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.


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 $= \int_0^2 (4-h)(62.4\pi (3 + h)^2) dh$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.