Topic 4: Triple Integration

TOPIC 4- TRIPLE INTEGRATION (ADAPTED FROM DR MOHSEN’S NOTES) LECTURE NOTES It should not be a surprise that we can also do triple integrals—integrals over a three-dimensional region. The simplest application allows us to compute volumes in an alternate way. To approximate a volume in three dimensions, we can divide the three-dimensional region into small rectangular boxes, each of size ∆x × ∆y × ∆z with volume ∆x∆y∆z. Then we add them all up and take the limit, to get an integral: Z x1 Z y1 Z z1 dz dy dx. x0 y0 z0 If the limits are constant, we are simply computing the volume of a rectangular box. Date: April 1, 2026. 1 2 LECTURE NOTES Example 1. Compute the volume of the box with opposite corners at (0, 0, 0) and (1, 2, 3). Solution by algebraic method: The volume of a rectangular box is given by the product of its side lengths. Here, the side lengths are: 1 − 0 = 1, 2 − 0 = 2, 3 − 0 = 3. Thus, the volume is V = 1 × 2 × 3 = 6. Solution by integration 1 Z 2 Z Z 3 dz dy dx. V = 0 0 0 The first inner integral is 3 Z dz = 3, 0 so for the second inner integral, we have Z 2 3 dy = 6, 0 and finally Z V = 1 6 dx = 6. 0 Remark 1. The average value of a function f (x, y, z) over a region R is defined to be ZZZ 1 fave = f (x, y, z) dV, V (R) R where V (R) is the volume of R. TOPIC 4- TRIPLE INTEGRATION (ADAPTED FROM DR MOHSEN’S NOTES) 3 Example 2. Suppose the temperature at a point is given by T (x, y, z) = xyz. Find the average temperature in the cube with opposite corners at (0, 0, 0) and (2, 2, 2). Solution Since 0 ≤ x, y, z ≤ 2, we have V (R) = 2 × 2 × 2 = 8. It means the average temperature is Tave = 1 V (R) ZZZ so f (x, y, z) dV = R 1 8 Z 2 Z 2 Z 2 (xyz) dz dy dx, 0 0 0 Z Z Z 1 2 2 1 2 (xy) dy dx = x dx = 1. 4 0 0 2 0 It means that the average temperature at each point of this cube is equal to 1. Tave = 4 LECTURE NOTES Example 3. Evaluate Z 1 Z x (x+y) Z
2x + y − 1 dz dy dx. I= 0 0 0 Solution The first inner integral is Z (x+y)
2x + y − 1 dz = (2x + y − 1) (x+y) Z dz, 0 0 = (2x + y − 1)(x + y), = 2x2 + 3xy − y 2 − x − y, = (2x2 − x) + (3x − 1)y − y 2 . Solution The first inner integral is Z (x+y)
2x + y − 1 dz = (2x + y − 1) (x+y) Z 0 dz, 0 = (2x + y − 1)(x + y), = 2x2 + 3xy − y 2 − x − y, = (2x2 − x) + (3x − 1)y − y 2 . Solution (Cont.) So for the second inner integral, we have Z x
(2x2 − x) + (3x − 1)y − y 2 dy 0 x Z 2 Z (2x − x) dy + = 0 x Z x (3x − 1)y dy − 0 = y(2x2 − x) + 21 y 2 (3x − 1) − 13 y 3 y 2 dy 0 y=x y=0 = 37 x3 − 32 x2 . And finally Z I= 0 1 7 3 3x
