Triple integrals are defined for functions of three variables as the limit of a sum, just as the single and double integrals are defined respectively for functions of one and two variables. The Box interval B
B = { x, y, z | a = x = b, c = y = d, and p = z = q} where the function f(x, y, z) is defined is partitioned into lmn sub boxes by dividing the intervals [a, b], [c, d] and [p. q] into l, m and n sub intervals of equal widths $\Delta$x, $\Delta$y and $\Delta$z for respective partitions.

Triple Integral

Each sub box has thus a volume $\Delta$V = $\Delta$x. $\Delta$y. $\Delta$z. Analogically with single and double integrals, the triple integral of f(x, y, z) over the Box is defined as the limit of the triple Riemann sum, if the limit exists.$\int \int_{B} \int f(x,y,z)dV$ = $\lim_{l,m,n \to \infty }\sum_{i=1}^{l}\sum_{j=1}^{m}\sum_{k=1}^{n}f(x_{ijk}y_{ijk}z_{ijk})\Delta V$.

For practical purpose, Fubini's theorem on triple integral is used to evaluate triple integrals.
If f(x, y, z) is continuous on a rectangular Box B = [a, b] x [c, d] x [p, q], then
$\int \int_{B} \int f(x,y,z)dV$ = $\int_{p}^{q}\int_{c}^{d}\int_{a}^{b}f(x,y,z)dxdydz$
If the solid region considered for triple integrals happens to lie between the graphs two continuous functions of any two of the variables, the formula used is modified to include a double integral as follows:
$\int \int_{E} \int f(x,y,z)dV$ = $\int \int_{D}$$[\int_{u_{1}(x,y)}^{u_{2}(x,y)}f(x,y,z)dz]dA$
where the region E is bounded by the continuous functions z = u1 (x,y) and z =u2 (x, y).
Similar formulas are also used when the solid region for the triple integrals is bounded by functions of other two variable combinations like yz and zx.

The triple integral cannot be interpreted geometrically as the function f in three variables represents a hyper surface in four dimensional space and cannot be visualized.

Suppose the solid region on which the triple integral is defined is
E = {(x, y, z) | (x, y) ∈ D, u1 (x,y) ≤ z ≤ u2 (x,y)} where the plane region D is defined in polar coordinates as
D = { (r, θ) | α ≤ θ ≤ β, h1 (θ) ≤ r ≤ h2 (θ) }.
The triple integral in rectangular coordinates
$\underset{E}{\int \int \int }f(x,y,z)dV$ = $\underset{D}{\int \int }[\int_{u_{1}(x,y)}^{u_{2}(x,y)}f(x,y,z)dz]dA$
can be transformed for cylindrical coordinates system as
$\underset{E}{\int \int \int }f(x,y,z)dV$ = $\int_{\alpha }^{\beta }\int_{h_{1}(\theta )}^{h_{2}(\theta )}\int_{u_{1}(rcos\theta ,rsin\theta )}^{u_{2}(rcos\theta ,rsin\theta )}f(rcos\theta ,rsin\theta ,z)rdzdrd\theta $.

Remember the transformations required for cylindrical coordinate system are x = r cos θ, y = r sin θ, z = z and
dV = r dz dr dθ.
This transformation is useful when the solid region E can be easily described in cylindrical coordinates and the f (x, y, z) contains expressions of the form x2 + y2.
The relationships between the rectangular coordinates and spherical coordinates are as follows:

x = ρ sin Φ cos θ, y = ρ sin Φ sin θ and z = ρ cos Φ

The counter part of the Box in spherical coordinate system is the spherical wedge
E = { (ρ, θ, Φ ) | p ≤ ρ ≤ q, α ≤ θ ≤ β, a ≤ Φ ≤ b}
The triple integration formula in Spherical coordinates defined over the spherical wedge E is
$\underset{E}{\int \int \int }f(x,y,z)dV$ = $\int_{a}^{b}\int_{\alpha }^{\beta }\int_{p}^{q}f(\rho sin\phi cos\theta ,\rho sin\phi sin\theta ,\rho cos\phi )\rho ^{2}sin\phi d\rho d\theta d\phi $

Along with the usual transformations for rectangular coordinates, dV is replaced by ρ sin Φ dρ dθ dΦ to evaluate a triple integral in spherical coordinate system.

