Solving for Work Using Calculus

[Markovnikov]

Homework Statement

A parabolic reflector is being used as a water tank. The tank follows y = √(60) * x². There is a 12 meter pipe extending from the top of the tank. The width of the top of the tank should be 2 meters in length. Calculate the work done removing the water.

Now invert the tank, calculate the work done removing the water.

Homework Equations

Work = (Density)*∫A(y)D(y)dy

The Attempt at a Solution

Part 1.

A(y) = π*r² or also π*x² here.
If y = √(60) * x² then x² = y/√(60).
Hence, A(y) = π*(y/√(60)).

D(y) = Distance water travels = (12+√60-y)

My limits of integration are the "distance" of water or 0 to √60. Density = 1000 kg/m³

Integral: 1000π ∫(y/√(60))*(12+√60-y)dy from 0 to √60

Part 2.

A(y) = π*r² or also π*x² here.
Inverting the tank: Take the negative of the function and add back what is needed to set the intersection of the y-axis back to where it was.
If y = -√(60)*x² + √60 then x² = (y -√60)/(-√60).
Hence, A(y) = π*(y -√60)/(-√60)

D(y) = Distance water travels = (12+√60-y)

My limits of integration are the "distance" of water or 0 to √60. Density = 1000 kg/m³

Integral: 1000π ∫((y -√60)/(-√60))*(12+√60-y)dy from 0 to √60

I don't think it is necessary to solve the integrals, but for some reason that I am not seeing...I am not getting the same answer. If it is the same tank, just inverted, it should be the same amount of work, right?

Can anyone help me in seeing (what I've been trying to see for...oh...4 hours ) what I'm not seeing?

Thanks,
- Chemistry Major attempting to do Physics -

P.S. I'm new here, so if I did anything wrong, let me know and sorry!

 

[HallsofIvy]

Homework Statement

A parabolic reflector is being used as a water tank. The tank follows y = √(60) * x². There is a 12 meter pipe extending from the top of the tank. The width of the top of the tank should be 2 meters in length. Calculate the work done removing the water.

Now invert the tank, calculate the work done removing the water.

Homework Equations

Work = (Density)*∫A(y)D(y)dy

The Attempt at a Solution

Part 1.

A(y) = π*r² or also π*x² here.
If y = √(60) * x² then x² = y/√(60).
Hence, A(y) = π*(y/√(60)).

D(y) = Distance water travels = (12+√60-y)

My limits of integration are the "distance" of water or 0 to √60. Density = 1000 kg/m³

Integral: 1000π ∫(y/√(60))*(12+√60-y)dy from 0 to √60

Part 2.

A(y) = π*r² or also π*x² here.
Inverting the tank: Take the negative of the function and add back what is needed to set the intersection of the y-axis back to where it was.
If y = -√(60)*x² + √60 then x² = (y -√60)/(-√60).
Hence, A(y) = π*(y -√60)/(-√60)

D(y) = Distance water travels = (12+√60-y)

My limits of integration are the "distance" of water or 0 to √60. Density = 1000 kg/m³

Integral: 1000π ∫((y -√60)/(-√60))*(12+√60-y)dy from 0 to √60

I don't think it is necessary to solve the integrals, but for some reason that I am not seeing...I am not getting the same answer. If it is the same tank, just inverted, it should be the same amount of work, right?

NO! The inverted parabola has more of its water lower, that has to be lifted a greater height so requires more work than the upright parabola.

Can anyone help me in seeing (what I've been trying to see for...oh...4 hours ) what I'm not seeing?

Thanks,
- Chemistry Major attempting to do Physics -

P.S. I'm new here, so if I did anything wrong, let me know and sorry!

 

[Markovnikov]

I thought that the upright parabola had more water at the top, less water at the bottom. The less bit of water took more time to travel and the large amount of water took less time to travel.

Compared to the parabola upside down, the large amount of water took more time to travel and the less amount of water too less time to travel.

I thought these would sort of "cancel" each other out and give the same amount of work?

I could have swore my teacher said they would give the same amount of work - perhaps I'm worrying for nothing!

Thanks for the help!