 7 4 1 3 x=1 − 32 x2 dx = 12 x − 2 x x=0 = 1 12 . TOPIC 4- TRIPLE INTEGRATION (ADAPTED FROM DR MOHSEN’S NOTES) Example 4. Evaluate π/2 Z Z sin(θ) r cos(θ) Z r2 dz dr dθ. I= 0 0 0 Solution The first inner integral is Z r cos(θ) r2 dz = r2 r cos(θ) Z dz = r3 cos(θ), 0 0 so for the second inner integral, we have Z sin(θ) r3 cos(θ) dr = cos(θ) Z 0 sin(θ) r3 dr = 0 1 4 sin4 (θ) cos(θ). And finally Z I= 0 π/2 1 4 4
sin (θ) cos(θ) dθ = 1 4 π/2 Z sin4 (θ) cos(θ) dθ. 0 By changing the variable u = sin(θ)   du = cos(θ) dθ, θ = 0 ⇒ u = 0,   θ = π2 ⇒ u = 1, =⇒ we have I= 1 4 Z 0 1 u4 du = 1 20 . 5 6 LECTURE NOTES Example 5. Evaluate π Z sin(θ) Z Z r sin(θ) r cos2 (θ) dz dr dθ I= 0 0 0 Solution Z r sin(θ) 2 Z 2 r sin(θ) dz = r2 cos2 (θ) sin(θ). r cos (θ) dz = r cos (θ) 0 0 Z sin(θ) 2 2 Z 2 sin(θ) r cos (θ) sin(θ) dr = cos (θ) sin(θ) r2 dr 0 0 1 sin4 (θ) cos2 (θ) 3 = and finally Z π I= 0   Z 1 1 π 4 4 2 sin (θ) cos (θ) dθ = sin (θ) cos2 (θ) dθ. 3 3 0 sin4 (θ) cos2 (θ) = sin2 (θ) sin2 (θ) cos2 (θ) 2 = sin2 (θ) [sin(θ) cos(θ)] = sin2 (θ) = 1 4 1 2 sin(2θ) 2 sin2 (2θ) sin2 (θ). Solution(continue) Using the identity sin2 (x) = 21 [1 − cos(2x)], we have 1 sin2 (2θ) sin2 (θ) 4    1 1 1 = (1 − cos(4θ)) (1 − cos(2θ)) 4 2 2 sin4 (θ) cos2 (θ) = = 1 [(1 − cos(4θ))(1 − cos(2θ))] 16 1 [1 − cos(4θ) − cos(2θ) + cos(4θ) cos(2θ)] . 16 Solution(continue) By using the identity cos(x) cos(y) = 12 [cos(x + y) + cos(x − y)], we have = cos(4θ) cos(2θ) = = so sin4 (θ) cos2 (θ) = 1 [cos(4θ + 2θ) + cos(4θ − 2θ)] 2 1 [cos(6θ) + cos(2θ)] 2 1 [1 − cos(4θ) − cos(2θ) + cos(4θ) cos(2θ)] 16 TOPIC 4- TRIPLE INTEGRATION = (ADAPTED FROM DR MOHSEN’S NOTES)  1  1 − cos(4θ) − cos(2θ) + 12 (cos(6θ) + cos(2θ)) 16 1 1 1 1 cos(6θ) − cos(4θ) − cos(2θ) + . = 32 16 32 16 Solution (Cont.) Now by substituting this value sin4 (θ) cos2 (θ) = 1 1 1 1 cos(6θ) − cos(4θ) − cos(2θ) + 32 16 32 16 into the I= π Z π  4  sin (θ) cos2 (θ) dθ 0  1 1 1 1 cos(6θ) − cos(4θ) − cos(2θ) + dθ 32 16 32 16 0 Z π Z π Z π Z π 1 1 1 1 = cos(6θ) dθ − cos(4θ) dθ − cos(2θ) dθ + dθ 96 0 48 0 96 0 48 0  θ=π 1 1 1 1 = sin(6θ) − sin(4θ) − sin(2θ) + θ 576 192 192 48 θ=0 π = . 48 1 = 3 Z  1 3 7 8 LECTURE NOTES Example 6. 2 Z y2 Z ln(y+z) Z ex dx dz dy I= 1 0 y Solution The first inner integral is ln(y+z) Z x=ln(y+z) ex dx = [ex ]x=0 0 =y+z−1 so for the second inner integral, we have y2 Z y  z=y2 (y + z − 1) dz = yz + 12 z 2 − z z=y = y 3 + 12 y 4 − 25 y 2 + y. 2 Z
