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Finding integral of x2 and y2

  1. Jun 17, 2012 #1
    1. The problem statement, all variables and given/known data
    I would like to know how to integrate a function that has y^2 in it.


    2. Relevant equations
    eg: (x^2)/5 + (y^2) +900 = 0


    3. The attempt at a solution
    y = ((x^2)/5 - 900)^(1/2)
    Area = ∫((x^2)/5 - 900)^(1/2)
    Chain rule in reverse somehow?
    = (((x^2)/5 - 900)^(3/2))/(1.5)
    = No idea!
     
    Last edited: Jun 17, 2012
  2. jcsd
  3. Jun 17, 2012 #2
    Hi Mathpower,
    there must be something wrong in your relevant equations
    the attempted solution is wrong respect to it, but just looking at your initial equation, you have a sum of squares + 900 = 0, that can't be solved since the left side is >=900

    Cheers
     
  4. Jun 17, 2012 #3
    OK thank you for your help! What I am attempting to do is to learn how to do implicit differentiation in reverse.
    Let me change the question:


    1. The problem statement, all variables and given/known data
    I would like to know how to integrate a function that has y^2 in it.


    2. Relevant equations
    eg: (x^2)/5 + (y^2) -1 = 0


    3. The attempt at a solution
    y = ((x^2)/5 - 1)^(1/2)
    Area = ∫((x^2)/5 - 1)^(1/2)
    Chain rule in reverse somehow?
    = (((x^2)/5 - 1)^(3/2))/(1.5)
    = No idea
     
  5. Jun 17, 2012 #4
    it is hard to figure out what you wanted to do. did you want to calculate the area? where? in the first quadrant?

    Try a change of variables [STRIKE]u^2=x^2/5[/STRIKE] (see comment below), then you'll have a nice circle. then you could do a change of variable to polar coordinates.

    If you don't want to do that, then you can use trig subsitution when you have roots and squared terms.
     
    Last edited: Jun 17, 2012
  6. Jun 17, 2012 #5
    That should be,

    [tex]y = \pm \sqrt{1-\frac{x^2}{5} }[/tex]

    So you would basically have two integrals to do,

    [tex]\int y dy= \int \sqrt{1-\frac{x^2}{5} } dx[/tex]

    And,

    [tex]\int y dy= \int -\sqrt{1-\frac{x^2}{5} } dx[/tex]

    Which one you have to integrate, and with what limits, depends on what you wish to find out, as algebrat observed.
     
  7. Jun 17, 2012 #6
    EGADS!!! I should have said u=x/√5 me thinks!
     
  8. Jun 17, 2012 #7
    Oops I kind of mixed up what I wanted to learn with the actual test question that I had. My apologies!
    If possible please show me how to do the indefinite integral (not the definite integral as my working had previously suggested) of (x^2)/5 + (y^2) -1 = 0

    I am sure that I can pick up how to find the area after that. Oh...and let me know how I can post a message like Infinitum (ie: the math writing style...if you know what I mean)
     
  9. Jun 17, 2012 #8
    For area, integrate each of those expressions above, separately. Use the substitution, [itex]u= \frac{x}{\sqrt{5}}[/itex] You will have two answers, opposite in sign.

    For only the indefinite integral, you can integrate the expression,

    [tex]\int y^2 dy = \int (1-\frac{x^2}{5})dx[/tex]

    But this, you cannot use to find area.

    As for the math, its called LaTeX. You can learn about it here : https://www.physicsforums.com/showthread.php?t=546968
     
    Last edited: Jun 17, 2012
  10. Jun 17, 2012 #9
    WOW! Thank you so much for your help! Amazing that I am getting help from someone in USA!
    oh...so:
    Let: [tex]u^2 = (x^2 /5)[/tex]
    [tex]\sqrt{1-u^2}[/tex]
    Indefinite integral:
    [tex]\int\sqrt{1x-u^2}[/tex]
    =[tex]2 (1x-u^2)^(-0.5) . (2u) [/tex]
    =[tex]4u/\sqrt(1x-u^2)[/tex]
    =[tex](4x^2/5)/(1x-(x^2/5))^0.5[/tex]
    =4x^2/(5^(0.5)(5x-x^2))

    I get the feeling I haven't done it correctly! The problem is I haven't been taught the whole integration topic yet...but I do know differentiation...and that doesn't really help in this situation!
     
    Last edited: Jun 17, 2012
  11. Jun 17, 2012 #10
    Your steps after [itex]\int \sqrt{1-u^2}[/itex] aren't clear..

    When you substitute a variable, you need to change the differential coefficient accordingly. You will have,

    [tex]u = \frac{x}{\sqrt{5}}[/tex]

    [tex]du = \frac{dx}{\sqrt{5}}[/tex]

    Hence,

    [tex]dx = \sqrt{5} du[/tex]

    Which you need to use in place of dx.

    Also, once you reached [tex]\int \sqrt{1-u^2} du[/tex]
    You can do another substitution, [itex]t=sin(u)[/itex] and apply a similar technique as above. Then put back the t and u, so that you get x.
     
  12. Jun 17, 2012 #11

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    Why would you say that? I had not noticed that people from the USA refused to help others.

     
  13. Jun 17, 2012 #12
    Haha, maybe Mathpower just meant that they are far away from the USA, and it is amazing that the internet is bringing everyone together.

    (You have assumed the negation is as you said.)

    Perhaps we should start another thread on whether the USA is really helping anyone.
     
  14. Jun 18, 2012 #13
    Oops I made a typo... I have corrected it in the above quote. Is it correct now?
     
    Last edited: Jun 18, 2012
  15. Jun 18, 2012 #14
    Hold on...hold on... Talk about jumping to conclusions my friend!
    I never even remotely meant neither suggested what you interpreted.
    What I meant to say was that, it is quite amazing how someone 1000 's of kilometers away can help me!
     
  16. Jun 18, 2012 #15
    Why the negative exponent? And how did you get the 'x' term from?? You substituted u to get rid of x... Also, you should read post #10 again :smile:
     
  17. Jun 18, 2012 #16
    Ok ...woah ..what have I done!
    I look back at my working now and I am shocked! I somehow managed to mix up differentiation with integration...ie: when differentiating we subtract 1 off the power, hence 0.5-1=-0.5
    Lol!
    But atleast I know why I did it and our class got taught integrating by substituting today! So I do know how to do it now...a very big thank you for your help.

    Oh and just as a side note: I guess Americans really are so patriotic after all!
     
  18. Jun 18, 2012 #17
    Let u=x/sqrt(5). Then use trig substitution.
     
  19. Jun 19, 2012 #18
    Good to know you got it clear now :smile:
     
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