Calculus II Class 02/26

The figure below represents a dam.  We are given that the force on a portion of the dam is equal to the product of average pressure and area, and that pressure at depth y is 64 lbs/ft^3 * h.  The water level is assumed to be even with the top of the dam.   We wish to find the total force on the dam.

• We begin by considering a region on which the pressure is nearly constant.
• Nearly constant pressure will occur on any thin horizontal strip.  So we consider a horizontal strip of width `dy, located at vertical position y relative to the bottom of the dam.
• We need to find the dimensions of this strip so we can find its area, and the pressure at this vertical location.  We will then multiply pressure by area to find the force on the strip.

Due to the straight sides of the dam the width will increase linearly with y.   Since width increases by 100 ft as altitude increases by 100 ft, we conclude that width increases by 1 ft per foot of altitude.  We quickly conclude that the width function is width = 200 + y.

The width of the dam will be the length of the thin horizontal strip.

The strip therefore has length 200 + y and width `dy so its area is

• `dA = strip length * strip width = ( 200 + y ) * `dy.

At the position of this strip water depth is 100 - y.  Pressure at this depth is 64 * depth = 64 * (100 - y).

The force contribution of this strip is therefore

• `dF = pressure * area = 64 ( 100 - y) * (200  + y) * `dy.

Adding up force contributions we obtain the sum and integral shown below.

• Note that horizontal strips experiencing water pressure extend from y = 0 to y = 100.
• Note also that the integrand ( 100 - y) * (200  + y) multiplies out into an easily integrated polynomial.

For a semicircular dam we use the same strategy, with the following differences in details:

• We measure y downward from the top of the dam, so that y will be the depth.
• We must use the geometry of the circle and the Pythagorean Theorem to determine the length of the horizontal strip.

We obtain the sum and the integral shown below.  Note that strips experiencing pressure extend from the top at y = 0 to the lowest point at y = 150.

If the profile of the dam consists of the bottom 100 ft of a circle, and the top has a width of 300 ft, this determines the position of the center of the circle.  This information will allow us to then determine the width of a horizontal strip at a specified vertical location.

Knowing that the radius extends from the center to an extreme point at the top of the dam, and also from the center to the bottom of the dam, we first use the Pythagorean Theorem to find that the distance from the center to the top of the dam must be sqrt(r^2 - 150^2).

We then observe that this distance plus 100 must be equal to the radius r itself.   This gives us an equation we easily enough solve for r.

Knowing r we determine the width of a typical strip at distance y below the center of the circle.

Noting that the water begins 62.5 ft below the center we see that the depth at this vertical position is y - 62.5.  This allows us to find the pressure at this depth.

We then proceed to find the force contribution of this horizontal strip.

Summing the forces and taking the limit as `dy -> 0 we obtain the integral in the figure below.

• Note that the horizontal strips experiencing pressure run from y = 62.5 at the water surface to y = 162.5 at the lowest point, so these y values are the limits on the integral.