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Math Help - Use a double integral to find the area of the region whose upperbound is the function

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    Use a double integral to find the area of the region whose upperbound is the function

    Use a double integral to find the area of the region whose upperbound is the function y = (x+3)^1/2 the lower boundary is y = (x+3)/3

    I got 4.5
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    MHF Contributor Amer's Avatar
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    Quote Originally Posted by ewkimchi View Post
    Use a double integral to find the area of the region whose upperbound is the function y = (x+3)^1/2 the lower boundary is y = (x+3)/3

    I got 4.5
    find the point of the intersection

    \sqrt{x+3}=\frac{x+3}{3}

    9(x+3) = x^2 + 6x+9

    x^2 -3x-18 = 0

    (x-6)(x+3) = 0 \Rightarrow x=6,x=-3



    \int_{6}^{-3} \int_{\frac{x+3}{3}}^{\sqrt{x+3}} dy dx

    I got 4.5 to
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