0
$\begingroup$

A conical tank, full of water, has a radius of 10 feet at the top and an altitude of 8 feet. Find the work done in pumping all the water out of the tank through a spout 6 feet above the top of the tank.

$$62.5\pi \int_0^8 (10-\frac54y)^2 (y+6) \,dy= (400000\pi)/3$$

This is how I set up the integral and then my answer, I think its wrong but I am unsure of how to properly set up this problem.

Please help me, any help is appreciated.

$\endgroup$
6
  • $\begingroup$ Can you explain why and how you came to this integral? $\endgroup$ Commented Mar 21, 2020 at 20:29
  • $\begingroup$ I used the formula W = F * D I used this formula because we are suppose to figure out the parts for each and then use the integral to find the work done $\endgroup$
    – wave
    Commented Mar 21, 2020 at 20:29
  • $\begingroup$ Why do you integrate from $0$ to $2$? $\endgroup$ Commented Mar 21, 2020 at 20:33
  • $\begingroup$ Oh, my bad that is suppose to be 0 to 8 $\endgroup$
    – wave
    Commented Mar 21, 2020 at 20:34
  • $\begingroup$ And the radius of the cone at the altitude $-8$ feet should be equal to zero. It is not so in your proposal. And what is the $62.5$ coefficient? $\endgroup$ Commented Mar 21, 2020 at 20:38

1 Answer 1

1
$\begingroup$

The tank is an upside down cone. Let's choose $y=0$ to be at the vertex. At $y=8$ the radius of the horizontal cross section is $10$, so at any intermediate $y$ the radius is $$r=y\frac{10}8$$ The volume of a disk full of water at height $y$ and thickness $dy$ is then $$dV=\pi r^2(y) dy=\pi\frac {25}{16}y^2dy$$ The work done to move this water is $$dW=\rho g dV\cdot(6+(8-y))$$ Here $\rho g$ is the weight density. You need to move the water to the top of the tank $(8-y)$ feet, then extra $6$ feet to the spout. So $$W=\int_0^8dW=62.5\pi\frac{25}{16}\int_0^8y^2(14-y)dy$$

Note: I think yours is similar, except that you chose $y=0$ at the top of the tank, and $y$ increasing towards the bottom.

$\endgroup$
1
  • $\begingroup$ Note that I've had a small mistake, where I've put 10 instead of 8. The numberical answer is same as yours $\endgroup$
    – Andrei
    Commented Mar 21, 2020 at 21:05

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .