Page 262 - Calculus Demystified

Applications of the Integral
CHAPTER 8
is8 feet. Refer to Fig. 8.38. The pool isfull. Calculate the total force on one 249
of the long sides of the pool.
20
10
8

Fig. 8.38

SOLUTION
We let the independent variable h denote depth, measured vertically down
from the surface of the water. Since the pool is rectangular with vertical sides,
w(h) is constantly equal to 20 (because we are interested in the long side). We
use 62.4 pounds per cubic foot for the density of water. According to (∗), the
total force on the long side is
8 8

h · 62.4 · w(h) dh = h · 62.4 · 20 dh = 39936 lbs.
0 0
You Try It: A tank full of water is in the shape of a cube of side 10 feet. How
much force is exerted against one wall of the tank between the depths of 3 feet and
6 feet?

EXAMPLE 8.27
A tank has vertical cross section in the shape of an inverted isosceles tri-
angle with horizontal base, as shown in Fig. 8.39. Notice that the base of the
tank haslength 4 feet and the height is9 feet.The tank isfilled with water to
a depth of 5 feet. Water hasdensity 62.4 poundsper cubic foot. Calculate
the total force on one end of the tank.

4 ft

9 ft

5 ft

Fig. 8.39