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Math Help - Geometric Interpretation of Integral

  1. #1
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    Question Geometric Interpretation of Integral

    Could someone please explain to me what it means when I have to give a geometric interpretation of each term of the integral, say, \sqrt{a^2-x^2} between x = b and x = 0?

    From what I can make of the question, all I can think of is that a is the radius of the semicircle.

    Please help?
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    Quote Originally Posted by xwrathbringerx View Post
    Could someone please explain to me what it means when I have to give a geometric interpretation of each term of the integral, say, \sqrt{a^2-x^2} between x = b and x = 0?

    From what I can make of the question, all I can think of is that a is the radius of the semicircle.

    Please help?
    Are you sure it is not x = a, as opposed to x = b? What is b?
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  3. #3
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    Ooops sori

    I forgot to mention:

    This is part of a question. The first part required me to find between x = b and x = 0 using the substitution  u =\arcsin\frac{x}{a}, a > b > 0.
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    Geometric Interpretations of Integrals

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    Last edited by Veraltus; November 15th 2009 at 03:14 PM. Reason: Someone posted something to completely disregard the information
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    Quote Originally Posted by xwrathbringerx View Post
    Ooops sori

    I forgot to mention:

    This is part of a question. The first part required me to find between x = b and x = 0 using the substitution  u =\arcsin\frac{x}{a}, a > b > 0.
    Ok, well, we know that we can interpret the integral, geometrically, to mean the net area under the curve... what can you do with this?
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    Exclamation

    .... I have no idea.
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    Are you asking how to solve the integral?
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  8. #8
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    The integral is just \frac{a^2}{2}[cosb - 1] rite?
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    Quote Originally Posted by xwrathbringerx View Post
    .... I have no idea.
    the inetgral gives the area under the semi-circle and above the x-axis between x = 0 and x = b

    Quote Originally Posted by Veraltus View Post
    Are you asking how to solve the integral?
    it does not seem that way

    Quote Originally Posted by xwrathbringerx View Post
    The integral is just \frac{a^2}{2}[cosb - 1] right?
    actually, no. if you differentiate that, you do not get the original integrand...

    why do you not follow the hint? (write it as x = asin(u) though...)
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  10. #10
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    the inetgral gives the area under the semi-circle and above the x-axis between x = 0 and x = b
    Hmmmm but does that cover "each term of the integral" since the question asked for the geometric interpretation of all the terms?
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    Quote Originally Posted by xwrathbringerx View Post
    Hmmmm but does that cover "each term of the integral" since the question asked for the geometric interpretation of all the terms?
    will you please post the question EXACTLY as it was stated, giving any preliminary info that may have been there also
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  12. #12
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    Attached Thumbnails Attached Thumbnails Geometric Interpretation of Integral-untitled.bmp  
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    Quote Originally Posted by xwrathbringerx View Post
    .
    an awkward way to ask the question. but i stand by my answer.
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    I've attached it. It's a thumbnail.

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  15. #15
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    Well using the standard trigonometric substitution method...and going back to our original integral i got..


    \int \sqrt{a^2 -x^2}dxfrom x = 0 to x = b...

    so use the substitution x=asin(\theta) and  dx = acos(\theta)d\theta

    so the integral becomes

    \int \sqrt{a^2 - a^2 sin^2(\theta)}acos(\theta)d\theta which becomes \int \sqrt{a^2 (1-sin^2 (\theta))}acos(\theta)d\theta

    so we finally get \int acos^2(\theta)d\theta

    We have to also change our limits of integration... from x = 0 to \theta = 0 and we get that x = b turns to \theta = arcsin(\frac {b}{a})

    so my final answer was... \frac {a^2sin^{-1}( \frac {b}{a})}{2}  + \frac {a^2 sin (2*sin^{-1} (\frac {b}{a}))}{4}
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