By the Pythagorean theorem,the height of the dam is $\sqrt{{15}^2-5^2}\=\sqrt{200}\=10\sqrt{2}$. (See Figure 5.9.2(a).)

so that $\mathrm{dA}\=\left(\frac{y}{\sqrt{2}}\+15\right)\mathrm{dy}$. The force is the product of the pressure $P\=62.5 d$ and the area, or

  ${\int }_0^{10\sqrt{2}}P d \mathrm{dA}\=\frac{125}{2}{\int }_0^{10\sqrt{2}}\left(10\sqrt{2}-y\right) \left(\frac{y}{\sqrt{2}}\+15\right) \mathit{\ⅆ}y\=$$\frac{343750}{3}$ ≐$114583.33$ lbs

Alternatively, obtain the hydrostatic force as the product of the area and the pressure at the centroid of the dam.

From Example 5.7.5, $\stackrel{\&conjugate0;}{y}$ = $65\sqrt{2}\/12$, and $A\=200\sqrt{2}$.

Consequently, the hydrostatic force is $\left(h-\stackrel{\&conjugate0;}{y}\right)A \mathrm{\δ}$ = $62.5\times \left(10\sqrt{2}-65\sqrt{2}\/12\right)\times 200\sqrt{2}$$\doteq$$1.145833333{10}^5$

Alternatively, $F\=\left(h-\stackrel{\&conjugate0;}{y}\right)A \mathrm{\δ}$ =   $62.5\times \left(10\sqrt{2}-65\sqrt{2}\/12\right)\times 200\sqrt{2}$$\doteq$$1.145833333{10}^5$.