The ends of a "parabolic" water tank are the shape of the region inside the graph of $y = x^2$ for $0 \leq y \leq 9$; the cross sections parallel to the top of the tank (and the ground) are rectangles.

At its center the tank is $9$ feet deep and $6$ feet across. The tank is $9$ feet long.

Rain has filled the tank and water is removed by pumping it up to a spout that is $2$ feet above the top of the tank.

Set up a definite integral to find the work $W$ that is done to lower the water to a depth of 4 feet and then find the work.

My integral:

$$W = \int_4^9 62.5\times9\times2\times\sqrt{9-y^2}\times9\times(11 - y) \ dy $$

I did a cross-sectional slice of rectangles so $A(x) = L\times W\times H$

Length and height were givens, I found width using triangles and using the Pythagorean theorem.

Why is this wrong. I've seen that it's suppose to be: [ 62.5] [ 18 ] [ 11 √y - y^(3/2) ] dy over y in [ 4 , 9 ]

But not sure why.

Please help Thank you

  • $\begingroup$ Draw a picture of the tank. Where are there triangles? $\endgroup$ Commented May 10, 2017 at 15:49
  • $\begingroup$ The question is a bit ambiguous if you are not familiar with "parabolic" tanks. It is shaped like a trough every vertical left to right cross section is the same parabolic shape. From front to back the tank does not taper or modify at all. I think you are picturing it as bowl shape. $\endgroup$
    – fleablood
    Commented May 10, 2017 at 16:13
  • $\begingroup$ @MatthewLeingang can't we use Pythagorean on, spherical and hemispherical tanks though? $\endgroup$
    – yre
    Commented May 10, 2017 at 17:22

1 Answer 1


The height of the water, $y$ (not the height of the tank) varies as water is pumped out of the tank.

The width of the tank for any $y$ is $2\sqrt y$

This comes from $y=x^2$ and considering the cross section of that parabola.

The length of tank is 9 feet.

The area of each cross section is $18\sqrt y$

The spout is 2 feet above the top of the tank (or 11 feet above the bottom).

$\rho$ is the density of water.

And we only need to pump the water down to the 4 foot mark.

$\rho\int_4^9 (11-y)(18\sqrt y)\ dy$

  • $\begingroup$ Is this because we are taking cross sections of circles, so pi*r^2? and here radius is sqrt(y) = x, so then pi*(sqrt(y))^2 = x or rather pi*(sqrt(y)) = x. But I see now pi in your answer. What are we taking cross sections of? $\endgroup$
    – yre
    Commented May 11, 2017 at 13:48
  • $\begingroup$ Or maybe we take cross sections of rectangles? So A = LWH, hence A = 9*(2(sqrt(y))*dy? Is this right? $\endgroup$
    – yre
    Commented May 11, 2017 at 13:50

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