Okay, in this case we cannot just determine a force function, \(F\left( x \right)\) that will work for us. So, we are going to need to approach this from a different standpoint.

Let’s first set \(x = 0\) to be the lower end of the tank/cone and \(x = 15\) to be the top of the tank/cone. With this *definition* of our \(x\)’s we can now see that the water in the tank will correspond to the interval \(\left[ {0,12} \right]\).

So, let’s start off by dividing \(\left[ {0,12} \right]\) into \(n\) subintervals each of width \(\Delta x\) and let’s also let \(x_i^*\) be any point from the \(i\)^{th} subinterval where \(i = 1,2, \ldots n\). Now, for each subinterval we will approximate the water in the tank corresponding to that interval as a cylinder of radius \({r_i}\) and height \(\Delta x\).

Here is a quick sketch of the tank. Note that the sketch really isn’t to scale and we are looking at the tank from directly in front so we can see all the various quantities that we need to work with.

The red strip in the sketch represents the “cylinder” of water in the \(i\)^{th} subinterval. A quick application of similar triangles will allow us to relate \({r_i}\) to \(x_i^*\) (which we’ll need in a bit) as follows.

\[\frac{{{r_i}}}{{x_i^*}} = \frac{4}{{15}}\hspace{0.25in} \Rightarrow \hspace{0.25in}{r_i} = \frac{4}{{15}}x_i^*\]

Okay, the mass, \({m_i}\), of the volume of water, \({V_i}\), for the \(i\)^{th} subinterval is simply,

\[{m_i} = {\mbox{density }} \times \,\,{V_i}\]

We know the density of the water (it was given in the problem statement) and because we are approximating the water in the \(i\)^{th} subinterval as a cylinder we can easily approximate the volume as,

\[{V_i} \approx \pi {\left( {{\mbox{radius}}} \right)^2}\left( {{\mbox{height}}} \right)\]

We can now approximate the mass of water in the \(i\)^{th} subinterval,

\[{m_i} \approx \left( {1000} \right)\left[ {\pi r_i^2\Delta x} \right] = 1000\pi {\left( {\frac{4}{{15}}x_i^*} \right)^2}\Delta x = \frac{{640}}{9}\pi {\left( {x_i^*} \right)^2}\Delta x\]

To raise this volume of water we need to overcome the force of gravity that is acting on the volume and that is, \(F = {m_i}g\), where \(g = 9.8\,m/s^{2}\) is the gravitational acceleration. The force to raise the volume of water in the \(i\)^{th} subinterval is then approximately,

\[{F_i} = {m_i}g \approx \left( {9.8} \right)\frac{{640}}{9}\pi {\left( {x_i^*} \right)^2}\Delta x\]

Next, in order to reach to the top of the tank the water in the \(i\)^{th} subinterval will need to travel approximately \(15 - x_i^*\) to reach the top of the tank. Because the volume of the water in the \(i\)^{th} subinterval is constant the force needed to raise the water through any distance is also a constant force.

Therefore, the work to move the volume of water in the \(i\)^{th} subinterval to the top of the tank, *i.e.* raise it a distance of \(15 - x_i^*\), is then approximately,

\[{W_i} \approx {F_i}\left( {15 - x_i^*} \right) = \left( {9.8} \right)\frac{{640}}{9}\pi {\left( {x_i^*} \right)^2}\left( {15 - x_i^*} \right)\Delta x\]

The total amount of work required to raise all the water to the top of the tank is then approximately the sum of each of the \({W_i}\) for \(i = 1,2, \ldots n\). Or,

\[W \approx \sum\limits_{i = 1}^n {\left( {9.8} \right)\frac{{640}}{9}\pi {{\left( {x_i^*} \right)}^2}\left( {15 - x_i^*} \right)\Delta x} \]

To get the actual amount of work we simply need to take \(n \to \infty \). *I.e.* compute the following limit,

\[W = \mathop {\lim }\limits_{n \to \infty } \sum\limits_{i = 1}^n {\left( {9.8} \right)\frac{{640}}{9}\pi {{\left( {x_i^*} \right)}^2}\left( {15 - x_i^*} \right)\Delta x} \]

This limit of a summation should look somewhat familiar to you. It’s probably been some time, but recalling the definition of the definite integral we can see that this is nothing more than the following definite integral,

\[\begin{align*}W &= \int_{0}^{{12}}{{\left( {9.8} \right)\frac{{640}}{9}\pi {x^2}\left( {15 - x} \right)\,dx}} = \left( {9.8} \right)\frac{{640}}{9}\pi \int_{0}^{{12}}{{15{x^2} - {x^3}\,dx}}\\ & = \left. {\left( {9.8} \right)\frac{{640}}{9}\pi \left( {5{x^3} - \frac{1}{4}{x^4}} \right)} \right|_0^{12} = 7,566,362.543\,{\mbox{J}}\end{align*}\]