A tank in the shape of an inverted right circular cone has height 12 meters and radius 11 meters. It is filled with 6 meters of hot chocolate. Find the work required to empty the tank by pumping the hot chocolate over the top of the tank. Note: the density of hot chocolate is $\delta = 1500 kg/m^3$

I have tried several different things but it's obvious I am overlooking something important.

This is my latest logic (method):

Volume of each slice: $V = \frac{1}{3}\pi r^2 h = \frac{1}{3}\pi (\frac{11y}{12})^2 dy = \frac{121\pi y^2}{432}dy $

Mass of each slice: $ M = \delta V = \frac{1500kg}{3m^3} *\frac{121\pi y^2}{432}dy = \frac{15125\pi y^2}{36}dy$

Force of each slice: $ F = Mg = 9.8*M = 9.8\frac{15125\pi y^2}{36}dy = \frac{148225\pi y^2}{36}dy$

$ \int_{6}^{12} \frac{148225\pi y^2}{36}(12-y) dy = 4891425\pi $

I've tried all sorts of different stuff. This was my latest method.


$$\text{Force = area} \cdot \text{density} \cdot \text{depth}$$

Think of the top of the cone as $y=0$ and the bottom as $y=12$. The vertex of the cone is $(0,12)$ and the point at the far right edge of the top of the cone is $(11,0)$. Using these points, you can find the equation of the line that is the edge of the cone \begin{align*} m &= -\frac{12}{11},\\ y &= -\frac{12}{11}x + 12. \end{align*} Convert it to $x$ because we will be integrating with respect to $y$ $$x = -\frac{11}{12y} + 11.$$ This is also the radius, $r$.

We will integrate using cylinders of face area $\pi r^2$ and depth $dy$. Density is $1500$, as given.

Because work is $\text{force}\cdot \text{distance}$, we must take into account the distance the cocoa must be lifted. Since we pump to the top of the container, $y=0$, this is simply our current depth, $y$.

We will take the integral, but only for the levels that we actually have cocoa to pump at.

Therefore work equals the integral $$\int_6^{12} \pi\left(-\frac{11}{12}y + 12\right)^2 1500 y\, dy$$

This comes out to $712748.833$


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