There is a cylindrical tank lying on its side filled completely with liquid. The diameter of the tank is 4ft, and the length of the tank is 10 ft.
Here is a terrible drawing of it but close enough:
I have to set up an integral to find the work required to pump all the liquid to a level 1 ft above the top of the tank.
So I have my formula:
Work = π (weight density) ∫ (area)(displacement)(thickness)
Someone else did this problem and obtained a bounds of 2 to -2, a displacement of (3-y), an area of 2x, and a thickness of dy. Where did the 2 to -2, (3-y), and 2x come from? This is probably intuitive but I don't see it.
Best Answer
(Where did $\pi$ come from?)
Look at the tank straight on, so that you only see the circular side. Imagine that circle on the plane, with the center at $(0,0)$. Since the circle has diameter $4$, the bottom of the circle is at $(0,-2)$, and the top of the circle is at $(0,2)$. You want to lift the liquid to one foot above the top of the tank, so you are trying to lift it to the line $y=3$.
Now, take a horizontal slice of that circle of thickness $\Delta y$, at height $y_i$, where $y_i$ is somewhere between $-2$ and $2$. This slice corresponds to a volume of liquid that is approximately equal to the area of this slice, times $10$ (to account for the length of the tank).
To find the area, note that the circle has equation $x^2+y^2 = 4$, so that the $x$ coordinate is given by $x=\pm\sqrt{4-y^2}$. So the width of the slice is going to be $2x=2\sqrt{4-(y_i)^2}$. The thickness is $\Delta y$. So the volume of the slab of liquid is approximately equal to $$20\sqrt{4 - y_i^2}\Delta y\text{ cubic feet.}$$ Now, you want to lift it all the way to $y=3$; how far from $y=3$ are you? Since the slab is at height $y_i$, the distance to $y=3$ is $3-y_i$.
So the work done in lifting this one slice of liquid is approximately: $$\begin{align*} \text{Work}&=\text{Weight}\times\text{Displacement}\\ &= \left(\text{Volume}\times\text{Density}\right)\times\text{Displacement}\\ &\approx \left(20\sqrt{4-y_i^2}\Delta y\right)\times\text{Density}\times(3-y_i)\\ &= \text{Density}\times 20(3-y_i)\sqrt{4-y_i^2}\Delta y. \end{align*}$$ (Be sure to get the units right; the density should be in pounds per cubic feet, in which case the units of work will be ft-lbs. )
Now, to get the total work, you slice up the tank into these thin slices, figure out the work for each, and add it all up: $$\text{Work}\approx \sum_{i=1}^n \text{Density}\times 20(3-y_i)\sqrt{4-y_i^2}\Delta y.$$ Taking the limit as $n\to\infty$, this becomes an integral. The integrand is $$\text{Density}\times 20(3-y)\sqrt{4-y^2}\,dy.$$ (The $\sqrt{4-y^2}$ is the value of $x$ at the slice).
What are the limits of integration? Going back to the picture, what are the possible values of $y$? Because $y$ denotes where you are in the tank, and the tank extends from $y=-2$ to $y=2$, then the limits of integration go from $-2$ to $2$.