Chapter 9 - Further Applications of the Integral and Taylor Polynomials - 9.2 Fluid Pressure and Force - Exercises - Page 475: 13

$$\frac{815,360}{3} \mathrm{N}\approx 271786.7~N$$

We calculate the force as: \begin{align*} F&=w \int_{0}^{2} y(y) d y+w \int_{2}^{6} 2 y d y\\ &=\left.w \frac{y^{3}}{3}\right|_{0} ^{2}+\left.w y^{2}\right|_{2} ^{6}\\ &=\frac{8 w}{3}+32 w\\ &=\frac{104 w}{3}\\ &=\frac{104}{3} (800)(9.8)\\ &=\frac{815,360}{3} \mathrm{N} \end{align*}