Putting It Together: Applications of Integration

Finding Hydrostatic Force

We now return our attention to the Hoover Dam, mentioned at the beginning of this chapter. The actual dam is arched, rather than flat, but we are going to make some simplifying assumptions to help us with the calculations. Assume the face of the Hoover Dam is shaped like an isosceles trapezoid with lower base 750 ft, upper base 1250 ft, and height 750 ft (see the following figure).

This figure has two images. The first is a picture of a dam. The second image beside the dam is a trapezoidal figure representing the dimensions of the dam. The top is 1250 feet, the bottom is 750 feet. The height is 750 feet.

Figure 1.

When the reservoir is full, Lake Mead’s maximum depth is about 530 ft, and the surface of the lake is about 10 ft below the top of the dam (see the following figure).

This figure is a trapezoid with the longer side on top. There is a smaller trapezoid inside the first with height labeled 530 feet. It is also 10 feet below the top of the larger trapezoid.

Figure 2. A simplified model of the Hoover Dam with assumed dimensions.

  1. Find the force on the face of the dam when the reservoir is full.
  2. The southwest United States has been experiencing a drought, and the surface of Lake Mead is about 125 ft below where it would be if the reservoir were full. What is the force on the face of the dam under these circumstances?

 

  1. We begin by establishing a frame of reference. As usual, we choose to orient the [latex]x\text{-axis}[/latex] vertically, with the downward direction being positive. This time, however, we are going to let [latex]x=0[/latex] represent the top of the dam, rather than the surface of the water. When the reservoir is full, the surface of the water is 10 ft below the top of the dam, so [latex]s(x)=x-10[/latex] (see the following figure).
    This figure is a trapezoid with the longer side on top. There is a smaller trapezoid inside the first with height labeled s(x)=x-10. It represents the depth of the water. It is also 10 feet below the top of the larger trapezoid. The top of the larger trapezoid is at x=0.

    Figure 3. We first choose a frame of reference.

    To find the width function, we again turn to similar triangles as shown in the figure below.

    This figure has two images. The first is a trapezoid with larger side on the top. The length of the top is divided into 3 measures. The first measure is 250 feet, the second is 750 feet, and the third is 250 feet. The height of the trapezoid is 750 feet. The length of the bottom is 750 feet. Inside of the trapezoid the width is labeled w(x). Inside if one of the triangular sides is the width r. The second image is the same trapezoid. It has the height labeled as 750 feet. Inside the trapezoid it has the height divided into two segments. The first is labeled x, and the second is labeled 750-x. On the side of the trapezoid a triangle has been formed by a vertical line from the bottom side to the top. Inside of the triangle is a horizontal line segment labeled r.

    Figure 4. We use similar triangles to determine a function for the width of the dam. (a) Assumed dimensions of the dam; (b) highlighting the similar triangles.

    From the figure, we see that [latex]w(x)=750+2r.[/latex] Using properties of similar triangles, we get [latex]r=250-(1\text{/}3)x.[/latex] Thus,

    [latex]w(x)=1250-\frac{2}{3}x\text{(step 2).}[/latex]

    Using a weight-density of 62.4 lb/ft3 (step 3) and applying (Figure), we get

    [latex]\begin{array}{cc}\hfill F& ={\displaystyle\int }_{a}^{b}\rho w(x)s(x)dx\hfill \\ & ={\displaystyle\int }_{10}^{540}62.4(1250-\frac{2}{3}x)(x-10)dx=62.4{\displaystyle\int }_{10}^{540}-\frac{2}{3}\left[{x}^{2}-1885x+18750\right]dx\hfill \\ & =-62.4(\frac{2}{3}){\left[\frac{{x}^{3}}{3}-\frac{1885{x}^{2}}{2}+18750x\right]|}_{10}^{540}\approx 8,832,245,000\text{lb}=4,416,122.5\text{t}\text{.}\hfill \end{array}[/latex]

Note the change from pounds to tons [latex](2000[/latex] lb = 1 ton) (step 4). This changes our depth function, [latex]s(x),[/latex] and our limits of integration. We have [latex]s(x)=x-135.[/latex] The lower limit of integration is 135. The upper limit remains 540. Evaluating the integral, we get

[latex]\begin{array}{cc}\hfill F& ={\displaystyle\int }_{a}^{b}\rho w(x)s(x)dx\hfill \\ & ={\displaystyle\int }_{135}^{540}62.4(1250-\frac{2}{3}x)(x-135)dx\hfill \\ & =-62.4(\frac{2}{3}){\displaystyle\int }_{135}^{540}(x-1875)(x-135)dx=-62.4(\frac{2}{3}){\displaystyle\int }_{135}^{540}({x}^{2}-2010x+253125)dx\hfill \\ & =-62.4(\frac{2}{3}){\left[\frac{{x}^{3}}{3}-1005{x}^{2}+253125x\right]|}_{135}^{540}\approx 5,015,230,000\text{lb}=2,507,615\text{t}\text{.}\hfill \end{array}[/latex]