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Math Help - Area

  1. #1
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    Area

    Find c>0 such that the area of the region enclosed by the parabolas y = x^2 - c^2 and y = c^2 -x^2 is 230.
    c =?
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by qbkr21 View Post
    Find c>0 such that the area of the region enclosed by the parabolas y = x^2 - c^2 and y = c^2 -x^2 is 230.
    c =?
    you are integrating with respect to x, so you cannot integrate the c terms like that. you would just attach x to them. by the way, when i did it your way i got 0 = 230, which means you probably made another mistake somewhere
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  3. #3
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    Re:

    Could you trying working it out? or at least some of it... so that I can get a feel for what you mean...

    Sure appreciate it....


    -qbkr21
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by qbkr21 View Post
    Could you trying working it out? or at least some of it... so that I can get a feel for what you mean...

    Sure appreciate it....


    -qbkr21
    \int_{-c}^{c}\left[ c^2 - x^2 - \left( x^2 - c^2 \right) \right]dx

    =2 \int_{0}^{c} \left( 2c^2 - 2x^2 \right) dx

    = 4 \int_{0}^{c} \left( c^2 - x^2 \right)dx

    = 4 \left[ c^2 x - \frac {1}{3}x^3 \right]_{0}^{c}

    = \frac {8}{3}c^3

    check my computation, i was in a rush. i have to leave for a little while

    did you see the difference with how i integrated c? i treated it as a constant--which it is
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  5. #5
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    Re:

    Yes you were right on the money as always!!

    c = 4.41828

    Thanks so much!!!

    -qbkr21
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    find c

    Quote Originally Posted by Jhevon View Post
    \int_{-c}^{c}\left[ c^2 - x^2 - \left( x^2 - c^2 \right) \right]dx

    =2 \int_{0}^{c} \left( 2c^2 - 2x^2 \right) dx

    = 4 \int_{0}^{c} \left( c^2 - x^2 \right)dx

    = 4 \left[ c^2 x - \frac {1}{3}x^3 \right]_{0}^{c}

    = \frac {8}{3}c^3

    check my computation, i was in a rush. i have to leave for a little while

    did you see the difference with how i integrated c? i treated it as a constant--which it is
    I think you are right
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    Last edited by curvature; June 3rd 2007 at 11:00 PM.
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by curvature View Post
    I think you are right
    i know, qbkr21 already confirmed it. i believe he has the answer to the question in the back of his text. i hope he gets why i could have integrated between 0 and c though and then multiply by two, furthermore, i hope he gets why i would want to do that in the first place
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  8. #8
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    Quote Originally Posted by Jhevon View Post
    \int_{-c}^{c}\left[ c^2 - x^2 - \left( x^2 - c^2 \right) \right]dx

    =2 \int_{0}^{c} \left( 2c^2 - 2x^2 \right) dx
    hi
    what did you do to the 2 outside to make the lower integral -c 0?
    can it be any constants?
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  9. #9
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    Quote Originally Posted by ^_^Engineer_Adam^_^ View Post
    hi
    what did you do to the 2 outside to make the lower integral -c 0?
    can it be any constants?
    Note that your integrand is an even function: f(-x) = f(x). In this case
    \int_{-c}^c dx \, f(x) = \int_{-c}^0 dx \, f(x) + \int_0^c dx \, f(x)

    Use the substitution y = -x in the first integral:

    = \int_{c}^0 (-dy) \, f(-y) + \int_0^c dx \, f(x)

    = -\left ( -\int_0^c dy \, f(-y) \right ) + \int_0^c dx \, f(x)

    But f(-y) = f(y) when f is an even function:

    = \int_0^c dy \, f(y) + \int_0^c dx \, f(x)

    and now just replace the dummy variable y in the first integration with x:

    = \int_0^c dx \, f(x) + \int_0^c dx \, f(x)

    = 2 \int_0^c dx \, f(x)

    -Dan
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