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Thread: Area enclosed by two curves

  1. #1
    Junior Member bijosn's Avatar
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    Area enclosed by two curves

    Is my solution correct


    QUESTION:
    Calculate the area of the region enclosed by the curves$\displaystyle x = 3y$
    and $\displaystyle x=2+y^2$




    this is my solution, i ommited the intermidiete steps...
    coverted them to a more familiar functions of y
    $\displaystyle y=\frac{3}{x} $
    $\displaystyle y = \sqrt{x - 2}$

    the curves intercept at the two points, (3,1) and (6,2),



    $\displaystyle \int_3^6\left[ \sqrt{x - 2} - \frac{3}{x} \right] dx $
    $\displaystyle \int_3^6 \left[ \left( x - 2 \right)^\frac{1}{2} - 3x^{-1} \right] dx $
    = 2.58722??
    Last edited by bijosn; Feb 6th 2011 at 11:03 AM.
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  2. #2
    MHF Contributor alexmahone's Avatar
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    Quote Originally Posted by bijosn View Post
    Is my solution correct


    QUESTION:
    Calculate the area of the region enclosed by the curves$\displaystyle x = 3y$
    and $\displaystyle x=2+y^2$




    this is my solution, i ommited the intermidiete steps...
    coverted them to a more familiar functions of y
    $\displaystyle y=\frac{3}{x} $
    $\displaystyle y = \sqrt{x - 2}$

    the curves intercept at the two points, (3,1) and (6,2),



    $\displaystyle \int_3^6\left[ \sqrt{x - 2} - \frac{3}{x} \right] dx $
    $\displaystyle \int_3^6 \left[ \left( x - 2 \right)^\frac{1}{2} - 3x^{-1} \right] dx $
    = 2.58722??
    $\displaystyle y=\displaystyle \frac{x}{3}$ and $\displaystyle y=\displaystyle \sqrt{x-2}$

    $\displaystyle A=\displaystyle \int_3^6(\sqrt{x-2}-\frac{x}{3})dx$

    = $\displaystyle \displaystyle \frac{(x-2)^{3/2}}{3/2}-\frac{x^2}{6})\big|_3^{6}$

    = $\displaystyle \displaystyle \frac{2}{3}(4^{3/2}-1)-\frac{1}{6}\times 27$

    = $\displaystyle \displaystyle \frac{2}{3}\times 7-\frac{9}{2}$

    = $\displaystyle \displaystyle \frac{28-27}{6}$

    = $\displaystyle \displaystyle \frac{1}{6}$
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  3. #3
    MHF Contributor
    skeeter's Avatar
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    Quote Originally Posted by bijosn View Post
    Is my solution correct


    QUESTION:
    Calculate the area of the region enclosed by the curves$\displaystyle x = 3y$
    and $\displaystyle x=2+y^2$




    this is my solution, i ommited the intermidiete steps...
    coverted them to a more familiar functions of y
    $\displaystyle y=\frac{3}{x} $
    $\displaystyle y = \sqrt{x - 2}$

    the curves intercept at the two points, (3,1) and (6,2),



    $\displaystyle \int_3^6\left[ \sqrt{x - 2} - \frac{3}{x} \right] dx $
    $\displaystyle \int_3^6 \left[ \left( x - 2 \right)^\frac{1}{2} - 3x^{-1} \right] dx $
    = 2.58722??
    no.

    $\displaystyle 3y = 2 + y^2$

    $\displaystyle y = 2$ , $\displaystyle y = 1$

    $\displaystyle \displaystyle A = \int_1^2 3y - (2+y^2) \, dy$

    try again.
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  4. #4
    Junior Member bijosn's Avatar
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    Quote Originally Posted by alexmahone View Post
    $\displaystyle y=\displaystyle \frac{x}{3}$ and $\displaystyle y=\displaystyle \sqrt{x-2}$

    $\displaystyle A=\displaystyle \int_3^6(\sqrt{x-2}-\frac{x}{3})dx$

    = $\displaystyle \displaystyle \frac{(x-2)^{3/2}}{3/2}-\frac{x^2}{6})\big|_3^{6}$

    = $\displaystyle \displaystyle \frac{2}{3}(4^{3/2}-1)-\frac{1}{6}\times 27$

    = $\displaystyle \displaystyle \frac{2}{3}\times 7-\frac{9}{2}$

    = $\displaystyle \displaystyle \frac{28-27}{6}$

    = $\displaystyle \displaystyle \frac{1}{6}$
    thank you, I made a silly mistake with the first equation, putting it as $\displaystyle y=\frac{3}{x}$ instead of $\displaystyle y =\frac{x}{3}$
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  5. #5
    Junior Member bijosn's Avatar
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    Quote Originally Posted by skeeter View Post
    no.

    $\displaystyle 3y = 2 + y^2$

    $\displaystyle y = 2$ , $\displaystyle y = 1$

    $\displaystyle \displaystyle A = \int_1^2 3y - (2+y^2) \, dy$

    try again.
    thank you

    $\displaystyle \displaystyle A = \int_1^2 3y - 2 - y^2 \, dy$
    $\displaystyle \displaystyle = ( \frac{3y^2}{2} - 2y - \frac{y^3}{3} ) \big|_1^{2}$
    $\displaystyle = \frac{1}{6} $
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