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Multivariable Calculus: Triple Integrals

Introduction to Triple Integrals

Triple integrals extend the concept of integration to functions of three variables over three-dimensional regions. Just as single integrals accumulate values along a line and double integrals accumulate values over a two-dimensional area, triple integrals accumulate values throughout volumes.

The concept of triple integration is fundamental to many areas of mathematics and physics, including fluid dynamics, electromagnetism, probability theory, and engineering calculations involving mass, momentum, and energy distribution through three-dimensional bodies.

Geometric Interpretation

Geometrically, if f(x,y,z) = 1, the triple integral over a region E represents the volume of that region. More generally, the triple integral E f(x,y,z) dV can be interpreted as accumulating the weighted "content" throughout the region, where each infinitesimal volume element dV is weighted by the function value f(x,y,z).

Notation and Setup

A triple integral is typically written as:

E f(x,y,z) dV

Where:

  • E is the three-dimensional region of integration
  • f(x,y,z) is the function being integrated
  • dV represents an infinitesimal volume element

Setting Up the Integral

In Cartesian coordinates, the triple integral becomes:

ab g(x)g(x) h(x,y)h(x,y) f(x,y,z) dz dy dx

The limits of integration define the region E. The outer integral's limits are constants, while the inner integrals' limits may be functions of the outer variables.

Evaluating Triple Integrals

Triple integrals are evaluated by performing successive integrations, starting from the innermost integral and working outward. Each integration treats all variables except the one being integrated as constants.

Basic Example

Evaluate 01 02 03 (x+y+z) dz dy dx

First, integrate with respect to z: (x+y+z) dz = [xz + yz + z/2] = 3x + 3y + 9/2

Next, integrate with respect to y: (3x + 3y + 9/2) dy = [3xy + 3y/2 + (9/2)y] = 6x + 6 + 9 = 6x + 15

Finally, integrate with respect to x: (6x + 15) dx = [3x + 15x] = 3 + 15 = 18

Coordinate Systems for Triple Integrals

Cartesian Coordinates

In Cartesian coordinates, the volume element is simply dV = dx dy dz, and the integral takes the form shown above. This is the most straightforward coordinate system when working with rectangular regions.

A simple 3D coordinate system showing the x, y, and z axes meeting at the origin.

3D Cartesian coordinate system

Cylindrical Coordinates

Cylindrical coordinates (r,,z) are useful when dealing with problems involving cylindrical symmetry. The relationship to Cartesian coordinates is:

x = r cos(), y = r sin(), z = z

With the volume element dV = r dr d dz, the triple integral becomes:

E f(r cos(), r sin(), z) r dr d dz

A cylindrical coordinate system showing the radial distance r from the z-axis, angle from the x-axis, and height z.

Cylindrical coordinate system

Spherical Coordinates

Spherical coordinates (,,) are ideal for problems with spherical symmetry, such as when integrating over spheres or portions of spheres. The relationship to Cartesian coordinates is:

x = sin() cos(), y = sin() sin(), z = cos()

With the volume element dV = sin() d d d, the triple integral becomes:

E f(,,) sin() d d d

A spherical coordinate system showing the radial distance from the origin, polar angle from the positive z-axis, and azimuthal angle in the xy-plane.

Spherical coordinate system

Applications of Triple Integrals

Volume Calculation

The simplest application of triple integrals is finding the volume of a three-dimensional region. If f(x,y,z) = 1, then E dV gives the volume of region E.

Example: Volume of a Sphere

Find the volume of a sphere with radius R using spherical coordinates.

The region is 0 R, 0 , 0 2.

Volume = R sin() d d d

= [/3]R sin() d d

= (R/3) sin() d d

= (R/3) [-cos()] d

= (R/3) 2 d

= (2R/3) d

= (2R/3) 2

= (4/3)R

Which matches the well-known formula for the volume of a sphere!

