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Section 13.8: Triple Integrals Consider a function f of three variables defined on a closed box B = [a, b] × [c, d] × [r, s] = {(x, y, z)|a ≤ x ≤ b, c ≤ y ≤ d, r ≤ z ≤ s}. Partition the intervals [a, b], [c, d], and [r, s] as a = x0 < x1 < · · · < xi−1 < xi < · · · < xl = b, c = y0 < y1 < · · · < yj−1 < yj < · · · < ym = d, r = z0 < z1 < · · · < zk−1 < zk < · · · < zn = s. This forms a partition, P , of the box B into lmn sub-boxes Bijk = [xi−1 , xi ] × [yj−1 , yj ] × [zk−1 , zk ]. Let ∆xi = xi − xi−1 , ∆yj = yj − yj−1 , and ∆zk = zk − zk−1 . Then the area of the box Bijk is ∆Vijk = ∆xi ∆yj ∆zk . ∗ ∗ ) in each sub-box Bijk and form the triple Rie, zijk Choose a representative point (x∗ijk , yijk mann sum l X m X n X ∗ ∗ f (x∗ijk , yijk , zijk )∆Vijk . i=1 j=1 k=1 Let ||P || denote the norm of the partition P which is the length of the longest diagonal of all sub-boxes Bijk . Definition: The triple integral of f over B is ZZZ f (x, y, z)dV = lim ||P ||→0 B l X m X n X ∗ ∗ f (x∗ijk , yijk , zijk )∆Vijk i=1 j=1 k=1 provided that the limit exists. If the limit exists, then f is called integrable. Note: The practical method for evaluating triple integrals is to express them as iterated integrals. Theorem: (Fubini’s Theorem) If f is continuous on the box B = [a, b] × [c, d] × [r, s], then ZZZ Z sZ dZ b f (x, y, z)dV = f (x, y, z)dxdydz. B r c a Note: There are five other possible orders of integration which all give the same result. Example: Evaluate the triple integral ZZZ 8x2 yz 3 dV, B where B = [−1, 2] × [0, 1] × [1, 2]. By Fubini’s Theorem, ZZZ 2 3 8x yz dV 2 Z Z 1 Z 2 8x2 yz 3 dzdydx = −1 2 B Z 0 Z 1 1 = Z −1 0 2 Z 1 = −1 2 Z = Z −1 2 2 2x yz dydx 2 4 1 30x2 ydydx 0 1 15x y dx 2 2 0 15x2 dx −1 2 = 5x3 −1 = 45. = Definition: There are three types of solid regions. • A solid region E is said to be of type 1 if E = {(x, y, z)|(x, y) ∈ D, g1 (x, y) ≤ z ≤ g2 (x, y)}, where D is the projection of E onto the xy-plane. • A solid region E is said to be of type 2 if E = {(x, y, z)|(y, z) ∈ D, g1 (y, z) ≤ x ≤ g2 (y, z)}, where D is the projection of E onto the yz-plane. • A solid region E is said to be of type 3 if E = {(x, y, z)|(x, z) ∈ D, g1 (x, z) ≤ y ≤ g2 (x, z)}, where D is the projection of E onto the xz-plane. Theorem: Suppose that f is a continuous function over some region E. 1. If E is a type 1 region, then ZZZ Z Z "Z f (x, y, z)dV = E D f (x, y, z)dz dA. g1 (x,y) 2. If E is a type 2 region, then ZZZ Z Z "Z f (x, y, z)dV = E D # g2 (y,z) f (x, y, z)dx dA. g1 (y,z) 3. If E is a type 3 region, then ZZZ Z Z "Z f (x, y, z)dV = E # g2 (x,y) D # g2 (x,z) f (x, y, z)dy dA. g1 (x,z) In particular, if E is a type 1 region and its projection D onto the xy-plane is a type I region, then ZZZ Z Z Z b h2 (x) g2 (x,y) f (x, y, z)dV = f (x, y, z)dzdydx. E a h1 (x) g1 (x,y) ZZZ zdV , where E is the solid region bounded by Example: Evaluate the triple integral E the planes x = 0, y = 0, z = 0 and 2x + y + z = 2. The solid region E and its projection D onto the xy-plane are shown in the figure below. The region E can be expressed as E = {(x, y, z)|0 ≤ x ≤ 1, 0 ≤ y ≤ 2 − 2x, 0 ≤ z ≤ 2 − 2x − y}. Then, ZZZ 1 Z zdV Z 2−2x Z = E zdzdydx 0 1 = 2 = = = = 2−2x−y Z 0 0 1 Z 2−2x (2 − 2x − y)2 dydx 0 0 2−2x Z 1 1 3 − (2 − 2x − y) dx 6 0 0 Z 1 1 (2 − 2x)3 dx 6 0 1 1 4 − (2 − 2x) 48 0 1 . 