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.

From Figure 5.9.1(a), the width of a horizontal strip across the porthole is $w\=2 x$, where

$x^2\=r^2-{\left(r-y\right)}^2\=2 r y-y^2$ 

so the area of this strip is $\mathrm{dA}\=w \mathrm{dy}$, or

$\mathrm{dA}\=2 \sqrt{2 r y-y^2}\mathrm{dy}$ 

This strip is at a depth of

$d\=20\+2 r-y$ 

The force on this strip is

$P d\=62.5 d \mathrm{dA}$

The total force on the porthole is given by the integral

$\frac{125}{2}{\int }_0^{2 r}\left(20\+2 r-y\right) \left(2 \sqrt{2 r y-y^2}\right)\mathit{\ⅆ}y$ 

whose value is $\frac{125}{2}\mathrm{\π} r^2\left(r\+20\right)$.  At $r\=1\/2$, this evaluates to $\frac{5125}{16}\mathrm{\π}\doteq 1006.29$ lbs.

Alternatively, the area of the porthole is $\mathrm{\π} {\left(1\/2\right)}^2$, 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 $1\/2$.)

(For some integrals, Maple interprets "simplify" to mean "evaluate" because the evaluated form is simpler than the unevaluated form. Because the Context Panel provides for assumptions only with the Simplify option, the integral has been so evaluated in order to apply the assumption that $r\>0$.)$$