Mathematical Methods for Physics and Engineering : A Comprehensive Guide

MULTIPLE INTEGRALS

z

0 2 y

dV=dx dy dz

z=2y

z=x^2 +y^2

x

Figure 6.4 The region bounded by the paraboloidz=x^2 +y^2 and the plane

z=2yis divided into vertical slabs, the slabs into horizontal strips and the

strips into boxes.

unit length anddsis an element of arc length along the wire. When evaluating

the required integral, we are free to divide up the body into mass elements in

the most convenient way, provided that over each mass element the density is

approximately constant.

Find the mass of the tetrahedron bounded by the three coordinate surfaces and the plane

x/a+y/b+z/c=1, if its density is given byρ(x, y, z)=ρ 0 (1 +x/a).

From (6.8), we can immediately write down the mass of the tetrahedron as

M=

∫

R

ρ 0

(

1+

x

a

)

dV=

∫a

0

dx ρ 0

(

1+

x

a

)∫b−bx/a

0

dy

∫c( 1 −y/b−x/a)

0

dz,

where we have taken the density outside the integrations with respect tozandysince it
depends only onx. Therefore the integrations with respect tozandyproceed exactly as
they did when finding the volume of the tetrahedron, and we have

M=cρ 0

∫a

0

dx

(

1+

x

a

)(bx 2

2 a^2

−

bx

a

+

b

2

)

. (6.9)

We could have arrived at (6.9) more directly by dividing the tetrahedron into triangular
slabs of thicknessdxperpendicular to thex-axis (see figure 6.3), each of which is of
constant density, sinceρdepends onxalone. A slab at a positionxhas volumedV=
1
2 c(1−x/a)(b−bx/a)dxand massdM=ρdV=ρ^0 (1 +x/a)dV. Integrating overxwe
again obtain (6.9). This integral iseasily evaluated and givesM= 245 abcρ 0 .