(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 $2$.
The tank as described is $2$-dimensional, and will not hold much water.
So we assume the tank is obtained by rotating $y=x^2$ about the $y$-axis. In that case the radius of cross-section at height $y$ is $|x|$, where $y=x^2$. So the area of cross-section is $\pi x^2$, which happens to be $\pi y$.
Remark: The rest is a matter of units. The integral $\int_0^4 k \pi(11-y)y\,dy$ is right. If you are using foot-pounds, the constant is right.
Best Answer
The simplest method is to treat it as moving the total mass of the rope a height from the cg of the rope to the edge. That is, mgh where $m = 50\cdot 40/1000$ kg, $g = 9.81$ and $h = 25$ m.
If you want to go the calculus route, set it up as a series of infinitely thin discs being raised different heights:
Mass of each disc is $\pi\cdot .004^2\cdot \frac{.04}{\pi\cdot .004^2}dx = .04 \ dx$
Distance each disc travels is $x$ from $0$ to $50$.
$$W = \int_0^{50} mgx \ dx$$ $$W = \int_0^{50} .04\cdot 9.81x \ dx$$
$W = .02\cdot 9.81(x)^2$
$W = .02\cdot 9.81(50)^2$
$W = 490.5$ Joules