Here is the problem:
A tank in the shape of two foot high frustrum of a cone has base radius of three feet, and a radius at the top of the five feet, is filled with water which weighs $62.4$ pounds per cubic feet. How much work is required to pump all of the water to a height of two feet above the frustrum?
The first thing I did was analyze the problem:
We need to consider the water to be subdivided into disk of thickness $∧y$ and radius $x$. Because the increment of each disk is given by its weight we have :
$$∧F = weight = 62.4~lbs-ft^3*(Volume)$$
we can tell that the frustrum is made by cutting the top end of a cone therefore we have that the volume of the frustrum is made of the top and the botton, for the bottom we have 1/3PIR^2H for the top of the frustrum we have
$1/3\pi r^2h$ therefore the volume of the frustrum is $1/3\pi R^2H – 1/3\pi r^2h $.
so we have that $∧F = 62.4 lbs-ft^3*(1/3\pi R^2H – 1/3\pi r^2h)$.
This is where I got lost,
work = force*Distance
and the integral is from a to b of F(x)dx
have I done everything correct till now?
I would really appreciate a feedback on how can solve this problem. and thank you in advance.
Best Answer
the radius of the tank as a function of h. $r = 3 + h$
The volume of each disk $\pi r^2 \, dh = \pi (3 + h)^2 \, dh$
The weight of each disk equals the volume times the density $= 62.4\pi (3 + h)^2\,dh$
and the distance that that water must be lifted. 2 feet above the top of the tank, 4 feet above the base of the tank, $d = (4 - h)$
I think we have enough to set up the integral
work $= \int_0^2 (4-h) 62.4\pi (3 + h)^2 dh = \frac{278\pi}{3}\rho$ with $\rho=62.4$.