Page 261 - Calculus Demystified

Applications of the Integral
CHAPTER 8
248 Assume that the width of the tank at depth h is w(h).The portion of the wall between
h = h j−1 and h = h j is then approximated by a rectangle R j of length w(h j ) and
width 8h = h j − h j−1 (Fig. 8.37).

h = a
h = b

Fig. 8.36

w(h j )

h
Fig. 8.37
TEAMFLY
Now we have the following data:
Area of Rectangle = w(h j ) · 8h square feet
Depth of Water ≈ h j feet
Density of Liquid = ρ pounds per cubic foot.

It follows that the force exerted on this thin portion of the wall is about
P j = h j · ρ · w(h j ) · 8h.

Adding up the force on each R j gives a total force of

k
k
P j = h j ρw(h j )8h.
j=1 j=1
But this last expression is a Riemann sum for the integral
b
hρw(h) dh. (∗)
a
EXAMPLE 8.26
A swimming pool is rectangular in shape, with vertical sides. The bottom
of the pool hasdimensions10 feet by 20 feet and the depth of the water

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Team-Fly