(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 $2$.
Basically, the hard part of this question would be finding the surface area of the trough at a certain depth of the water. So let's say the height of the water is $y$ from the bottom, while the depth of the water is $1-y$. Now at whatever height $y$ you are at, the length will always be $3$. The width, however varies between $0$ and $2$. Notice that the equation of the parabola is $y=x^2$. If you were to put a vertical axis in between the parabola, you would see that the width of the trough at that height is twice the distance from the vertical axis to either side of the parabola. The distance between the vertical axis to the parabola is $x=\sqrt{y}$, so the total width of the parabola at height $y$ is $2\sqrt{y}$. That means at height $y$, the top surface area is $2\sqrt{y}*3 = 6\sqrt{y}$.
Now all that is left is your integral. You need to do an integral from 0 to 1 of (the weight of water)(the depth of the water)(the area of the water at the given depth)(d-height of water) =
$\int_0^1 (62)(1-y)(6\sqrt{y})dy = 99.2$ foot-pounds.
Best Answer
The work done to pump out liquid from the top outlet is different for the liquid at different depths.
Volume of infinitely thin layer of liquid ($dx$) at depth $x$ from top is,
$dV = \pi \cdot 4^2 \cdot dx = 16 \pi ~ dx ~ $ and this liquid needs to be taken $x$ distance against gravity.
So work done in pumping out the liquid in unit ft-lb is,
$ \displaystyle \int_4^8 \rho~ x ~ dV $
where $\rho = 12 $ lb / ft$^3$