Question

A conical tank that is 5 meters high has a radius of 2 meters, and is filled with a liquid that weights 800 kg per cubic meter. How much work is done in discharging all the liquid at point 3 meters above the top of the tank?

Answer 1

Let we have an inverted cone.

Put the origin of the coordinate system at the center of the top of the tank. Have coordinates measure depth (so y = 0 at the center of the top of the tank, y = 5 at the bottom of the tank).

From similar triangles

$${x \over 5-y}={2 \over 5}=>x=2-{2 \over 5}y$$

Segment the liquid into a stack of "disks", each with thickness $dy$. One such "disk" at height $y$ from the bottom of the tank has Volume

$$dV=\pi x^2dy=\pi(2-{2 \over 5}y)^2 dy$$

The mass of this "disk" is

$$dm=\rho dV=800\pi(2-{2 \over 5}y)^2dy=\pi (3200-1280y+128y^2)dy$$

The vertical distance this disk must travel is $(y+3)$

The weight (force due to gravity) of the disk is $dm\cdot g,$ where $g$ is the gravitational constant 

$g=9.81\ m/s^2.$

$$Work=\pi\displaystyle\int_{0}^59.81\cdot(3200-1280y+128y^2)(y+3)dy=$$

$$=1255.68\pi\big[{y^4 \over 4}-{7y^3 \over 3}-{5y^2 \over 2}+75y\big]\begin{matrix} 5 \\ 0 \end{matrix}=44472\pi\approx139713(J)$$