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Math Help - Use a triple integral to find the volume of the solid bounded.

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    Use a triple integral to find the volume of the solid bounded.

    z=9-x^3, y=-x^2+2, y=0, z=0, x is greater than and equal to 0.
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    Senior Member AllanCuz's Avatar
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    Quote Originally Posted by asdf122345 View Post
    z=9-x^3, y=-x^2+2, y=0, z=0, x is greater than and equal to 0.
    The definition of a triple integral is

    \int_a^b dx \int_{g_1(x)}^{g_2(x)} dy \int_{h_1(x,y)}^{h_2(x,y)} dz

    By following this we get,

    \int_0^{\sqrt{2}} dx \int_0^{-x^2 + 2} dy \int_0^{9-x^3} dz
    Last edited by AllanCuz; April 22nd 2010 at 05:31 PM.
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    Where the radical x come from?
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    Quote Originally Posted by asdf122345 View Post
    Where the radical x come from?
    y=-x^2+2

    This gives our XY domain. To find our max value of x, set y=0. If you re-arrange such that

    x^2 = 2 - y

    Clearly, x is max when y=0.
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    So shouldn't it be radical 2 instead of radical x? Or am I thinking something wrong.
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    Senior Member AllanCuz's Avatar
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    Quote Originally Posted by asdf122345 View Post
    So shouldn't it be radical 2 instead of radical x? Or am I thinking something wrong.
    Yeah, i just wrote it wrong. It's radical 2
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