A tank, shaped like a cone has a height of $15$ meters and a base radius of $6$ meters. The cone is standing up, with the circular part as its base. It is full of water, and we want to pump all of it out through a pipe that is at the tip of the cone. Assume that the density of water is $\rho = 1000~\text{kg}/\text{m}^3$ and the acceleration due to gravity is $g = 9.8~\text{m}/\text{s}^2$ .
Set up the integral of the work done to pump all the water out of the tank. DO NOT EVALUATE THE INTEGRAL. [Note: You can use ρ and g in your answer; you do not have to use 1000 and 9.8.]
My attempt
$$\int _{ 0 }^{ 15 }{ (1000)(9.8)\left(\frac { 4\pi y^ 2 }{ 25 }\right)(15-y)dy }$$
but the answer was $$\int _{ 0 }^{ 15 }{ (1000)(9.8)\left[\frac { 2(15-y) }{ 5 }\right]^ 2(15-y)dy }$$
Can anyone please explain this?
It looks like the $A(y)$ seems to be the problem
The tank is a cone, so the radius of the cross section increases linearly with $h$
At $y=0$, $r=6$ and at $y=15$, $r=0$
So
$$r(y)=\frac{6}{15}(15-y)$$
$$A(y)=\pi \left[\frac{6(15-y)}{15}\right]^2=\pi \left[\frac{2(15-y)}{5}\right]^2$$