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As per Figure 5.9.1(a), let the origin of a Cartesian coordinate system be at the bottom of the porthole, and let $y\ge 0$ measure distance upwards. The surface of the water is then at $y\=21$ since there are 20 ft of water above the top of the porthole.

Figure 5.9.1(a) Porthole under hydrostatic pressure



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From Figure 5.9.1(a), the width of a horizontal strip across the porthole is $w\=2x$, where
${x}^{2}\={r}^{2}{\left(ry\right)}^{2}\=2ry{y}^{2}$
so the area of this strip is $\mathrm{dA}\=w\mathrm{dy}$, or
$\mathrm{dA}\=2\sqrt{2ry{y}^{2}}\mathrm{dy}$
This strip is at a depth of
$d\=20\+2ry$
The force on this strip is
$Pdequals;62.5d\mathrm{dA}$
The total force on the porthole is given by the integral
$\frac{125}{2}{\int}_{0}^{2r}\left(20plus;2ry\right)\left(2\sqrt{2ry{y}^{2}}\right)\mathit{DifferentialD;}y$
whose value is $\frac{125}{2}\mathrm{\π}{r}^{2}\left(rplus;20\right)$. At $r\=1\/2$, this evaluates to $\frac{5125}{16}\mathrm{\π}\doteq 1006.29$ lbs.
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Alternatively, the area of the porthole is $\mathrm{\π}{\left(1sol;2\right)}^{2}$, and its centroid (the center) is at a depth of $20.5$ ft. The product of the area and the pressure at the centroid is then $62.5\times 20.5\times \mathrm{\π}\/4$$\doteq$${1006.291397}$ lbs.
(Note: There was no mathematical reason for keeping the radius of the porthole as the symbolic $r$. It could just as well have been set immediately to $1\/2$ for computational purposes. However, the letter $r$ typesets more compactly than the fraction $1\/2$.)