Applications of surface integration

4 Applications of surface integration

4.1 Force on a dam

Section 27.1 considered the force on a rectangular dam of width 100 m and height 40 m. Instead, imagine that the dam is not rectangular in profile but instead has a width of 100 m at the top but only 80 m at the bottom. The top and bottom of the dam can be given by line segments $y = 0$ (bottom) and $y = 40$ while the sides are parts of the lines $y = 40 - 4 x$ i.e. $x = 10 - \frac{y}{4}$ (left) and $y = 40 + 4 (x - 100) = 4 x - 360$ i.e. $x = 90 + \frac{y}{4}$ (right). (See Figure 20).

Figure 20

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Thus the dam exists at heights $y$ between 0 and 40 while for each value of $y$ , the horizontal coordinate $x$ varies between $x = 10 - \frac{y}{4}$ and $x = 90 + \frac{y}{4}$ . Thus the surface integral representing the total force i.e.

$I = \int_{A} 1 0^{4} (40 - y) d A$ becomes the double integral $I = \int_{0}^{40} \int_{10 - \frac{y}{4}}^{90 + \frac{y}{4}} 1 0^{4} (40 - y) d x d y$

which can be evaluated as follows

\begin{array}{rcl} I & = & \int_{0}^{40} \int_{10 - \frac{y}{4}}^{90 + \frac{y}{4}} 1 0^{4} (40 - y) d x d y \\ = & 1 0^{4} \int_{0}^{40} {[(40 - y) x]}_{10 - \frac{y}{4}}^{90 + \frac{y}{4}} d y = 1 0^{4} \int_{0}^{40} [(40 - y) (90 + \frac{y}{4}) - (40 - y) (10 - \frac{y}{4})] d y \\ = & 1 0^{4} \int_{0}^{40} [(40 - y) (80 + \frac{y}{2})] d y = 1 0^{4} \int_{0}^{40} [3200 - 60 y - \frac{y^{2}}{2}] d y \\ = & 1 0^{4} {[3200 y - 30 y^{2} - \frac{1}{6} y^{3}]}_{0}^{40} = 1 0^{4} [(3200 \times 40 - 30 \times 4 0^{2} - \frac{1}{6} 4 0^{3}) - 0] \\ = & 1 0^{4} \times \frac{208000}{3} \approx 6.93 \times 1 0^{8} N \end{array}

i.e. the total force is just under 700 meganewtons.

4.2 Centre of pressure

A plane area in the shape of a quadrant of a circle of radius $a$ is immersed vertically in a fluid with one bounding radius in the surface. Find the position of the centre of pressure.

Figure 21

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Note: In subsection 6 of Section 27.1 it was shown that the coordinates of the centre of pressure of a (thin) object are

$x_{p} = \frac{\int_{A} x y d A}{\int_{A} y d A} and y_{p} = \frac{\int_{A} y^{2} d A}{\int_{A} y d A}$

\begin{array}{rcl} \int_{A} y d A & = & \int_{0}^{\frac{π}{2}} \int_{0}^{a} r^{2} sin θ d r d θ = \int_{0}^{\frac{π}{2}} {[\frac{1}{3} r^{3} sin θ]}_{0}^{a} d θ \\ = & \frac{1}{3} a^{3} \int_{0}^{\frac{π}{2}} sin θ d θ = \frac{1}{3} a^{3} [- cos θ]_{0}^{\frac{π}{2}} = \frac{1}{3} a^{3} \\ \int_{A} x y d A & = & \int_{0}^{\frac{π}{2}} \int_{0}^{a} r^{3} cos θ sin θ d θ = \int_{0}^{\frac{π}{2}} {[\frac{1}{4} r^{4} cos θ sin θ]}_{0}^{a} d θ \\ = & \frac{1}{4} a^{4} \int_{0}^{\frac{π}{2}} sin θ cos θ d θ = \frac{1}{4} a^{4} {[\frac{1}{2} {sin}^{2} θ]}_{0}^{\frac{π}{2}} = \frac{1}{8} a^{4} \\ \int_{A} y^{2} d A & = & \int_{0}^{\frac{π}{2}} \int_{0}^{a} r^{3} {sin}^{2} θ d r d θ = \int_{0}^{\frac{π}{2}} {[\frac{1}{4} r^{4} {sin}^{2} θ]}_{0}^{a} d θ \\ = & \frac{1}{4} a^{4} \int_{0}^{\frac{π}{2}} {sin}^{2} θ d θ = \frac{1}{4} a^{4} \int_{0}^{\frac{π}{2}} \frac{1}{2} (1 - cos 2 θ) d θ = \frac{1}{8} a^{4} {[θ - \frac{1}{2} sin 2 θ]}_{0}^{\frac{π}{2}} = \frac{1}{16} π a^{4} \end{array}

Then $x_{p} = \frac{\int_{A} x y d A}{\int_{A} y d A} = \frac{\frac{1}{8} a^{4}}{\frac{1}{3} a^{3}} = \frac{3}{8} a$ and $y_{p} = \frac{\int_{A} y^{2} d A}{\int_{A} y d A} = \frac{\frac{1}{16} π a^{4}}{\frac{1}{3} a^{3}} = \frac{3}{16} π a$ .

The centre of pressure is at $(\frac{3}{8} a, \frac{3}{16} π a)$ .