1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Integration problem

  1. Sep 1, 2005 #1
    Please click here for the problem link

    I'm having a hard time trying to figure out how the last two steps were taken to obtain the answer. I actually have an exam in couple hours and I would really appreciate if someone could explain carefully as soon as possible. Thanks in advance.
     
  2. jcsd
  3. Sep 1, 2005 #2
    In the first line, the x integration is being done first. The whole square root bit is a function of z and not x, so you just treat it like a constant. So, when you do the x integral, all you get is

    [tex]x\sqrt{17 + 4z^2}[/tex]

    Next you have to put in the x limits: 0-1. Clearly, putting in x = 0 will give you nothing and putting in x = 1 gives you

    [tex]\sqrt{17 + 4z^2}[/tex]

    This leaves you with the z integral at the end of the first line.

    As for the second line, I'm a bit stuck right now :/.
     
    Last edited: Sep 1, 2005
  4. Sep 1, 2005 #3
    lol thanks for the reply but the second line is what i'm having a problem with..
     
  5. Sep 1, 2005 #4

    arildno

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    Dearly Missed

    The logarithmic expression is nothing else than the cosh or sinh inverse 8don't remember which)
    In order to crack the integral, remember that [tex]Cosh^{2}y=1+Sinh^{2}y[/tex]
    Hence, make the change of varibles: [tex]z=a*Sinh(y)[/tex], where "a" is a constant to be fitted in the nicest manner possible.
     
  6. Sep 1, 2005 #5
    I still don't understand how the first line became the second line.. Also, how would i use the expression Cosh^2y = 1 + Sinh^2y there? I'd really appreciate if you could tell me step by step.. I just need to know how the first line became the second line.. This is not a homework problem as you can tell the answer is already there.. I'm just studying for a test.
     
  7. Sep 1, 2005 #6
    I know it's very early but is there anyone who can help me?
     
  8. Sep 1, 2005 #7

    arildno

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    Dearly Missed

    OK, let's rewrite the integral in the first line as:
    [tex]\int_{0}^{1}\sqrt{17+4z^{2}}dz=\sqrt{17}\int_{0}^{1}\sqrt{1+(\frac{2z}{\sqrt{17}})^{2}}dz[/tex]
    Thus, the change of variables [tex]\frac{2z}{\sqrt{17}}=Sinh(y)[/tex] is rather natural..
     
  9. Sep 1, 2005 #8
    Thank you for the help.. I guess I just don't remember the properties of sinh & cosh.. I thought there was an easier way to do it because I never used cosh or sinh to solve an integral.. I guess I will get this problem wrong if it comes out on the test.. Thanks though.
     
  10. Sep 2, 2005 #9
    my math exam is over so i don't need to know how to solve it any more but can anybody solve this thing step by step? when i usually post a problem, many people respond, but it seems like not many people know how to solve this..
     
  11. Sep 2, 2005 #10

    arildno

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    Dearly Missed

    Well, you now have [tex]z=\frac{\sqrt{17}}{2}Sinh(y)[/tex], which implies that:
    [tex]dz=\frac{\sqrt{17}}{2}Cosh(y)dy[/tex]
    Along with the new limits [tex]0\leq{y}\leq{Sinh}^{-1}(\frac{2}{\sqrt{17}})[/tex]
    Therefore, we get:
    [tex]\sqrt{17}\int_{0}^{1}\sqrt{1+(\frac{2z}{17})^{2}}dz=\frac{17}{2}\int_{0}^{{Sinh}^{-1}(\frac{2}{\sqrt{17}})}Cosh^{2}ydy[/tex]

    All right?
     
  12. Sep 2, 2005 #11

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    C'mon- in the very first post, Nylex told you how to go from the first line to the second- how the integration was done- and you replied "lol thanks for the reply but the second line is what i'm having a problem with..".
    When it was pointed out that the second line was just a matter of evaluating functions you said "I still don't understand how the first line became the second line..", so we're back to what was said in the first post!
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Integration problem
  1. Integration problem (Replies: 8)

  2. Integral problem. (Replies: 2)

  3. Integration problem (Replies: 3)

  4. Integration problem (Replies: 4)

  5. Integration problem (Replies: 3)

Loading...