Triple Integrals

Math 202 

Lia Vas 

Triple Integrals 

Let f(x, y, z) be a function of three variables defined on a solid region 

E = { (x, y, z) | a ≤ x ≤ b, c(x) ≤ y ≤ d(x), g(x, y) ≤ z ≤ h(x, y) } 

The triple integral of f over E is 

∫ ∫ ∫ 

∫ ( b ∫ ( d(x) ∫ h(x,y) 

) 

f(x, y, z) dx dy dz = 

f(x, y, z) dz 

E 

a c(x) g(x,y) 

dy 

) 

dx 

Applications of triple integral: 

1. Volume. The volume of the solid region E as above is 

∫ ∫ ∫ 

∫ ∫ 

V (E) = dx dy dz = (h(x, y) − g(x, y)) dx dy 

where D is the projection of E onto the xy-plane. 

2. The average value of function f(x, y, z) over the solid region E is 

fave = 1 ∫ ∫ ∫ 

f(x, y, z) dx dy dz 

V (E) E 

where V (E) is the volume of the solid region E. 

E 

3. Suppose that a solid object occupies the region E and its density at point (x, y, z) is given 

with ρ(x, y, z). The total mass of the object is 

∫ ∫ ∫ 

m = ρ(x, y, z) dx dy dz 

The coordinates (x, y, z) of the center of mass of the object are given by 

x = 1 ∫ ∫ ∫ 

x ρ(x, y, z) dx dy dz y = 1 ∫ ∫ ∫ 

y ρ(x, y, z) dx dy dz 

m E m E 

z = 1 ∫ ∫ ∫ 

z ρ(x, y, z) dx dy dz 

m E 

If the solid region E is as follows 

E 

E = { (x, y, z) | c ≤ y ≤ d, a(y) ≤ x ≤ b(y), g(x, y) ≤ z ≤ h(x, y) } 

D 

then the triple integral of a function of three variables f(x, y, z) over E is 

∫ ∫ ∫ 

∫ ( d ∫ ( b(y) ∫ ) 

h(x,y) 

f(x, y, z) dx dy dz = 

f(x, y, z) dz 

E 

c 

a(y) 

g(x,y) 

dx 

) 

dy

Practice problems. 

1. Evaluate the triple integral 

a) ∫ ∫ ∫ 

E 

x 3 y 2 z dx dy dz 

where E = { (x, y, z) | 1 ≤ x ≤ 2, 0 ≤ y ≤ x, 0 ≤ z ≤ y 2 } 

b) ∫ ∫ ∫ 

E 

2x dx dy dz 

where E = { (x, y, z) | 0 ≤ y ≤ 2, 0 ≤ x ≤ √ 4 − y 2 , 0 ≤ z ≤ y } 

c) ∫ ∫ ∫ 

E 

6xy dx dy dz 

where E lies under the plane z = x + y + 1 and above the region in the xy-plane bounded 

by the curves y = √ x, y = 0 and x = 1. 

d) ∫ ∫ ∫ 

E 

xy dx dy dz 

where E is the solid tetrahedron with vertices (0,0,0), (1, 0, 0), (0, 2, 0) and (0, 0, 3). 

2. Find the volume of the tetrahedron bounded by the coordinate planes and the plane 2x + 3y + 

6z = 12. 

3. Find the average value of the function f(x, y, z) = xyz over the cube with side length 4 that 

lies in the first octant with one vertex in the origin and edges parallel to the coordinate axes. 

4. Find the mass and the center of mass of the solid E given in problem 1c) and that has the 

density function ρ(x, y, z) = 2. 

Solutions. 

1. a) ∫ 2 

∫ x 

1 0 

∫ y 2 

0 x 3 y 2 z dx dy dz = ∫ 2 

1 x3 dx ∫ x 

0 y2 dy ∫ y 2 

0 z dz = ∫ 2 

1 x3 dx ∫ x 

0 y2 dy z2 

2 |y2 0 = ∫ 2 

1 x3 dx ∫ x 

0 

y 6 

2 dy = ∫ 2 

1 x3 dx y7 

14 |x 0 = ∫ 2 

1 x10 

14 

dx = 

x11 

14(11) |2 1 = 13.29. 

b) Here you have to evaluate the integral with respect to y last since all the other variables 

have y in the bounds. ∫ 2 

0 dy ∫ √ 4−y 2 

0 2x dx ∫ y 

0 dz = ∫ 2 

0 dy ∫ √ 4−y 2 

0 2x dx y = ∫ √ 

2 

0 y dy x2 4−y 

| 

2 

0 = 

∫ 2 

0 y(4 − y2 ) dy = (2y 2 − y4 

4 )|2 0 = 8 − 4 = 4. 

c) Sketch the region in xy-plane first. The x-bounds are 0 ≤ x ≤ 1. The y-bounds are 