y 3 + 12 y 4 − 52 y 2 + y dy I= 1 Z 2 = 2 Z 3 1 4 2y y dy + 1 Z 2 = 1 y 3 dy + 1 = 1 4y 1 2 4 Z 2 Z 5 2 2y dy − 1 2 y 4 dy − 1 5 2 Z 1 2 y 2 dy + 1 1 5 10 y − 56 y 3 + 12 y 2 = 38 15 1 60 = 151 60 . y dy dy + + − 2 Z Z y dy 1 y=2 y=1 2 TOPIC 4- TRIPLE INTEGRATION (ADAPTED FROM DR MOHSEN’S NOTES) UVW-subs in multiple integrals ZZZ ZZZ f (x, y, z) dx dy dz = f (u, v, w) |J(u, v, w)| du dv dw D V such that ∂x ∂x ∂u ∂v ∂y ∂y J(u, v, w) = ∂u ∂v ∂z ∂z ∂u ∂v Cylindrical coordinates is a special case with x = r cos(θ), y = r sin(θ), ∂x ∂w ∂y ∂w ∂z ∂w z = z, and the Jacobian is J(r, θ, z) = r. Then the triple integral takes the form ZZZ ZZZ
f (x, y, z) dx dy dz = r f r cos(θ), r sin(θ), z dr dθ dz. D V 9 10 LECTURE NOTES Example 7 Evaluate ZZZ I= y dV E where E is the region that lies below the plane z = x + 2, above the xy-plane, and between the cylinders x2 + y 2 = 1 and x2 + y 2 = 4. Solution Using the cylindrical coordinates, w have ZZZ Z 2π Z 2 Z r cos(θ)+2 y dV = E r sin(θ) r dz dr dθ 0 0 0 The first inner integral is r cos(θ)+2 Z r sin(θ) r dz = 21 r3 sin(2θ) + 2r2 sin(θ). 0 as for the second integral we have 2 Z 1 1 2r 3  sin(2θ) + 2r2 sin(θ) dr = 15 8 sin(2θ) + 14 3 sin(θ) and finally Z I= 0 2π  15 8 sin(2θ) + 14 3  sin(θ) dθ = 15 8 Z 2π sin(2θ) dθ + 0 θ=2π = − 15 16 cos(2θ) − 14 3 cos(θ) θ=0 = 0. 14 3 Z 2π sin(θ) dθ 0 TOPIC 4- TRIPLE INTEGRATION (ADAPTED FROM DR MOHSEN’S NOTES) Example 8 Convert Z Z √1−y2 Z √x2 +y2 1 xyz dz dx dy −1 x2 +y 2 0 into an integral in cylindrical coordinates. Solution Using cylindrical coordinates, we have Z Z √ 2Z √ 2 x +y 2 1−y 1 xyz dz dx dy I= −1 Z π/2 x2 +y 2 0 1 Z Z r (r cos θ)(r sin θ)z dz dr dθ = −π/2 Z r2 0 π/2 Z 1 Z r = −π/2 0 r2 1 2 2r z sin(2θ) dz dr dθ. Solution (Cont.) The first inner integral is Z r Z 1 3 1 3 2 r z sin(2θ) dz = 2 r sin(2θ) r 1 4 z dz = r2 r2
r5 − r7 sin(2θ) so for the second inner integral, we have Z 0 1 1 4 5 r −r 7
 sin(2θ) dr = = 1 96 1 4 1 Z
r5 − r7 dr sin(2θ) 0 sin(2θ). Solution (Cont.) and finally Z π/2 I= −π/2  1 96  sin(2θ) dθ = 1 = − 192 cos(2θ) = 0. 1 96 Z π/2 sin(2θ) dθ −π/2 θ=π/2 θ=−π/2 11 12 LECTURE NOTES Definition 1 In geometry, a tetrahedron, also known as a triangular pyramid, is a polyhedron composed of four triangular faces, six straight edges, and four vertex corners. The tetrahedron is the simplest of all the ordinary convex polyhedra and the only one that has fewer than five faces. Remark 2 Equation of plane With given points P (x0 , y0 , z0 ), Q(x1 , y1 , z1 ), and R(x2 , y2 , z2 ), the