4

5 xample valuate x + z dv, where is the region bounded by the paraboloid y x + z and the plane y 4. is of Type III. The projection of onto the xz plane is the disk x + z 4. Thus, 4 x 4 x + z dv x + z dydzdx x +z 4 x 4 x π (4 x y ) x + z dzdx 4 x [ π ] [ ] 4 r r rdrdθ dθ (4r r 4 )dr [ 4r π r5 5 ] 18π 15 where we used the polar coordinates, r x + z, x r cos θ, z r sin θ We summarize our findings from these examples: The outer limits have to be constant. They cannot depend on any of the variables. The middle limits can depend on the variable from the outer integral only. They cannot depend on the variable from the inner integral. The inner limits can depend on the variable from the outer integral and the variable from the middle integral. Applications of Triple Integrals If f(x, y, z) 1, the triple integral dv is just the volume of the solid region. xample Find the volume of the tetrahedron bounded by the planes x + y + z, x y, x and z. The projection of onto the xy plane is the triangle with vertices (, ), (, 1), and (1, 1 ). In this case, x/ x y x/ V dv dzdydx ( x y)dydx 1 x/ All the applications from physics discussed in Section 1.4 can be extended to triple integrals. Namely, Total mass: m ρ(x, y, z)dv. 5 x/

6 Total electric charge: Moment about xy plane: Moment about xz plane: Moment about yz plane: Center of Mass: Q M xy M xz M yz σ(x, y, z)dv. zρ(x, y, z)dv. yρ(x, y, z)dv. xρ(x, y, z)dv. x M yz m, y M xz m, z M xy m. Moment of Inertia about the x axis: I x (y + z )ρ(x, y, z)dv. Moment of Inertia about the y axis: I y (x + z )ρ(x, y, z)dv. Moment of Inertia about the z axis: I z (x + y )ρ(x, y, z)dv. xample Find the center of mass of the solid bounded by the parabolic cylinder x y and the planes x z, z, and x 1. Assume the solid has constant density ρ. 6

7 We have m ρ ρ M xy ρ M xz M yz ρ x y ρdzdxdy ρ xdxdy y [ ] x x1 dy ρ (1 y 4 )dy xy ] 1 (1 y 4 )dy ρ [y y5 4ρ 5 5 x [ z zρdzdxdy ρ y x y y x dxdy ρ [ x Hence, the center of mass is ] x1 xρdzdxdy ρ xy dy ρ y ] zx z (1 y 6 )dy ρ 7 x 5 7, y. z 5 14 y x dxdy (1 y 6 )dy 4ρ 7. dxdy 7