(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

A parabolic reflector is being used as a water tank. The tank follows y = √(60) * x^{2}. 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.

2. Relevant equations

Work = (Density)*∫A(y)D(y)dy

3. The attempt at a solution

Part 1.

A(y) = π*r^{2}or also π*x^{2}here.

If y = √(60) * x^{2}then x^{2}= 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/m^{3}

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

Part 2.

A(y) = π*r^{2}or also π*x^{2}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)*x^{2}+ √60 then x^{2}= (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/m^{3}

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 :rofl:) 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! :tongue2:

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# Solving for Work Using Calculus

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