Page 424 - Advanced engineering mathematics

404 CHAPTER 12 Vector Integral Calculus

so

∇· FdV = 3dV
M M
= 3[volume of the cone of height 1, radius 1]

1
= 3 π = π.
3
EXAMPLE 12.25

Let be the piecewise smooth surface of the cube having vertices
(0,0,0),(1,0,0),(0,1,0),(0,0,1)

(1,1,0),(0,1,1),(1,0,1),(1,1,1).

2
2
2
Let F(x, y, z) = x i + y j + z k. We want to compute the flux of F across . This flux is

F · ndσ. We can certainly evaluate this integral, but this will be tedious because has

six smooth faces. It is easier to use the triple integral of the divergence theorem. Compute
∇· F = 2x + 2y + 2z.
Then

flux = F · ndσ

= ∇· FdV = (2x + 2y + 2z)dV
M M
1
1
1
= (2x + 2y + 2z)dz dy dx
0 0 0
1
1
= (2x + 2y + 1)dy dx
0 0
1
(2x + 2)dx = 3.
=
0
12.8.1 Archimedes’s Principle
Archimedes’s principle is that the buoyant force a fluid exerts on a solid object immersed in it,
is equal to the weight of the fluid displaced. An aircraft carrier floats in the ocean if it displaces
a volume of seawater whose weight at least equals that of the carrier. We will use the divergence
theorem to derive this principle.
Imagine a solid object M bounded by a piecewise smooth surface .Let ρ be the constant
density of the fluid. Draw a coordinate system as in Figure 12.22 with M below the surface.
Using the fact that pressure equals depth multiplied by density, the pressure ρ(x, y, z) at a point
on is p(x, y, z) =−ρz, the negative sign because z is negative in the downward direction and
we want pressure to be positive. Now consider a piece j of . The force of the pressure on
j is approximately −ρz multiplied by the area A j of j .If n is the unit outer normal to j ,
then the force caused by the pressure on j is approximately ρznA j . The vertical component of
this force is the magnitude of the buoyant force acting upward on j . This vertical component is

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October 14, 2010 14:53 THM/NEIL Page-404 27410_12_ch12_p367-424