Chapter 9 - Further Applications of the Integral and Taylor Polynomials - 9.2 Fluid Pressure and Force - Exercises - Page 475: 22

$$595350 \mathrm{N}$$

Since $$ \frac{\Delta y}{\sin 30^{\circ}}=2 \Delta y $$ and the area of each strip is $$ A=(3)(2)(\Delta y)=6 \Delta y $$ Therefore, \begin{aligned} F=\rho g \int_{a}^{b} y f(y) d y &=9800 \int_{6}^{7.5} 6 y d y \\ &=9800\left[3 y^{2}\right]_{6}^{7.5} \\ &=9800[168.75-108] \\ &=9800[60.75] \\ &=595350 \mathrm{N} \end{aligned}