0 ≤ y ≤ √ x. The z-bounds are determined by the plane z = x + y + 1 and the xy-plane 

and so 0 ≤ z ≤ x + y + 1. So ∫ 1 ∫ √ x ∫ x+y+1 

0 0 0 6xy dx dy dz = ∫ 1 

0 6xdx ∫ √ x 

0 y dy ∫ x+y+1 

0 dz = 

∫ 1 

0 6xdx ∫ √ x 

0 y dy(x + y + 1) = ∫ 1 

y2 

0 6xdx (x + y3 

+ y2 

2 3 2 )|√ x 

0 = ∫ 1 

0 6xdx (x x + x3/2 + x) = 65. 

2 3 2 28

d) First, find the equation of the plane determined by P = (1, 0, 0), Q = (0, 2, 0) and R = 

(0, 0, 3). Vectors −→ P Q = (−1, 2, 0) and −→ P R = (−1, 0, 3) are in the plane so the vector perpendicular 

to the plane can be taken to be the cross product −→ P Q× −→ 

⃗i ⃗j ⃗ k 

P R = 

−1 2 0 

= (6, 3, 2). The 

∣ −1 0 3 ∣ 

equation of plane, using point P for example, is 6(x − 1) + 3y + 2z = 0 ⇒ 6x + 3y + 2z = 6 ⇒ 

z = 3 − 3x − 3y. 

2 

Thus, the upper bound for z is 3 − 3x − 3 y. The lower bound for z is 0. In xy-plane we have 

2 

a triangle with vertices (0,0), (1,0) and (0,2). So, the bounds for x are 0 ≤ x ≤ 1. The lower 

bound for y is 0 and the upper bound is the line passing (0,2) and (1,0). The equation of this 

line is y = −2x + 2. Thus, 

∫ ∫ ∫ ∫ xy dx dy dz = 1 

0 xdx ∫ −2x+2 

0 ydy ∫ 3−3x− 3 2 y 

0 dz = ∫ 1 

0 xdx ∫ −2x+2 

0 ydy (3 − 3x − 3y) = 2 

∫ 1 

0 

xdx (3 

y2 

= 1 

10 . 

2 

− 3x 

y2 

2 − y3 

2 )|−2x+2 

0 = ∫ 1 

0 

xdx (3 

(−2x+2)2 

2 

− 3x (−2x+2)2 

2 

− (−2x+2)3 

2 

) = use calculator 

2. The upper bound for z can be obtained by solving 2x + 3y + 6z = 12 for z. So, 0 ≤ z ≤ 

2 − 1x − 1 y. The projection of the relevant region in xy-plane is a triangle determined by the 

3 2 

coordinate axes and by the line 2x + 3y + 6(0) = 12 ⇒ y = 4 − 2 x (alternatively, find the 

3 

equation of the line passing (6,0) and (0,4)). The bounds for x are 0 ≤ x ≤ 6. The volume is 

V = ∫ 6 

0 dx ∫ 4− 2 3 x 

0 dy ∫ 2− 1 3 x− 1 2 y 

0 dz = ∫ 6 

0 dx ∫ 4− 2 3 x 

0 dy (2− 1x− 1y) = ∫ 6 

3 2 0 dx (2y − 1 y2 

xy − 2 3 4 )|4− 3 x 

0 = 

∫ 6 

0 dx (2(4 − 2x) − 1x(4 − 2x) − (4− 2 3 x)2 

) = use calculator = 8. 

3 3 3 4 

3. When integrating over the cube, the bounds for all three variables are 0 and 4. The volume of 

the cube of side 4 is 4 3 = 64 (which agrees with the fact that V = ∫ 4 ∫ 4 ∫ 4 

0 0 0 dxdydz = x|4 0y| 4 0z| 4 0 = 

64). 

The average value is f ave = 1 

64 

∫ 4 ∫ 4 ∫ 4 

0 0 0 

1 x 

xyz dxdydz = 2 

64 2 |4 y 2 

0 2 |4 0 z2 

2 |4 0 = 1 

64 83 = 8. 

4. The bounds are the same as in problem 1c. The mass is m = ∫ 1 

0 

∫ 1 

0 2 dx ∫ √ x 

0 (x + y + 1)dy = ∫ 1 

∫ √ x ∫ x+y+1 

0 0 2 dx dy dz = 

0 2(x√ x + x + √ x) dx = 2.633. 

2 

The x-coordinate is x = 1 ∫ √ 

2.633 int1 x ∫ x+y+1 

0 0 0 2xdx dy dz = 1 

2.633 int1 02x(x √ x + x + √ x) dx = 

2 

= 0.647. Similarly you find that y = 0.418, and z = 1.032. 

1.705 

2.633

Triple Integrals

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