Finding Hydrostatic Force
A cylindrical sewer pipe of radius 2 m is half full of water. A gate used to seal off the sewer is placed perpendicular to the pipe opening. Find the hydrostatic force exerted on one side of the gate.
The equation of the circle with center at \((0,0)\) and radius \(2\) is \(x^{2}+y^{2}=4\). Then \[ x=g(y)=-\sqrt{4-y^{2}}\qquad \hbox{and }\qquad x=f(y)=\sqrt{4-y^{2}} \]
The height of the water is at \(H=0\). Since the water fills the cylinder from \(y=-2\) to \(y=0,\) the hydrostatic force \(F\) exerted on one side of the gate by the pressure of the water is \[ \begin{array}{rcl@{\qquad}l} F &=&\int_{c}^{d}\rho g( {H-y}) [ f(y)-g(y)] ~{\it dy} & \\[3pt] &=&\int_{-2}^{0}\rho g(0-y)\big[ \sqrt{4-y^{2}}-\big( -\!\sqrt{4-y^{2}}\big) \big] ~{\it dy} &\color{#0066A7}{H=0; c=-2; d=0.} \\ &=&9800\int_{-2}^{0}-2y\sqrt{4-y^{2}}~{\it dy} &\color{#0066A7}{\rho =1000\hbox{ kg}/\hbox{m}^{3}; g=9.8\ \hbox{ m}/\hbox{s}^{2}.} \\ &=&9800\int_{0}^{4}\sqrt{u}~{\it du}=9800\left[ \frac{2u^{3/2}}{3}\right]_{0}^{4} &\color{#0066A7}{\hbox{Let \(u=4-y^{2} \); }} \\ &&&\color{#0066A7}{\hbox{then \(du=-2y~{dy} \).}}\\ &&&\color{#0066A7}{\hbox{When \(y=-2 \), then \(u=0 \);}} \\ &&&\color{#0066A7}{\hbox{when \(y=0 \), then \(u=4 \).}}\\[-3pc] \\ &=&9800\left( \dfrac{16}{3}\right) \approx 52{,}267 {N} \end{array}\]