## Calculus (3rd Edition)

$$940,800 \mathrm{N}$$
A horizontal strip at depth $y$ has length 6 and width $$\frac{\Delta y}{\sin 30^{\circ}}=2 \Delta y$$ Thus, \begin{align*} F&=2 \rho g \int_{0}^{4} 6 y d y\\ &=6\rho g y^2\bigg|_{0}^{4}\\ &=96 \rho g \end{align*} If distances are in feet, then $\rho g=w=62.5 \mathrm{lb} / \mathrm{ft}^{3}$ and $F=6000 \mathrm{lb}$; if distances are in meters, then $\rho g=9800 \mathrm{N} / \mathrm{m}^{3}$ and $$F=940,800 \mathrm{N}$$