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Work/Parametric Equations

  1. Mar 10, 2010 #1
    1. The problem statement, all variables and given/known data
    A dish, described by the equation x2+z2=ey for 0 < y < 10cm (or equal too) is filled 5cm high with milk. How much work does it take to spill all the milk on the floor? Take milk density to be 1030 Kg/m3 and gravitational acceleration to be 9.8 m/s2.

    2. Relevant equations
    I know that the work formula is
    W= int(0 to h): F(y)dy+(T-h)(F(h)), and F(y0)=(dens)(g.acc)(Volume above y0), but I am stuck at the volume part of the equation.
    It should be V(y)=int(0 to y): A(y), where the area is Pi(r)2. This is where I get stuck. How do I convert the x2+z2=ey into a radius function? So far, nobody's been able to explain parametrics to me very clearly even with just x and y, but now there's a z variable and I'm just clueless.

    3. The attempt at a solution
    I got y=ln(x2+z2) and x=sqrt(ey+z2) so far..
  2. jcsd
  3. Mar 10, 2010 #2


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    Isn't x^2+z^2=e^y a circle in the x-z plane with radius e^y?
  4. Mar 10, 2010 #3
    so then is it just the integral of pi(ey)2?

    In which case V(y)= Pi((1/2)(e2y)) from 0-5, and a constant Pi((1/2)(e2(5))) from 5-10cm?
  5. Mar 10, 2010 #4


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    Yes, the volume is integral of A(y)=pi*(e^y)^2. I'm having a some trouble following the rest the work calculation. T is 10cm, right? So isn't the formula just integral 0 to 5 (g.acc)*(density)*(T-y)*A(y)dy?
  6. Mar 10, 2010 #5
    Hm.. I'm not entirely sure, I just posted the formula straight off my formula sheet. The first part is the work of moving the varying amount of fluid, and the second just moving it out through the open space in the dish. The force is constant for the (T-h) (10-5cm), but we need to take a second integral for the force of moving the volume as it changes during the emptying. Thats what I understand from my notes, at least. If it helps I just finished the question and got this, basically:

    W = (integral(0-.05): 10094pi/2(e2y)) + (.1-.05)(10094pi/2)(e2(.05))

    Where the bold is the force to move the milk up until its surface in the dish and the regular face is just the force to move the then-constant volume from the 5cm up to the 10cm. I could be wrong though...
  7. Mar 10, 2010 #6


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    Ah, I see. The first part moves all the milk up to the 5cm point and the other part lifts it out. Ok. But your formulas still look pretty garbled. Partially because you haven't defined what F(y) is supposed to be or anything. I am thoroughly confused. Work is the integral of (g.acc)*(dens)*A(y)*(distance to lift from y)*dy. Try and formulate it from that. Your notes may be a little garbled.
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