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Finding area

  1. Oct 23, 2004 #1
    Find the area of the region bounded by the hyperbola [tex]9x^2-4y^2 = 36[/tex] and the line [tex]x = 3[/tex].

    I'm thinking that I have to integrate for x, so I'll have the sum of twice the area from [tex]2[/tex] to [tex]3[/tex].
    The function will be [tex] + \sqrt {\frac {9x^2-36}{4}}[/tex]

    hence, the integral will be[tex] 2\int_2^3 {\sqrt {\frac {9x^2-36}{4}}}dx [/tex]
    I just wanted to know if my reasoning is right

    Thanks in advance
    Last edited: Oct 23, 2004
  2. jcsd
  3. Oct 23, 2004 #2


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    It looks right to me. Go ahead with it.
  4. Oct 23, 2004 #3
    when I integrate [tex]\int \sqrt {\frac {9x^2-36}{4}}dx[/tex] I get as an answer [tex]\frac {3}{2} x{\sqrt {x^2-4}} - 6 ln({\frac {x}{2}}+\frac{\sqrt {x^2-4}}{2})[/tex] however maple gives me [tex]\frac {3}{2} x{\sqrt {x^2-4}} - 6 ln(x+\sqrt {x^2-4})[/tex]

    I used [tex]x=2sec(\theta)[/tex] hence [tex]\frac {x}{2}=sec(\theta)[/tex] so [tex]tan(\theta)=\frac{\sqrt{x^2-4}}{2}[/tex]

    which I substituted into [tex]6(sec(\theta)tan(\theta)-ln(sec(\theta)+tan(\theta)))[/tex]

    I know its Saturday night and any help will greatly be appreciated

  5. Oct 23, 2004 #4
    Weird, overall it gives the same answer, 4.2878 sq units, in both maple and on paper, however I don't understand how maple takes the 1/2 out of the ln.
  6. Oct 23, 2004 #5
    Ok, I quadrupled checked my integral and it is right, it seems that even though maple does not display the 1/2 it still accounts for it.
  7. Oct 24, 2004 #6
    i'm not sure but i don't think that the 1/2 over the guys inside the natural log have any impact on a definite integratl. combining the two you would get that sum of the numerators over 2. which would be the natural log of the top minus ln 2 -- which being a constant wouldn't be affected by variables.

    why it's there, i'm clueless. the only thing i could think of is the values determined by the trig functions of theta, but that doesn't seem to be the case.
    er edit: maybe maple simplified it out as it can be included in C indefinintly , and is extraneous in the case of a definant integral

    i quickly worked out the same answer you had, i haven't ever used maple so i don't know.
    Last edited: Oct 24, 2004
  8. Oct 24, 2004 #7


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    The only difference between

    [tex]ln({\frac {x}{2}}+\frac{\sqrt {x^2-4}}{2})[/tex]
    [tex]ln({x+ \srt{x^2- 4})[/tex]
    is -ln(2) which is a constant.
    Anti-derivatives can have any constant added and, anyway, cancels when you evaluate at the limits of integration.
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