# Notes on Volume Integrals

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```					      EEM3J – Analytical & Numerical Methods                                                       Page 1
Supplementary notes on Volume Integrals – MJR 6 December 2001

Notes on Volume Integrals

Let B denote the solid box in 3 dimensional space with opposite corners at the points
(0,0,0) and (X,Y,Z).
A volume V in R3 is parameterised by a function ϕ : B → R 3 .

Let f:R3→R be a scalar field. How do we calculate the volume integral               ∫∫∫ fdV ?
V

As with line integrals and surface integrals, we are interested in the volume V and not
the particular parameterisation ϕ. Hence the integrand needs to include a term which
ensures that the value of the integral is independent of ϕ.

ϕ                                               V

∆bn                                              ϕ (∆bn)

Imagine that the volume B is partitioned into N equal small boxes, ∆b1,…, ∆bN. Then
we can define the volume integral (compare with the previous line and surface
integrals) by:
N

∫∫∫    fdV = lim ∑ f i ϕ (∆bn )
N →∞
V                n =1

where fn denotes the value of f on ϕ(∆bn) (we’ll assume that ∆bn is so small that f can
assumed to be constant on ∆bn) and |ϕ(∆bn)| denotes the volume of ϕ(∆bn). As before,
we need to compensate for variation in |ϕ(∆bn)| due to the choice of ϕ. This is
achieved as follows:
Write ϕ (u , v, w) = (ϕ1 (u , v, w),ϕ 2 (u, v, w),ϕ 3 (u, v, w)) where each ϕj is a function from
R3 into R. The Jacobian matrix J of ϕ is given by:
 ∂ϕ1     ∂ϕ 2    ∂ϕ 3 
 ∂u       ∂u      ∂u 
 ∂ϕ      ∂ϕ 2    ∂ϕ 3 
J = 1                    
 ∂v       ∂v      ∂v 
 ∂ϕ1     ∂ϕ 2    ∂ϕ 3 
 ∂w
         ∂w      ∂w  
Now let |J| denote the absolute value of the determinant of J. The volume integral is
then defined as:
EEM3J – Analytical & Numerical Methods                                                 Page 2
Supplementary notes on Volume Integrals – MJR 6 December 2001

ZY X

∫∫∫ fdV = ∫ ∫ ∫ f (ϕ (u, v, w)) J dudvdw
V              0 0 0

Example
You already know that the volume of a cylinder with base radius R in the (x,y) plane
and height H is equal to HπR2. Let’s derive this expression by evaluating a volume
integral.
First parameterise the solid cylinder:
ϕ : [0, H ] × [0, R ] × [0,2π ] → R 3 , ϕ (h, r , u ) = (r cos u, r sin u, h )
Calculate the Jacobian:
0            0    1
J = cos u     sin u 0 = r cos 2 u + r sin 2 u = r
− r sin u r cos u 0

Hence the volume integral is:
2π R H            2π R            2π 2π       R                  2π
r2H          R2H       R2H 
∫∫∫
V
1dV = ∫∫ ∫ rdhdrdu = ∫∫ rHdrdu = ∫ 
0 0 0          0 0         0

2 0
du = ∫
0
2
du =     u  = πR 2 H
 2 0
Now let f be the scalar field f ( x, y, z ) = x + y − z . Let’s evaluate the volume integral
∫∫∫ fdV
V
over the same volume. We can use the same parameterisation, and hence the

same Jacobian matrix. Note that f (ϕ (h, r , u )) = r cos u + r sin u − h
2π R H

∫∫∫ fdV = ∫∫ ∫ (r cos u + r sin u − h )rdhdrdu
V               0 0 0
2π R                                    H
 2                    rh 2 
= ∫∫ hr cos u + hr sin u −
2
drdu
0 0
2 0
2π R
rH 2
=    ∫∫ Hr cos u + Hr sin u −
2          2
drdu
0 0
2
2π                                     R
 Hr 3                     2 2

= ∫      (cos u + sin u ) − H r  du
0
3                       4 0
2π
HR 3                     2 2
=    ∫ 3    (cos u + sin u ) − H R du
0
4
2π
 HR 3                     2 2

=      (sin u − cos u ) − H R u 
 3                         4 0
− 2πH 2 R 2 − H 2 R 2π
=               =
4           2

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