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Solid bounded by different region

  1. Oct 26, 2016 #1
    1. The problem statement, all variables and given/known data
    By using cylindrical coordinates , evaluate the volume of solid bounded on top of sphere (x^2) + (y^2) + (z^2) = 9 and it's sides by (x^2) + (y^2) = 4x .



    2. Relevant equations


    3. The attempt at a solution
    I have sketched out the diagram , but i dun know which part is the solid formed.... Can someone highlight the solid formed ?
     

    Attached Files:

  2. jcsd
  3. Oct 26, 2016 #2

    Ray Vickson

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    Is the wording you wrote the exact statement of the problem? If so, it is confusing, and in English may easily be misinterpreted. There is a difference in meaning between saying "bounded on top of sphere.... "; the "on top" says to me that the region lies above the sphere---it is on top of the sphere. However, maybe it means "bounded on top BY the sphere..." which means the region lies below the sphere: the sphere is the top of the region. I suspect that the second interpretation is intended, but the wording of the problem could be taken to imply the opposite.

    Anyway, what about a lower limit? is the opposite side of the sphere also the bottom of the region? If we don't specify a bottom, we can have ##z## extending to ##-\infty##, giving an infinite volume.
     
  4. Oct 27, 2016 #3
    sorry , i made a typo . The original question is evaluate the volume of solid bounded on top by the sphere (x^2) + (y^2) + (z^2) = 9........
    Here's after correction , the cylinder span from z = -3 to 3 ...
    What is the portion of the solid bounded on top of sphere (x^2) + (y^2) + (z^2) = 9 and it's sides by (x^2) + (y^2) = 4x ?
    I couldnt imagine
     

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  5. Oct 28, 2016 #4
    Anyone can response ?
     
  6. Oct 28, 2016 #5
    or there's something wrong with my diagram ?
     
  7. Oct 28, 2016 #6

    Mark44

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    Neither of your diagrams is very helpful, and both are incorrect. You have the circular cylinder located in the wrong place in both diagrams, and your drawing of the sphere just looks like a circle. Here's a quick sketch I did that attempts to show the portion of a sphere in the first octant.
    sphere.png
     
  8. Oct 28, 2016 #7

    LCKurtz

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    @chetzread: In the ##xy## plane, draw the two circles ##x^2+y^2 = 9## (which is where the sphere cuts the ##xy## plane) and ##x^2+y^2 = 4x## (after you get it right). That picture should help you figure out the ##xy## region that your cylindrical coordinate ##r,\theta## region needs to describe. Hopefully you will see that you need two different integrals to set it up.
     
  9. Oct 28, 2016 #8
    @Mark44 , @LCKurtz , is my cylinder located at the wrong position ?

    So , i have chnaged the location ... Pls refer to the latest diagram i uploaded here

    Here's what i did :

    ##x^2+y^2 = 4x##

    ##x^2-4x+y^2 = 0##

    ##(x-2)^2 -4 +y^2 =0##

    ##x^2-4x+4-4+y^2 =0##

    ##x^2+y^2=4x##

    So , the circular part of cylinder located at x =2 , right ?
     

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    Last edited: Oct 28, 2016
  10. Oct 28, 2016 #9
     
  11. Oct 28, 2016 #10

    Mark44

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    The central axis of the cylinder goes through (2, 0, 0), yes, and is vertical. Your diagram is close to correct, except the sphere is slightly too large.
     
  12. Oct 28, 2016 #11
    Ok, my problem now is which is the region that the solid formed? Can you sketch the part that the solid form using my diagram? I have problem visualizing it...
     
  13. Oct 28, 2016 #12

    Mark44

    Staff: Mentor

    Draw a 2-D sketch of the projections of the two objects in the x-y plane.
     
  14. Oct 28, 2016 #13
    do you mean the green part is the solid formed ?
     

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  15. Oct 28, 2016 #14

    Mark44

    Staff: Mentor

    That's not what I had in mind. It's not a two-dim. sketch. Also, the solid is bounded on top by the sphere. As I read the problem description, the solid is inside the sphere, not outside it, as you show in your sketch.
     
  16. Oct 28, 2016 #15

    LCKurtz

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    We have both suggested this. You need to do it to make progress on this problem.
     
  17. Oct 28, 2016 #16
    here it is ....
     

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  18. Oct 28, 2016 #17

    LCKurtz

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    About time to hit the sack here. Where is the origin ##x=0## and ##x=1## on your sketch? Get it correct, then we can continue tomorrow.
     
  19. Oct 28, 2016 #18
    here it is x = 0 ...
     

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  20. Oct 29, 2016 #19

    Mark44

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    Now, how about another two-dim. sketch, this time looking at the x-z plane?
     
  21. Oct 29, 2016 #20
    here
     

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