# A tank is part of a cone with a 10 foot radius on top, 4 foot radius on bottom 12 feet below the top Water in the tank has depth 5 feet

A tank is part of a cone with a 10 foot radius on top and a 4 foot radius on bottom, 12 feet below the top. Water in the tank has depth 5 feet. Provide an integral for the work done pumping the water out the top of the tank.

UNITS CAN BE KEPT THE SAME. Not sure if this is supposed to use the surface area of a bound surface formula or a different method.

Each little bit of water has to be lifted up to the edge. The water at the bottom has to be lifted $12$ feet while the water at the top only has to be lifted $7$ feet. If we let the $z$ axis be vertical with $0$ at the bottom of the tank a slice of water that starts at position $z$ of thickness $dz$ has mass $A(z)\rho dz$ where $A(z)$ is the area of the tank in the horizontal direction at position $z$ and $\rho$ is the density of water.
• You need to use the dimensions of the cone to get $A(z)$. $A(0)=\pi 4^2$ because the radius is $4$ feet at the bottom. – Ross Millikan Mar 9 '18 at 0:13