3 We could have also projected this region onto the xz- or yz-planes. ZZZ p Example: Evaluate the triple integral y 2 + z 2 dV , where E is the solid region bounded E by the paraboloid x = y 2 + z 2 and the plane x = 9. The solid region E and its projection D onto the yz-plane are shown in the figure below. The region E can be expressed as E = {(x, y, z)|(y, z) ∈ D, y 2 + z 2 ≤ x ≤ 9}, where D is the disk y 2 + z 2 ≤ 9. In polar coordinates, D = {(r, θ)|0 ≤ θ ≤ 2π, 0 ≤ r ≤ 3}. Then, Z Z Z ZZZ p y 2 + z 2 dV 9 p = E Z = y 2 + z 2 dx dA D y 2 +z 2 2π Z 3 2 r (9 − r2 )drdθ 0 Z 0 3 (9r2 − r4 )dr 0 3 1 4 3 = 2π 9r − r 5 0 1944π = . 5 = 2π We could have also projected this region onto the xy- or xz-planes. Theorem: (Volume as a Triple Integral) The volume of the solid region E is ZZZ dV. V = E Example: Find the volume of the tetrahedron T bounded by the plane 2x + 2y + z = 12 and the coordinate planes. The tetrahedron T can be expressed as T = {(x, y, z)|0 ≤ x ≤ 6, 0 ≤ y ≤ 6 − x, 0 ≤ z ≤ 12 − 2x − 2y}. Thus, the volume of the tetrahedron is Z 6 Z 6−x Z 12−2x−2y 1dzdydx V = = 0 0 0 Z 6 Z 6−x (12 − 2x − 2y)dydx = 0 0 6−x Z 6 2 [(12 − 2x)y − y ] dx = 0 0 Z 6 = [(12 − 2x)(6 − x) − (6 − x)2 ]dx Z0 6 (x2 − 12x + 36)dx = 0 6 1 3 2 = x − 6x + 36x 3 0 = 72. Note: The applications of double integrals in Section 13.6 can be extended to triple integrals. Theorem: (Center of Mass of a Solid) Suppose a solid occupies some region E in space and its density at a point (x, y, z) ∈ E is given by ρ(x, y, z). Then the mass of the solid is given by ZZZ m= ρ(x, y, z)dV. E The coordinates (x̄, ȳ, z̄) of the center of mass of the solid are ZZZ ZZZ ZZZ 1 1 1 xρ(x, y, z)dV ȳ = yρ(x, y, z)dV z̄ = zρ(x, y, z)dV. x̄ = m m m E E E Example: Find the mass and center of mass of the solid E bounded by the parabolic cylinder z = 1 − y 2 and the planes x + 5z = 5, x = 0, and z = 0 with constant density ρ(x, y, z) = 2. The solid region E can be described as E = {(x, y, z)|0 ≤ x ≤ 5 − 5z, 0 ≤ z ≤ 1 − y 2 , −1 ≤ y ≤ 1}. Thus, the mass of the solid is 1 Z Z 1−y 2 Z 5−5z dxdzdy m = 2 −1 Z 1 0 Z 0 1−y 2 (5 − 5z)dzdy = 2 −1 1−y2 1 2 z− z 10 dy 2 −1 0 Z 1 1 2 2 4 (1 − y ) − (1 − 2y + y ) dy 10 2 −1 Z 1 (1 − y 4 )dy 5 −1 1 5y − y 5 −1 8. Z = = = = = 0 1 Using the formulas from the previous theorem, ZZZ 1 x̄ = 2xdV m E Z Z 2 Z 5−5z 1 1 1−y = xdxdzdy 4 −1 0 0 Z Z 2 1 1 1−y (5 − 5z)2 dzdy = 8 −1 0 Z Z 2 25 1 1−y (1 − z)2 dzdy = 8 −1 0 1−y2 Z 25 1 3 = − (1 − z) dy 24 −1 0 Z 25 1 (1 − y 6 )dy = 24 −1 1 25 1 7 = y− y 24 7 −1 = 25 . 14 Similarly, and 1 ȳ = 4 ZZZ 1 z̄ = 4 ZZZ ydV = 0, E 2 zdV = . 7 E