# Pumping Water out of Parabolic tank?

First of all, I understand how to do the integration part of this problem, but I am confused about the setup.

Here is the question:

Use integration to find the work done pumping all the water out of the parabolic tank represented by $y = x^2$ for $x \in [0, 2]$, to a point $7$ft above the top of the tank.

If I am not mistaken, the height should be $h(x) = 11 - y$. As the top of the tank is $4$ feet tall from the equation $y = x^2$. Correct?

Now this is where I get confused. Should the area of a slice be $A(x) = \pi y$?

Or should it be $A(x) = \pi y^2$?

If my first answer is correct, would the total work be represented by this expression? $$\operatorname{Work} = 62.4 \int_0^4 (11 - y) y \,dy$$

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.
• Oops, they are the $y$ values! Commented Dec 6, 2013 at 20:42
• Yes, it doesn't matter, it is absorbed into the $k$. But to prevent confusion, I will put a $\pi$ in. Commented Dec 6, 2013 at 20:45