   Chapter 16.9, Problem 32E

Chapter
Section
Textbook Problem

A solid occupies a region E with surface S and is immersed in a liquid with constant density ρ. We set up a coordinate system so that the xy-plane coincides with the surface of the liquid, and positive values of z are measured downward into the liquid. Then the pressure at depth z is p = ρgz, where g is the acceleration due to gravity (see Section 8.3). The total buoyant force on the solid due to the pressure distribution is given by the surface integral F   =   − ∬ S p n   d S   where n is the outer unit normal. Use the result of Exercise 31 to show that F = - Wk, where W is the weight of the liquid displaced by the solid. (Note that F is directed upward because z is directed downward.) The result is Archimedes’ Principle: The buoyant force on an object equals the weight of the displaced liquid.

To determine

To show: The expression F=SpndS is equals to the expression F=Wk .

Explanation

Given Data:

A solid occupies a region E with surface S and is immersed in a liquid with constant density ρ .

The pressure at the depth z is given as follows.

p=ρgz (1)

Here,

ρ is the constant density and

g is the acceleration due to the gravity.

Write the given surface integral that has to be evaluated as follows.

F=SpndS (2)

Formula used:

Write the expression to find flux of the vector field F(x,y,z) across the surface S .

SFndS=EdivFdV (3)

Write the expression to find divergence of vector field F(x,y,z)=Pi+Qj+Rk .

divF=xP+yQ+zR (4)

Write the expression for weight of the displace liquid when the solid immersed in the liquid.

W=ρgV(E) (5)

Here,

V(E) is the volume of the solid occupied region E .

Consider an arbitrary constant vector c as follows.

c=c1i+c2j+c3k

Consider the vector field F as follows.

F=fc

Substitute c1i+c2j+c3k for c ,

F=f(c1i+c2j+c3k)=fc1i+fc2j+fc3k

Calculation of divF :

Substitute fc1 for P , fc2 for Q , and fc3 for R in equation (4),

divF=x(fc1)+y(fc2)+z(fc3)=fxc1+fxc2+fxc3=divf(c1i+c2j+c3k)=divfc

Substitute (divfc) for divF in equation (3),

SFndS=E(divfc)dV (6)

If c=i , then the expression is written as follows.

SfindS=E(divfi)dV

Consider fi=fn1 and rewrite the expression as follows.

Sfn1ndS=E(fx)dV (7)

If c=j , then the expression is written as follows

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