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Why is the limit for theta = pi/2 instead of 2pi?

  1. Dec 16, 2015 #1
    1. The problem statement, all variables and given/known data
    use cylindrical coordinates to find the volume of the solid which is under z=xy, above xy-plane and inside the cylinder x^2+y^2=2x

    2. Relevant equations


    3. The attempt at a solution
    [tex] \int_{0}^{pi/2} \int_{0}^{2cos\theta} \int_{0}^{r^2\cos\theta\sin\theta} r\, dz \, dr \, d\theta [/tex]

    my professor had it pi/2, but when I do it myself, I dont see how its pi/2. 2pi would make more sense to me since it has a cylinder base.

    any help is appreciated
     
  2. jcsd
  3. Dec 16, 2015 #2

    blue_leaf77

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    The curve ##r = 2\cos \theta## is fully circulated when ##\theta## goes from 0 to ##\pi##, so I would actually expect the limit to be from 0 to ##\pi##.
     
  4. Dec 16, 2015 #3
    thank you ! Oh i see why its 0 to pi, i just graphed that curve, i see its not centered at (0,0), its in the first and fourth quadrant.
    so i must have made a mistake when I took my notes in the class then?
     
  5. Dec 16, 2015 #4

    blue_leaf77

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    Either ask one of your colleagues or confirm directly to the lecturer.
     
  6. Dec 16, 2015 #5
    no more classes left for this course though :(

    anyway,I just realized that if its 0 to pi, the volume become 0
     
  7. Dec 16, 2015 #6

    blue_leaf77

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    Ah I see why it should indeed be from 0 to ##\pi/2##. The cylinder occupies ##x>0## region of space, but in this region the surface ##z=xy## is positive if ##y>0## and negative if ##y<0##. Because the volume is required to be above xy plane where z is positive, the part of ##z=xy## surface which is negative (that with ##y<0## and ##x>0##) must be ruled out.
     
  8. Dec 16, 2015 #7
    oh interesting, I understand now. in order for z to be positive, y must be positive, which means first quadrant only.
    thx a lot !
     
  9. Dec 16, 2015 #8

    blue_leaf77

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    You are welcome.
     
  10. Dec 16, 2015 #9

    Ray Vickson

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    The third quadrant ##x<0## and ##y<0## would also be allowed by ##0 \leq z \leq xy##, but the cylindrical base ##x^2 + y^2 = 2x## lies completely in the first and fourth quadrants. Altogether, when you impose both conditions your region is restricted to the first quadrant.
     
  11. Dec 16, 2015 #10

    blue_leaf77

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    I was actually aware of that, that was infact what enables me to write post #2.
     
  12. Dec 16, 2015 #11

    Ray Vickson

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    Sorry: wrong response.

    I had meant to respond to the OP's post #7, which mistakenly stated that to have ##z > 0## it was necessary to have ##y > 0##.
     
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