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Cylindrical coordinates

  1. Oct 28, 2012 #1
    1. The problem statement, all variables and given/known data

    Let W= {(x,y,z)| x^2 + y^2 ≤ 1, -1 ≤ z ≤ 1} (W is a bounded cylindrical region)

    Evaluate the triple integral f(x,y,z)= z^2 x^2 + z^2 y^2 over W. Use cylindrical coordinates
    2. Relevant equations
    i dont see any relevant equations besides the obvious cylindrical coordinate equations.


    3. The attempt at a solution
    I have no clue what the problem is asking for to be honest? any help on how to get me started? thank you
     
  2. jcsd
  3. Oct 28, 2012 #2

    jedishrfu

    Staff: Mentor

    convert the function to cyl coordinates using x = r cos(theta) and y=r sin(theta) and z=z and what do you get?

    Can you simplify it?
     
  4. Oct 28, 2012 #3
    yes sir! im on it right now! thanks
     
  5. Oct 28, 2012 #4

    Dick

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    Express your function and limits in term of the cylindrical coordinates. Change (x,y,z) to (r,θ,z) and don't forget to change dxdydz to its cylindrical equivalent. Then integrate. That's what it is asking you.
     
  6. Oct 28, 2012 #5
    f(x,y,z)= z^2 (r^2 cos^2theta) + z^2 (r^2 sin^2 theta)
     
  7. Oct 28, 2012 #6
    thank you dick! im on it right now ! ill post results
     
  8. Oct 28, 2012 #7
    can i integrate the first limit then convert to polar? or do i have to flat out use cylindrical coordinates?
     
  9. Oct 28, 2012 #8

    Dick

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    There's not that much of a difference. "cylindrical" is just "polar" plus the extra coordinate z. Use what you know about polar.
     
  10. Oct 28, 2012 #9
    if i draw the xz and yz traces, will it be a horizontal line at 1 and at -1? then of course the xy trace is a circle with r=1
     
  11. Oct 28, 2012 #10
    i got the following limits, r ≤ 1,
    0≤ theta ≤ 2pi ,

    -1≤z≤1

    thats in polar right? now do i integrate? rdrdzdtheta?
     
  12. Oct 28, 2012 #11

    Dick

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    The intersection with the xz and yz planes are rectangles, as the problem said, it's a cylinder. I'm not sure why you worried about this.
     
  13. Oct 28, 2012 #12

    Dick

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    Those are the limits and that's the differential element alright. So what function will you integrate?
     
  14. Oct 28, 2012 #13
    the function i will integrate will be z^2 x^2 + x^2 y^2 which i need to convert so its
    z^2 (r^2 cos^2theta) + z^2 (r^2 sin^2 theta)

    i integrate that? if so do i integrate rdzdrdtheta or rdrdzdtheta? i cant find a limit for dz, is it just -1 to 1? i dont have to express in other variables ? if so then its

    ∫2pi-0 ∫1-0 ∫ 1-(-1) (the function) r dz dr dtheta?
     
  15. Oct 28, 2012 #14
    by the way thanks for helping dick
     
  16. Oct 28, 2012 #15

    Dick

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    No problem. But you can simplify the function you are integrating. Use a trig identity. It looks like you've got the rest ok.
     
  17. Oct 28, 2012 #16
    can i use sin^2+cos^2=1 then basically do z^2 r^2 + z^2 r^2 which is 2z^2r^2?
     
  18. Oct 28, 2012 #17

    Dick

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    You got the identity right. But the result came out wrong. Give that another thought.
     
  19. Oct 28, 2012 #18
    i conclude z^2 (r^2 cos^2 theta + r^2 sin^2 theta) so i can factor the r^2 as well and im left with (sin ^2 + cos ^2) and thats one

    so z^2 r^2???
     
  20. Oct 28, 2012 #19

    Dick

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    Much better. Now put it all together.
     
  21. Oct 28, 2012 #20
    will do ! ill post back with results thank you
     
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