Find a formula for the work required to empty a tank of water which is a solid of revolution about a vertical axis of symmetry.
Shell method: $2\pi\int_a^b xf(x)dx$
Work and force: $W=\int_a^b Fdx$
I know that to empty the tank I need to do it slab by slab from the bottom of the solid to the top because the work needed to take out a slab from the highest part of the solid is greater than one of the bottom, that's why it is not enough to get volume of the solid and multiply it by the heigth of itself.
In a previos example from the book(Calculus II, Marsden) a cone which also contains water in it, it was a lot easier to measure each slab and integrate all of them to get work.
The solution, which makes me even more confused, says:
$\rho\pi g\int_o^ax^2[ f(a) - f(x)]f'(x) dx$; the region is that under the graph $y = f(x), 0 < x < a$, revolved about the y-axis.
I don´t see how to get to that solution or even have another kind of solution. I´m very stuck here and I´d appreciate any kind of help. Thank you.
With the disk method, the volume of the slab at $x$ is $\pi x^2dy$ and work to lift it over the top of the tank is
$$dw=\rho \pi g x^2h(x)dy $$
where the height of the slab at $x$ is $h(x)=f(a)-f(x)$. Then, with $dy=f’(x)dx$, integrate to get the total work as
$$W=\rho\pi g\int_0^ax^2[ f(a) - f(x)]f'(x) dx$$