# Chapter 9 - Further Applications of the Integral and Taylor Polynomials - 9.2 Fluid Pressure and Force - Exercises - Page 475: 11

$$321,250,000 \mathrm{lb}$$

#### Work Step by Step

Let $f (y)$ denote the width of the dam wall at depth $y$ feet. Then the force on the dam wall is $$F=w\int_{0}^{100}yf(y)dy$$ Using the Trapezoidal Rule and the width and depth measurements in the figure, we get: \begin{aligned} F & \approx w \frac{20}{2}[0 \cdot f(0)+2 \cdot 20 \cdot f(20)+2 \cdot 40 \cdot f(40)+2 \cdot 60 \cdot f(60)+2 \cdot 80 \cdot f(80)+100 \cdot f(100)] \\ &=10 w(0+66,000+112,000+132,000+144,000+60,000)\\ &=321,250,000 \mathrm{lb} \end{aligned}

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