equation of the plane will be a(x − x0 ) + b(y − y0 ) + c(z − z0 ) = 0, −−→ −−→ ~ ~ ~ where ai + bj + ck = P Q × QR, such that ~k ~i ~j −−→ −−→ P Q × QR = x1 − x0 y1 − y0 z1 − z0 x2 − x1 y2 − y1 z2 − z1 Equation of line With given points P (x0 , y0 ) and Q(x1 , y1 ), the equation of the line will be y1 − y0 (x − x0 ). x1 − x0 To find the volume of a tetrahedron with given corner points, we can use a triple integral. For obtaining limit values of integration for x, y, and z, follow these steps: 1. Categorize corner points into two groups (each group with 3 points). Find the equation of the plane and thus the limit value of z for each group. 2. Remove the z value in each point and with the new 2-D points, find the equation of lines and thus the limit value of y for each group. 3. You will find the limit value of x if you draw all obtained points in step 2. Choosing the right category for points requires some skill, which you will develop with practice. y − y0 = TOPIC 4- TRIPLE INTEGRATION (ADAPTED FROM DR MOHSEN’S NOTES) 13 Example 9 Find the volume of the tetrahedron with corners at (0, 0, 0), (0, 3, 0), Solution To find the limits of integration   P (0, 0, 0) Q(0, 3, 0)  R(2, 3, 0) (2, 3, 0), (2, 3, 5). for z:     P Q(0, 3, 0) ⇒ QR(2, 0, 0)  ~i ~j ~k −−→ −−→ P Q × QR = 0 3 0 = 0~i + 0~j + 6~k. 2 0 0 Solution (Cont.) So a = 0, b = 0, c = 6. This means a(x−x0 )+b(y−y0 )+c(z−z0 ) = 0 ⇒ 0(x−2)+0(y−3)+6(z−0) = 0 ⇒ 6z = 0 ⇒ z = 0. To find the upper limit for z:     P (0, 0, 0)  P Q(0, 3, 0) Q(0, 3, 0) ⇒ QR(2, 0, 5)   R(2, 3, 5) Solution (Cont.) By removing the z value in each point, new 2-D points   3−0   P (0, 0)  PR : y − 0 = (x − 0) 2 −0 Q(0, 3) ⇒ 3−3    QR : y − 3 = (x − 0) R(2, 3) 2−0 By removing the z and y values in each point, we have So we have 2 Z V = 3 Z 3 2x 0 5 2x Z  are ⇒ y = 23 x ⇒ y=3 x = 0 and x = 2. dz dy dx. 0 Solution (Cont.) The first inner integral is Z 5 2x 0 dz = 52 x, so for the second inner integral, we have Z 3 5 x dy 3 2 x 2 = 15 2 x − 15 2 4 x , and finally Z V = 0 2 15 2 x − 15 2 4 x
dx =  15 4 x2 − 54 x3 2 0 = 5. 14 LECTURE NOTES Example 10 Find the volume of the tetrahedron with corners at (0, 0, 0), (0, 1, 0), Solution To find the limits of integration   P (0, 0, 0) Q(0, 1, 0)  R(1, 1, 0) (1, 1, 0), (0, 1, 1). for z:     P Q(0, 1, 0) ⇒ QR(1, 0, 0)  ~i ~j ~k −−→ −−→ P Q × QR = 0 1 0 = 0~i + 0~j − ~k. 1 0 0 Solution (Cont.) So a = 0, b = 0, c = −1. This means a(x − x0 ) + b(y − y0 ) + c(z − z0 ) = 0 0(x − 0) + 0(y − 0) − (z − 0) = 0 ⇒ z = 0. To find the upper limit for z:     P (0, 0, 0)  P Q(1, 1, 0) Q(1, 1, 0) ⇒ QR(−1, 0, 1)   R(0, 1, 1) Solution (Cont.) ~i ~j ~k 1 1 0 = ~i − ~j + ~k −1 0 1 So a = 1, b = −1, c = 1. This means −−→ −−→ P Q × QR = a(x − x0 ) + b(y − y0 ) + c(z − z0 ) = 0 (x − 0) − (y − 0) + (z − 0) = 0 ⇒ x − y + z = 0 ⇒ z = y − x. Solution (Cont.) By removing the z value in each point, new 2-D points are   1−0   P (0, 0)  PR : y − 0 = (x − 0) ⇒ y = x 1−0 Q(0, 1) ⇒ 1−1    [6pt]QR : y − 1 = (x − 0) ⇒ y = 1 R(1, 1) 1−0 By removing the z and y values in each point, we have x = 0 and x = 1. So we have Z 1 Z 1 Z V = dz dy dx. 0 Solution (Cont.) The first inner integral is y−x x 0 TOPIC 4- TRIPLE INTEGRATION Z (ADAPTED FROM DR MOHSEN’S NOTES) y−x dz = y − x, 0 so for the second inner integral, we have Z 1 (y − x) dy = x 1 2 2y − xy
y=1 y=x = 12 x2 − x + 21 , and finally Z V = 0 1 1 2 2x −x+ 1 2
dx = 1 3 6x − 21 x2 + 12 x
1 0 = 16 . 15 16 LECTURE NOTES Example 11 Find the volume of the tetrahedron bounded by the planes x + y + z = 5, Solution For z, we have z = 0 and x = 0, y = 0, z = 0. x + y + z = 5 ⇒ z = 5 − x − y, so 0 ≤ z ≤ 5 − x − y. For y, we have y = 0 and x + y = 5 ⇒ y = 5 − x, so 0 ≤ y ≤ 5 − x. Solution (Cont.) 0 ≤ y ≤ 5 − x, and finally for x, we have x = 0 and x = 5, so 0 ≤ x ≤ 5. It means that Z 5 Z 5−x 5−y−x Z V = Z 5 Z 5−x (5 − y − x) dy dx dz dy dx = 0 0 0 = 1 2 0 Z 0 0 5
x2 − 10x + 50 dx = 250 3 . TOPIC 4- TRIPLE INTEGRATION (ADAPTED FROM DR MOHSEN’S NOTES) Example 12 Find the volume of the tetrahedron enclosed by the coordinate planes 2x + y + z = 3. Solution For z, we have z = 0 and 2x + y + z = 3 ⇒ z = 3 − 2x − y, so 0 ≤ z ≤ 3 − 2x − y. For y, we have y = 0 and 2x + y = 3 ⇒ y = 3 − 2x, so 0 ≤ y ≤ 3 − 2x. Solution (Cont.) 0 ≤ y ≤ 3 − 2x, and finally for x, we have x = 0 and 2x = 3, so 0 ≤ x ≤ 23 . It means that Z V = 0 3 2 Z 3−2x Z 3−y−2x Z dz dy dx = 0 0 = 0 1 2 Z 0 3 2 3 2 Z 3−2x (3 − y − 2x) dy dx 0
4x2 − 12x + 9 dx = 94 . 17 18 LECTURE NOTES Definition 2 For a given object with density function f (x, y, z), ZZZ Total mass of a solid: M = f (x, y, z) dA R ZZZ Static moment about plane Oxy : Mxy = zf (x, y, z) dA R ZZZ Static moment about plane Oyz : Myz = xf (x, y, z) dA R ZZZ Static moment about plane Oxz : Mxz = yf (x, y, z) dA R  Center of gravity of the solid: (x̄, ȳ, z̄) = Myz Mxz Mxy , , M M M  . TOPIC 4- TRIPLE INTEGRATION (ADAPTED FROM DR MOHSEN’S NOTES) 19 Example 13 Suppose the density of an object is given by f (x, y, z) = xz, and the object occupies the tetrahedron with corners (0, 0, 0), (0, 1, 0), (1, 1, 0), (0, 1, 1). Find the mass and center of mass of the object. Solution To find the limits of integration for z:      P (0, 0, 0)  P Q(0, 1, 0) Q(0, 1, 0) ⇒ QR(1, 0, 0)   R(1, 1, 0) ~i ~j ~k −−→ −−→ P Q × QR = 0 1 0 = 0~i + 0~j − ~k. 1 0 0 Solution (Cont.) So a = 0, b = 0, c = −1. This means a(x − x0 ) + b(y − y0 ) + c(z − z0 ) = 0 0(x − 0) + 0(y − 0) − (z − 0) = 0 ⇒ z = 0. To find the upper limit for z:     P (0, 0, 0)  P Q(1, 1, 0) Q(1, 1, 0) ⇒ QR(−1, 0, 1)   R(0, 1, 1) Solution (Cont.) ~i ~j ~k 1 1 0 = ~i − ~j + ~k −1 0 1 So a = 1, b = −1, c = 1. This means −−→ −−→ P Q × QR = a(x − x0 ) + b(y − y0 ) + c(z − z0 ) = 0 (x − 0) − (y − 0) + (z − 0) = 0 ⇒ x − y + z = 0 ⇒ z = y − x. Solution (Cont.) By removing the z value in each point, new 2-D points are   1−0   P (0, 0) P R : y − 0 = (x − 0) ⇒ y = x 1 −0 ⇒ Q(0, 1) 1 −1    [6pt]QR : y − 1 = (x − 0) ⇒ y = 1 R(1, 1) 1−0 By removing the z and y values in each point, we have x = 0 and x = 1. So we have Z 1 Z 1 Z M= y−x Z 1 Z 1 Z f (x, y, z) dz dy dx = 0 x 0 y−x xz dz dy dx. 0 x 0 20 LECTURE NOTES Solution (Cont.) The first inner integral is y−x Z  1 xy − x2 , 2 0 so for the second inner integral, we have xz dz = 1 Z x   1 1 3 xy − x2 dy = x − 2x2 + x , 2 4 and finally Z M= 0 1
1 3 x − 2x2 + x dx = 4  1 4 1 3 1 2 x − x + x 16 6 8  Solution (Cont.) 1 Z x Z y−x Z Mxy = zf (x, y, z) dz dy dx 0 0 1 Z 0 x Z y−x Z xz 2 dz dy dx = 0 1 Z 0
− 31 x4 + x3 y − x2 y 2 + 31 xy 3 dy dx = 0 0 x Z 0 1 Z 1 5 12 x = 0 − 31 x4 + 12 x3 − 13 x2 + = 1 12 x
dx 1 360 . Solution (Cont.) 1 Z 1 Z y−x Z Mxz = yf (x, y, z) dz dy dx 0 x 1 Z 0 1 Z y−x Z = xyz dz dy dx 0 = 1 2 = Z 1 x 1 Z 0 0
x3 y − 2x2 y 2 + xy 3 dy dx x 1 Z 1 24
x5 + 6x3 − 8x2 + 3x dx 0 = 1 144 . Solution (Cont.) 1 Z 1 Z y−x Z Myz = xf (x, y, z) dz dy dx 0 Z x 1 Z 0 1 Z = 0 x 0 y−x x2 z dz dy dx 1 = 0 1 . 120 TOPIC 4- TRIPLE INTEGRATION = 1 2 Z 1 Z 0 = − 61 Z (ADAPTED FROM DR MOHSEN’S NOTES) 1
x4 − 2x3 y + x2 y 2 dy dx x 1
x5 − 3x4 + 3x3 − x2 dx 0 =  Center of mass: (x̄, ȳ, z̄) = 21 1 360 . Myz Mxz Mxy , , M M M   = 1/360 1/144 1/360 , , 1/120 1/120 1/120   = 1 5 1 , , 3 6 3 Solution (Cont.) So for the center of mass, we have   Myz Mxz Mxy , , (x̄, ȳ, z̄) = M M M   1/360 1/144 1/360 = , , 1/120 1/120 1/120   1 5 1 , , . = 3 6 3 Department of Mathematics and Computer Science, Lae, PNG University of Technology, Papua New Guinea Email address: john.lanta@pnguot.ac.pg  .