## Calculus (3rd Edition)

Published by W. H. Freeman

# Chapter 9 - Further Applications of the Integral and Taylor Polynomials - 9.2 Fluid Pressure and Force - Exercises - Page 474: 9

#### Answer

$$\frac{19,600}{3} r^{3}+4900 m r^{2} \mathrm{N}$$

#### Work Step by Step

Place the origin at the center of the semicircular plate with the positive $y$ -axis pointing downward. The water surface is then at $y=-m .$ Moreover, at location $y,$ the width of the plate is $2 \sqrt{r^{2}-y^{2}}$ and the depth is $y+m .$ Thus, $$F=2 \rho g \int_{0}^{r}(y+m) \sqrt{r^{2}-y^{2}} d y$$ Now, $$\int_{0}^{r} y \sqrt{r^{2}-y^{2}} d y=-\left.\frac{1}{3}\left(r^{2}-y^{2}\right)^{3 / 2}\right|_{0} ^{r}=\frac{1}{3} r^{3}$$ Geometrically, $\int_{0}^{r} \sqrt{r^{2}-y^{2}} d y$ represents the area of one quarter of a circle of radius $r,$ and thus has the value $\frac{\pi r^{2}}{4} .$ Bringing these results together, we find that $$F=2 \rho g\left(\frac{1}{3} r^{3}+\frac{\pi}{4} r^{2}\right)=\frac{19,600}{3} r^{3}+4900 m r^{2} \mathrm{N}$$

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