calculus
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.
The tank as described is $2$-dimensional, and will not hold much water.
So we assume the tank is obtained by rotating $y=x^2$ about the $y$-axis. In that case the radius of cross-section at height $y$ is $|x|$, where $y=x^2$. So the area of cross-section is $\pi x^2$, which happens to be $\pi y$.
Remark: The rest is a matter of units. The integral $\int_0^4 k \pi(11-y)y\,dy$ is right. If you are using foot-pounds, the constant is right.
I cannot get the specified answer also. My way of interpret it is:-
(1) Set up a representative disc as shown. It should be clear that its volume is $πx^2(dy)$.
(2) At that instant, $x = (3/8)y$.
(3) To empty all the water out, integrate $\int_0^8 πx^2(dy)$.
(4) Work done = 62.4 times the result of (3) and I got 4705.
Note:- If work done is equal to rate times amount of water, then it is equal to $62.4 [(1/3) π 3^2 (8)]$ directly.
Best Answer
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.