0
$\begingroup$

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

$\endgroup$
  • $\begingroup$ Draw a picture of the tank. Where are there triangles? $\endgroup$ – Matthew Leingang May 10 '17 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 May 10 '17 at 16:13
  • $\begingroup$ @MatthewLeingang can't we use Pythagorean on, spherical and hemispherical tanks though? $\endgroup$ – yre May 10 '17 at 17:22
1
$\begingroup$

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.

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

$\endgroup$
  • $\begingroup$ 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 May 11 '17 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 May 11 '17 at 13:50

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.