Mass and Density

If a three-dimensional object has a variable density function (x,y,z), then its mass can be found by integrating the density over the volume:

M = E (x,y,z) dV

Center of Mass

The coordinates of the center of mass (x, , ) of a three-dimensional object with density function (x,y,z) are:

x = (1/M) E x(x,y,z) dV
= (1/M) E y(x,y,z) dV
= (1/M) E z(x,y,z) dV

Moments of Inertia

Moments of inertia, important in rotational mechanics, can also be calculated using triple integrals:

Ix = E (y + z)(x,y,z) dV
Iy = E (x + z)(x,y,z) dV
Iz = E (x + y)(x,y,z) dV

Examples and Visualizations

Example 1: Rectangular Region

Evaluate E (x + y + z) dV where E is the rectangular box [0,1] [0,2] [0,3].

A 3D rectangular box showing the integrated region from x=[0,1], y=[0,2], and z=[0,3].

Rectangular region [0,1] [0,2] [0,3]

Solution:

(x + y + z) dx dy dz

First, integrate with respect to x: (x + y + z) dx = [x/3 + x(y + z)] = 1/3 + y + z

Next, integrate with respect to y: (1/3 + y + z) dy = [y/3 + y/3 + yz] = 2/3 + 8/3 + 2z = 10/3 + 2z

Finally, integrate with respect to z: (10/3 + 2z) dz = [10z/3 + 2z/3] = 10 + 18 = 28

Therefore, (x + y + z) dx dy dz = 28

Example 2: Cylinder Region

Evaluate E z dV where E is the region bounded by the cylinder x + y = 4 and the planes z = 0 and z = 5.

A 3D cylinder showing the integrated region bounded by x + y = 4 and z from 0 to 5.

Cylindrical region bounded by x + y = 4, 0 z 5

Solution: Using cylindrical coordinates:

The region is 0 r 2, 0 2, 0 z 5

zr dr d dz

First, integrate with respect to r: zr dr = z[r/2] = 2z

Next, integrate with respect to : 2z d = 2z[] = 4z

Finally, integrate with respect to z: 4z dz = 4[z/2] = 4(25/2) = 50

Therefore, E z dV = 50

Example 3: Spherical Region

Evaluate E (x + y + z) dV where E is the region enclosed by the sphere x + y + z = 9.

A 3D sphere showing the integrated region enclosed by x + y + z = 9.

Spherical region enclosed by x + y + z = 9

Solution: Using spherical coordinates:

The region is 0 3, 0 , 0 2

Since x + y + z = in spherical coordinates, we have f(,,) =

sin() d d d

First, integrate with respect to : d = [/5] = 243/5

Next, integrate with respect to : (243/5) sin() d = (243/5)[-cos()] = (243/5)(2)

Finally, integrate with respect to : (486/5) d = (486/5)[] = (486/5)(2)

Therefore, E (x + y + z) dV = (972/5)

Interactive Visualization

Explore different types of 3D regions and their integrals:

Different types of 3D regions for triple integration.

Select a region type:

Key Properties and Theorems

Linearity

Triple integrals satisfy linearity properties similar to single integrals:

E [cf + g] dV = cE f dV + E g dV

Additivity

If E can be decomposed into non-overlapping regions E, E, ..., E, then:

E f dV = i=1n E f dV

Fubini's Theorem

Fubini's Theorem extends to triple integrals, stating that if f is continuous on a rectangular region R = [a,b] [c,d] [e,f], then:

R f(x,y,z) dV = ef cd ab f(x,y,z) dx dy dz

Change of Variables

When changing variables from (x,y,z) to (u,v,w) via transformation x = x(u,v,w), y = y(u,v,w), z = z(u,v,w), the integral becomes:

E f(x,y,z) dV = T(E) f(x(u,v,w), y(u,v,w), z(u,v,w)) |J| du dv dw

Where |J| is the absolute value of the Jacobian determinant.

Divergence Theorem

The Divergence Theorem (Gauss's Theorem) connects triple integrals to surface integrals:

E (F) dV = E Fn dS

Where F is a vector field, F is its divergence, E is the boundary surface of region E, and n is the outward unit normal to E.

Key Takeaways

Triple integrals are powerful mathematical tools that allow us to:

  • Calculate volumes of complex three-dimensional regions
  • Determine mass, center of mass, and moments of inertia of three-dimensional objects with variable density
  • Solve problems in physics and engineering involving three-dimensional distributions
  • Apply different coordinate systems to simplify integration
  • Connect with surface integrals through the Divergence Theorem
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