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Volume of a cylindrical wedge

  1. Apr 15, 2012 #1

    RK7

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    1. The problem statement, all variables and given/known data

    Find the volume of the enclosed by the surfaces [tex]z=qx[/tex] [tex]z=0[/tex] and [tex]x²+y²=2ax[/tex]

    2. Relevant equations

    This is meant to be done with calculus but can verify my answer with simple geometry - should be [tex]\pi a^3q[/tex]

    3. The attempt at a solution
    So the top of the wedge will be when [tex]x=2a[/tex]

    Form rectangular slices of the wedge perpendicular to the x axis with area [tex]A=2yz=2\sqrt{x(2a-x)}qx[/tex] and volume [tex]2\sqrt{x(2a-x)}qx .dx[/tex]

    Then integrate this from x=0 to x=2a gives:
    V=[itex]\int ^{2a} _{0} 2q \sqrt{x^{3}(2a-x)} dx[/itex]


    I've checked this numerically with wolfram alpha for certain values of a and q but I haven't got a clue how to evaluate it.. the question said to use a double integral but I don't know what a suitable double integral would be...
     
  2. jcsd
  3. Apr 15, 2012 #2

    RK7

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    Hm I've come up with a double integral:

    Break it up into segments of the cirlce in the plane perpendicular to z axis

    The area of a segment at height z is [tex]\int^{2a} _{z/q} 2y dx = \int^{2a} _{z/q} 2 \sqrt{x(2a-x)} dx[/tex] and can integrate these from z=0 to 2aq giving:
    [tex]\int^{2aq} _{0} \int^{2a} _{z/q} 2 \sqrt{x(2a-x)} dx dz[/tex]
     
  4. Apr 15, 2012 #3

    RK7

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    Similar problem with evaluating the integral though..
     
  5. Apr 15, 2012 #4

    sharks

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    Are you sure the answer isn't [tex]\frac {\pi a^3q}{2}?[/tex]
     
  6. Apr 15, 2012 #5

    RK7

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    So the cylinder has radius a and touches x=0 and x=2a
    The cylinder goes up to a height of z=qx=2aq
    So the total volume of the cylinder is [tex]\pi a^{2} * 2aq = 2 \pi a^{3} q[/tex]
    So the volume of the half of the cylinder we're interested in is half that [tex]V=\pi a^3 q[/tex]

    Am I doing something wrong?
     
  7. Apr 15, 2012 #6

    sharks

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    What you just calculated in #5 is the volume of a normal vertically-halved cylinder of height z=2aq (upper surface being the horizontal plane z=2aq), when you should be calculating a cylindrical wedge.

    Let q=0. What does this tell you about the volume?
     
  8. Apr 15, 2012 #7

    RK7

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    What's wrong with the answer [tex] \pi a^3 q [/tex] ?

    And q=0 is just zero volume..
     

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  9. Apr 15, 2012 #8

    sharks

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    You need to divide your answer by 2 again. Why?

    1. You are dealing with half of a cylinder (vertically split in half).
    2. You are dealing with a wedge in the cylinder (diagonally split in half).
     
  10. Apr 15, 2012 #9

    RK7

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    I'm dealing with an entire cylinder split in two. The equation is x²+y²=2ax (equivalent ton (x-a)²+y²=a²) not x²+y²=a².
     
  11. Apr 15, 2012 #10

    sharks

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    That's what i meant. Enter all the equations in a 3D software to see the volume required.
     
  12. Apr 15, 2012 #11

    RK7

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    Can you upload a screenshot of what your software is showing?
     
  13. Apr 16, 2012 #12

    sharks

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    I have drawn it on paper as you would in the exams. I suggested using a graphing software to help you to visualize it in 3D.

    Here is the double integral (using polar coordinates):
    [tex]\int^{\theta = \frac{\pi}{2}}_{\theta = 0} \int^{r=2a\cos \theta}_{r=0} \left[ z \right]^{qr\cos \theta}_0\,.rdrd\theta[/tex]
     
    Last edited: Apr 16, 2012
  14. Apr 27, 2012 #13

    RK7

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    With respect, I'm fairly certain you're wrong and completely derailed the thread.
     
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