## Calculus (3rd Edition)

$$5.460393400 \times 10^{11} \mathrm{N}$$
Let $y=0$ be at the bottom of the dam, so that the top of the dam is at $y=185 .$ Then the width of the dam at height $y$ is $$2000+\frac{1000 y}{185} .$$ The dam is inclined at an angle of $55^{\circ}$ to the horizontal, so the height of a horizontal strip is $$\frac{\Delta y}{\sin 55^{\circ}} \approx 1.221 \Delta y$$ so that the area of such a strip is $$1.221\left(2000+\frac{1000 y}{185}\right) \Delta y$$ Then \begin{aligned} F &=\rho g \int_{0}^{185} 1.221 y\left(2000+\frac{1000 y}{185}\right) d y\\ &=\rho g \int_{0}^{185} 2442 y+6.6 y^{2} d y\\ &=\left.\rho g\left(1221 y^{2}+2.2 y^{3}\right)\right|_{0} ^{185} \\ &=55,718,300 \rho g\\ &=55,718,300 \cdot 9800\\ &=5.460393400 \times 10^{11} \mathrm{N} \end{aligned}