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How to integrate sqrt(16-r^2) from a to 4

  1. Mar 10, 2007 #1
    [tex]\int_a^4\sqrt{16-r^2}rdr[/tex]

    How do you integrate this?

    Is it by parts, and then by substitution for the root bit?

    Cheers Ash
     
    Last edited: Mar 10, 2007
  2. jcsd
  3. Mar 10, 2007 #2
    If [itex]\int_{a}^{4}{r\sqrt{16-r^2}}dr[/itex] is your integral, a substitution would suffice.
     
  4. Mar 10, 2007 #3
    which one though? u=16-r^2

    and then work r out in terms u, and then integrate by parts?

    Cheers
     
  5. Mar 10, 2007 #4
    Is the answer:
    [tex]\frac{1}{3}(16-r^2)^{\frac{3}{2}}[/tex]

    with limits a to 4
     
  6. Mar 10, 2007 #5
    Yes

    In a way, yes. What you actually do is differentiate the expression u = 16 - r2

    There is no need for that...follow the steps mentioned above and you'll know why.
     
  7. Mar 10, 2007 #6
    What so you just ignore the first r?
     
  8. Mar 10, 2007 #7
    That is correct.
     
  9. Mar 10, 2007 #8
    yes just worked that out on a website, but what happens to the first r then?

    cheers
     
  10. Mar 10, 2007 #9
    [tex]u = 16 - r^2[/tex]

    [tex]du = - 2rdr[/tex]

    Can you spot an rdr somewhere in the original integral? Also, remember, when you're making a change of variable, the limits change accordingly.
     
  11. Mar 10, 2007 #10
    sorry isnt it minus [tex]\frac{1}{3}(16-r^2)^{\frac{3}{2}}[/tex]
     
  12. Mar 10, 2007 #11
    yes done it now thanks, but how do you convert the limits?
     
  13. Mar 10, 2007 #12
    Yes, that's right. Sorry I missed that.

    It can also be +, but then, the limits have to interchanged.

    Well, you know how u and r are related. What values would u take when r has those values?
     
  14. Mar 10, 2007 #13
    oh so the upper limit is very u = 16 - 4^2 = 0

    and the lower limit is root(16 - u)
     
  15. Mar 10, 2007 #14
    Yes

    Why would it be that. Put r=a.
     
  16. Mar 10, 2007 #15
    or yes course

    so lower limit is 16 - a^2
     
  17. Mar 10, 2007 #16
    So is this right if someone doesnt mind checking.

    [tex]\int_a^4\sqrt{16-r^2}rdr[/tex]

    using [tex]u = 16 - r^2[/tex] which converts to

    [tex]du = - 2rdr[/tex]

    so -du/2r = dr

    so integral becomes

    the integral of - (root u)/2 with lower limit 16-a^2 and upper limit 0

    which converts to

    the integral of (root u)/2 with lower limit 0 and upper limit 16-a^2

    which becomes [((u^(3/2))/3] with lower limit 0 and upper limit 16-a^2

    which becomes ((16-a^2)^3/2)/3

    Is that right, I hope you all understand it,

    I dont understand that tex thing very well.

    cheers Ash
     
    Last edited: Mar 10, 2007
  18. Mar 10, 2007 #17
    That seems right.
     
  19. Mar 10, 2007 #18
    are you sure?????
     
  20. Mar 10, 2007 #19
    put r=4 sinx,then dr= 4 cosx dx,16-r2=16(1-sin{square}x).
    proceed.Work it out & tell me
     
  21. Mar 10, 2007 #20

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    Yes, he's sure! Aren't you? :approve:

    That leads to a more difficult integral Did you notice the "r" outside the square root? The way he did it to begin with was completely correct.
     
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