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Finding volume of a structure

  1. Mar 11, 2013 #1
    1. The problem statement, all variables and given/known data
    (see attachment)

    2. Relevant equations

    3. The attempt at a solution
    How the equation ##x^2+y^2-x=0## is that of a cylinder? That looks to me an equation of a circle. I haven't learnt about the different 3-D structures so I tried solving this using graphs. I tried to find the vertex of the paraboloid, it came out to be (0,0,1). Is this right? I am not sure how to proceed further from here.

    Any help is appreciated. Thanks!

    Attached Files:

  2. jcsd
  3. Mar 11, 2013 #2
    You get the cylinder by moving the circle along the z-axis. You can move the circle because z is not fixed by equation.

    You should definitely sketch the problem, in order to figure out the integration domain and the integrand.
  4. Mar 11, 2013 #3


    Staff: Mentor

    for the cylinder the axis is along the z-axis if the center is at x=0 y=0 then x^2 + y^2 = R^2 in a plane would be a circle but along z is a cylinder.

    Now, if the center of the cylinder is not at x but say at h then the equation would be rewritten as:

    (x-h)^2 + y^2 = R^2 and expanding the (x-h)^2 term you'd get x^2 -2hx + h^2

    and inserting that back into the equation gives you something like your equation x^2 + y^2 - x = 0
    so its a cylinder along the z-axis but not with a center at x=0 and y=0
  5. Mar 11, 2013 #4
    Thanks both of you but I am having trouble visualising the figure. I drew a sketch but I am unable to find the volume cut out by the parabola. The cylinder extends infinitely aand at some point cuts the parabola. I can't form the integral for finding out the volume.
  6. Mar 11, 2013 #5
    Let's start with the domain first. One of the limiting planes is z = 0, so your domain could be in the XY plane. What does it look like?
  7. Mar 11, 2013 #6
    Two circles. One with centre at (1/2,0) and radius 1/2, the other is a circle of radius 1 centred at origin formed due to paraboloid.
  8. Mar 11, 2013 #7
    What about the x = 0 and y = 0 planes?
  9. Mar 11, 2013 #8
    In both the cases, a parabola with vertex at (0,0,1).
  10. Mar 11, 2013 #9
    I did not make myself clear. In the XY plane, what do the x = 0 and y = 0 conditions mean with regard to the domain? You have already found two "nested" circles there - but where exactly is the structure located?
  11. Mar 11, 2013 #10
    Haven't I already said that its a parabola rotated about the z-axis? I am not able to understand what you mean when you say where it is located (and thats because of my poor English).
  12. Mar 11, 2013 #11
  13. Mar 11, 2013 #12
  14. Mar 11, 2013 #13
    The structure has a flat base at z = 0, i.e., it is in the XY plane. The contour of the base is somehow given by the four lines mentioned above. By integrating the structure's height within the domain described by the contour, you get the volume of the structure.
  15. Mar 11, 2013 #14
    What's the structure's height?

    Tell me if I am understanding the problem correctly. I need to find the volume of cylinder till the point it intercepts the paraboloid, right?
  16. Mar 11, 2013 #15


    Staff: Mentor

    I think thats right, you need to define an integral that is evaluated within the bounds of the circle base of the cylinder with the z ranging from 0 to whatever the parabola eqn says. The x would range from 0 to diameter of the cylinder and the y would range from 0 to whatever x^2 +y^2 -x = 0 defines it as and that by symmetry would give you have the volume.
  17. Mar 11, 2013 #16
    No, you are not understanding the problem correctly. The five surfaces must intersect to form some CLOSED volume. In the z direction it is simple: the bottom is the z = 0 plane, and the top is the paraboloid. What about the "sides" of the structure?
  18. Mar 11, 2013 #17
    Please see the attachment. Do I have to find the volume of the region filled with blue lines?

    Attached Files:

  19. Mar 11, 2013 #18


    Staff: Mentor

    Isn't the sides just the cylinder along the z-direction which is what the OP implied topped by the paraboloid surface?
  20. Mar 11, 2013 #19
    Yes, that is correct. You still need to find the XY domain, however.
  21. Mar 11, 2013 #20
    Then the problem specifies too many unnecessary conditions such as x = 0 and y = 0. Look at the drawing in #8. You are suggesting that the inner circle is the base of the structure. Then surely x = 0 and y = 0 are completely superfluous as conditions. Yet they are specified.
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