Page 260 - Calculus Demystified

CHAPTER 8
SOLUTION Applications of the Integral 247
We think of the curve as the graph of φ(y) = y 1/6 , 1 ≤ y ≤ 4096. Then the
formula for surface area is

4096

2 1/2

2π φ(y) 1 +[φ (y)] dy.
1

Calculating φ (y) and substituting, we find that the desired surface area is the
value of the integral

4096
2π y 1/6 1 + (1/6)y −5/6 2 1/2 dy.
1
You Try It: Write the integral that represents the surface area of a hemisphere of
radius one and evaluate it.

8.6 Hydrostatic Pressure

If a liquid sits in a tank, then it exerts force on the side of the tank. This force is
caused by gravity, and the greater the depth of the liquid then the greater the force.
Pascal’s principle asserts that the force exerted by a body of water depends on
depth alone, and is the same in all directions. Thus the force on a point in the side
of the tank is defined to be the depth of the liquid at that point times the density of
the liquid. Naturally, if we want to design tanks which will not burst their seams, it
is important to be able to calculate this force precisely.

Fig. 8.35

Imagine a tank of liquid having density ρ pounds per cubic foot as shown in
Fig. 8.35. We want to calculate the force on one flat side wall of the tank. Thus
we will use the independent variable h to denote depth, measured down from the
surface of the water, and calculate the force on the wall of the tank between depths
h = a and h = b (Fig. 8.36). We partition the interval [a, b]:

a = h 0 ≤ h 1 ≤ h 2 ≤ ··· ≤ h k−1 ≤ h k = b.