6.5 Physical applications  (Page 6/11)

1. We begin by establishing a frame of reference. As usual, we choose to orient the $x\text{-axis}$ vertically, with the downward direction being positive. This time, however, we are going to let $x=0$ represent the top of the dam, rather than the surface of the water. When the reservoir is full, the surface of the water is $10$ ft below the top of the dam, so $s(x)=x-10$ (see the following figure).
   

   

   
   To find the width function, we again turn to similar triangles as shown in the figure below.
   

   

   
   From the figure, we see that $w(x)=750+2r.$ Using properties of similar triangles, we get $r=250-\left(1\text{/}3\right)x.$ Thus,
   
   $w(x)=1250-\frac{2}{3}x\text{(step 2).}$
   
   Using a weight-density of $62.4$ lb/ft ³ (step 3) and applying [link] , we get
   
   $\begin{array}{cc}\hfill F& ={\displaystyle {\int }_a^b\rho }w(x)s(x)dx\hfill \\ & ={\displaystyle {\int }_{10}^{540}62.4\left(1250-\frac{2}{3}x\right)\left(x-10\right)dx}=62.4{\displaystyle {\int }_{10}^{540}-\frac{2}{3}\left[x^2-1885x+18750\right]}dx\hfill \\ & =\mathrm{-62.4}\left(\frac{2}{3}\right){\left[\frac{x^3}{3}-\frac{1885x^2}{2}+18750x\right]|}_{10}^{540}\approx \mathrm{8,832,245,000}\text{lb}=\mathrm{4,416,122.5}\text{t}\text{.}\hfill \end{array}$

Note the change from pounds to tons $(2000$ lb = $1$ ton) (step 4). This changes our depth function, $s(x),$ and our limits of integration. We have $s(x)=x-135.$ The lower limit of integration is $135.$ The upper limit remains $540.$ Evaluating the integral, we get