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? $$ \operatorname{Work} = 62.4 \int_0^4 (11 - y) y \,dy $$

Please help me!


1 Answer 1


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.

  • $\begingroup$ So the integral's bounds are x values and not y values? $\endgroup$
    – user97462
    Commented Dec 6, 2013 at 20:39
  • $\begingroup$ Oops, they are the $y$ values! $\endgroup$ Commented Dec 6, 2013 at 20:42
  • $\begingroup$ Also, shouldn't there be a pi in front of the k? Thanks. $\endgroup$
    – user97462
    Commented Dec 6, 2013 at 20:43
  • $\begingroup$ Yes, it doesn't matter, it is absorbed into the $k$. But to prevent confusion, I will put a $\pi$ in. $\endgroup$ Commented Dec 6, 2013 at 20:45

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