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Math Help - express the area in terms of an integral

  1. #1
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    express the area in terms of an integral

    express the area of the region in the x,y plane by the curve x^2/9+y^2/4=1 in terms of an integral

    i solved the equation for y to get

    y=sqrt(4-2/3x^2)

    and looking at the graph i think it would be
    3
    ∫sqrt(4-2/3X^2)
    -3

    can anyone tell me if this is at all right
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  2. #2
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    Quote Originally Posted by dat1611 View Post
    express the area of the region in the x,y plane by the curve x^2/9+y^2/4=1 in terms of an integral

    i solved the equation for y to get

    y=sqrt(4-2/3x^2)

    and looking at the graph i think it would be
    3
    ∫sqrt(4-2/3X^2)
    -3

    can anyone tell me if this is at all right
    \displaystyle\huge\frac{x^2}{9}+\frac{y^2}{4}=1

    \displaystyle\huge\frac{4}{9}x^2+y^2=4

    y^2=4-\frac{4}{9}x^2

    y=\pm\sqrt{4-\frac{4}{9}x^2}=\sqrt{4-\left(\frac{2}{3}x\right)^2}

    Looking at your original function, it's the equation of an ellipse,
    hence half of the graph lies below the x-axis.
    It's symmetrical, so you should double the integral, since you have only allowed for positive y.

    \displaystyle\huge\frac{x^2}{9}+\frac{y^2}{4}=1

    x=0,\ y=\pm2

    y=0,\ x=\pm3

    Area=2\displaystyle\huge\int_{x=-3}^{x=3}\sqrt{4-\left(\frac{2}{3}x\right)^2}\ dx
    Attached Thumbnails Attached Thumbnails express the area in terms of an integral-ellipse.jpg  
    Last edited by Archie Meade; August 9th 2010 at 02:49 PM.
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