Begin2.DVI

Example 7-34. Evaluate the integral

∫∫∫

V

f(x, y, z )dV, where f(x, y, z) = 6(x+y),

dV =dxdydz is an element of volume and V represents the volume enclosed by the

planes 2 x+ 2 y+z= 12, x= 0 ,y= 0 , and z= 0. The integration is to be performed

over this volume.

Solution

The figure on the left is a copy of the

figure 7-20 with summations of the element

dV illustrated. From this figure the limits of

integration can be determined. The volume

element is dV =dxdydz is placed at the gen-

eral point (x, y, z) within the volume. This

volume element can be visualized as a cube

inside the volume. Summation of these cu-

bic elements aids in determining the limits of integration for the integral to be

calculated. If this cube is summed in the z-direction, a parallelepiped is produced.

This parallelepiped has lower limit z = 0 and upper limit z = 12 − 2 x− 2 y. If the

parallelepiped is summed in the y-direction, then a triangular slab is formed with

lower limit y = 0 and upper limit y = 6 −x. Summing the triangular slabs in the

x-direction from x= 0 to x= 6 gives the limits of integration in the x-direction. At

each stage of the summation process, the volume element is weighted by the scalar

function f(x, y, z )giving the integral

∫∫∫

V

f(x, y, z )dV. From all this summation one

can verify the above integral can be expressed.

∫∫∫

V

f(x, y, z)dV =

∫x=6

x=0

∫y=6−x

y=0

∫ z=12− 2 x− 2 y

z=0

6(x+y)dz dy dx

=

∫ 6

0

[∫ 6 −x

0

6(x+y)(12 − 2 x− 2 y)dy

]

dx

=

∫ 6

0

(432 − 36 x^2 + 4x^3 )dx = 1296

This integral is calculated by first integrating in the z-direction holding the other

variables constant. This is followed by an integration in the y-direction holding x

constant. The last integration is then in the x-direction.