Solved Examples

Question 1: $\int_{0}^{1}\int_{x}^{2x}\int_{0}^{y}2xyzdzdydx$
Solution:
 
The integration is done in the following order. First with respect to z, second with respect to y and lastly with respect to
     x.
    $\int_{0}^{1}\int_{x}^{2x}\int_{0}^{y}2xyzdzdydx$ = $\int_{0}^{1}\int_{x}^{2x}[xyz^{2}]_{z=0}^{z=y}dydx$
                                                                              = $\int_{0}^{1}\int_{x}^{2x}xy^{3}dydx$

                                                                              = $\int_{0}^{1}[\frac{xy^{4}}{4}]_{y=x}^{y=2x}$dx

                                                                              = $\int_{0}^{1}\frac{15x^{5}}{4}$dx

                                                                              = $[\frac{15x^{6}}{24}]_{0}^{1}$

                                                                              = $[\frac{5x^{6}}{8}]_{0}^{1}$ = $\frac{5}{8}$.
 

Question 2: Evaluate the triple integral $\int \int_{E} \int xydV$ where E is the solid tetrahedron with vertices (0, 0, 0), (1, 0, 0), (0, 2, 0) and ( 0, 0, 3).
Simple Integral Example
Solution:
 
E is the region below the plane whose intercepts with the x, y and z axes are 1, 2 and 3.
    The equation to this plane is hence 6x + 3y + 2z = 6. The projection on the xy plane is
     bounded by x = 0 , y =0 and the line 6x + 3y = 6. Hence the triple integral is evaluated as follows:
     $\int \int_{E} \int xydV$
     = $\int_{0}^{1}\int_{0}^{2-2x}\int_{0}^{3-3x-\frac{3y}{2}}$$xydzdydx$

     = $\int_{0}^{1}\int_{0}^{2-2x}[xyz]_{z=0}^{z=3-3x-\frac{3y}{2}}$$dydx$                                  Integration with respect to z

     = $\int_{0}^{1}\int_{0}^{2-2x}(3xy-3x^{2}y-\frac{3}{2}xy^{2})$$dydx$

     = $\int_{0}^{1}[\frac{3}{2}xy^{2}-\frac{3}{2}x^{2}y^{2}-\frac{xy^{3}}{2}]_{y=0}^{y=2-2x}$                           Integration with respect to y

     = $\int_{0}^{1}$ (2x -6x2 + 6x3 - 2x4 ) dx

     = [x2 -2x3 + $\frac{3}{2}$x4 - $\frac{2}{5}$x5 ] $_{0}^{1}$                                                        Integration with respect to x

     = $\frac{1}{10}$
 

Question 3: Find the value of $\int \int_{E} \int$$\sqrt{x^{2}+y^{2}}dV$ where E is the region within the cylinder
x2 + y2 = 16 and between the planes z = 0 and z =4
Solution:
 
The solid region for integration E is defined in cylindrical coordinates as

E = { (r, θ, z) | 0 ≤ θ ≤ 2π, 0 ≤ r ≤ 4, 0 ≤ z ≤4 }

$\int \int_{E} \int$$\sqrt{x^{2}+y^{2}}dV$

= $\int_{0}^{2\pi }\int_{0}^{4}\int_{0}^{4}\sqrt{r^{2}}rdzdrd\theta $

= $\int_{0}^{2\pi }d\theta \int_{0}^{4}r^{2}dr\int_{0}^{4}dz$       Integration separated in order

= $\int_{0}^{2\pi }d\theta \int_{0}^{4}r^{2}[z]_{0}^{4}dr$            Integration with respect to z

= $\int_{0}^{2\pi }d\theta \int_{0}^{4}4r^{2}dr$

= $\int_{0}^{2\pi }[\frac{4}{3}r^{3}]_{0}^{4}d\theta$                       Integration with respect to r

= $\int_{0}^{2\pi }\frac{256}{3}d\theta $

= $[\frac{256}{3}\theta ]_{0}^{2\pi }$                               Integration with respect to θ

= $\frac{512}{3}$π