적분의 활용 전체 문제

(a) First, we set up axes and a sketch for one of the ends of the cylindrical tank. It is simplest to let ground level correspond to the (horizontal) $y$-axis with the positive $x$-axis pointing straight down, so that $x = 0$ corresponds to ground level and the value of $x$ tells us the depth of a particular location. The circular end of the tank has radius $3$ and is centered at the point $(5,0)$ since its top is $2$ feet underground. Thus, the equation of that circle is $$(x-5)^2 + y^2 = 3^2\,.$$ Solving for $y$, we find the equation of one half of the circle, the half that is pictured at right in the figure below: $y = f(x) = \sqrt{9 - (x-5)^2}$. Next, for a slice at depth $x$, its shape is a rectangular slab with thickness $\triangle x$. Its length is $10$, and its width is $2 f(x) = 2 \sqrt{9 - (x-5)^2}$. Thus, the volume of such a slice is $$V_{\text{slice}} \approx 10 \cdot 2 \sqrt{9 - (x-5)^2} \cdot \triangle x\,.$$ With the water weighing $62.4$ pounds per cubic foot, the weight of (and thus force to be exerted upon) a typical slice is $$F_{\text{slice}} \approx 62.4 \cdot 20 \sqrt{9 - (x-5)^2} \triangle x\,.$$ To move such a slice from its depth $x$ feet below the ground to a tank on a truck that's $4.5$ feet above ground, each slice must travel $x + 4.5$ feet, and thus the work to move the slice is $$W_{\text{slice}} \approx (x+4.5) 1248 \sqrt{9 - (x-5)^2} \triangle x\,.$$ Finally, to empty half the tank, we need to consider slices from $x = 2$ to $x = 5$, and thus using a definite integral to sum the work required for each slice, we get $$W = \int_2^5 1248 (x+4.5) \sqrt{9 - (x-5)^2} \, dx = 1404(19\pi - 8) \approx 72573.13$$ foot-pounds. (b) To find the hydrostatic force due to water pressure that is exerted on one end of the tank by the remaining water, we first find the area of a rectangular slice at a location $x$, which is $$A_{\text{slice}} \approx 2 f(x) \triangle x = 2 \sqrt{9 - (x-5)^2} \triangle x\,.$$ We note that such a slice lies between $x = 5$ and $x = 8$, which is the values of $x$ that represent the depths of remaining water from top to bottom. At an $x$ in this interval, the slice's depth below the surface of the water is $x - 5$. Thus, since pressure is weight-density times depth, it follows that $$P_{\text{slice}} \approx 62.4 (x-5)\,.$$ Finally, since $F = P \cdot A$, the total force water is exerting on this slice is $$F_{\text{slice}} = P_{\text{slice}} \cdot A_{\text{slice}} \approx 62.4 (x-5) \cdot 2 \sqrt{9 - (x-5)^2} \triangle x\,.$$ Integrating to sum the total force, we find $$F = \int_5^8 124.8 (x-5) \sqrt{9 - (x-5)^2} \, dx \approx 1123.2$$ pounds.