Pumping water out of a truncated cone using integration. So the issue I'm stuck with, is that I can do a cone, but I have no idea where to start with a cone that is truncated.
I have a truncated cone that has a base with radius of 3 meters and a top with a radius of 4 meters It's height is 4 meters and I want to pump the water out of a pipe that expands 1 meter above the truncated cone. I need to calculate the work required to pump out all the water.
I did a cylinder earlier and a cone, but I just can't seem to figure out this truncated cone.
Any advice on where to start with a problem like this?
 A: The tank is a truncated circular cone with a base of radius $r=3\,$m, a depth of $4\,$m and top radius of $r=4\,$m. Let $y$ denote the vertical distance from the bottom of the tank. Then $r=3+\frac{1}{4}y$ is the radius of the slice of water lying $y$ meters above the bottom of the tank.
    
Think of that slice of water as being solid rather than liquid, as if it were frozen. The amount of work done in pumping the water to a height of $5\,$m above the base is the same as if one had to move all those frozen slices to that height.
The volume of each slice is $\pi r^2\,dy$ with $dy$ representing the thickness. The mass of the slice (in the mks system) is found by multiplying the density $\rho=1000\,$ kg/m$^3$ times the volume, so 
\begin{equation}
M=1000\pi r^2\,dy=1000\pi\left(3+\frac{1}{4}y\right)^2dy
\end{equation}
Each of these slices of water must be moved upward a distance of $D=5-y$ meters against a gravitational force of $g=9.8\,$m/sec$^2$ resulting in the work for the slice at $y$ being
\begin{equation}
W_y=9800\pi\left(3+\frac{1}{4}y\right)^2(5-y)\,dy
\end{equation}
The total work to remove all the slices of water between $y=0$ and $y=4$ is
\begin{equation}
W=\int_{0}^{4}9800\pi\left(3+\frac{1}{4}y\right)^2(5-y)\,dy
\end{equation}
The rest is routine since the integrand is simply a third degree polynomial. The units will be joules.
A: In this case, you need to realise that one can obtain a cone by simply revolving/rotating a linear curve about the x-axis (or y-axis). You can then use the following integral. In general, the volume given by rotating the function $f(x)$ over an interval $[a,b]$ is given by:
$V(x)=\pi \int_a^b f(x)^2 \ dx$.
Combined with having a look at the following picture, I am sure you can calculate the volume of your truncated cone. 
(from: http://www.nabla.hr/DIASurfXFig.gif)

A: Suppose you had a tank in the shape of an inverted cone with a height of $16$ and radius $4$ on the top surface. Suppose you had to pump water out of that tank, but stop pumping when the depth of water is $12.$
In this modified problem, you will have pumped the water out of a truncated cone of height $4$ with top radius $4$ and bottom radius $3.$ This is just like the problem you were given, except that when you stop pumping, the surface is made of water rather than the solid material of the tank.
Can you solve the modified problem?
