# Find the work done in pumping all the water out of the tank through a spout 6 feet above the top of the tank.

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.

• Can you explain why and how you came to this integral? Commented Mar 21, 2020 at 20:29
• 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
– wave
Commented Mar 21, 2020 at 20:29
• Why do you integrate from $0$ to $2$? Commented Mar 21, 2020 at 20:33
• Oh, my bad that is suppose to be 0 to 8
– wave
Commented Mar 21, 2020 at 20:34
• 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? Commented Mar 21, 2020 at 20:38

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.