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Math Help - Help determining volume

  1. #1
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    Help determining volume

    I have the equation ...

    Determine the volume for f(x,y)=x^2+y^2+1; 0<x<3 (greater than or equal to on the <, I didn't know how to do that); 0<y<3 (also greater than or equal to)
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  2. #2
    Moo
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    Hello,

    Wouldn't it be \int_0^{2 \pi} \int_0^3 \int_0^3 x^2+y^2+1 \ dx dy dz ?
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    Quote Originally Posted by Moo View Post
    Hello,

    Wouldn't it be \int_0^{2 \pi} \int_0^3 \int_0^3 x^2+y^2+1 \ dx dy dz ?
    Yeah, that looks about right. Thanks, it's a cylinder it seems. Thanks for help with the setup I always stink at remembering form formulas
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  4. #4
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    Be careful of that z limit. It wouldn't be 2Pi, unless you're using polar.

    z=x^{2}+y^{2}+1

    Perhaps try this:

    \int_{0}^{3}\int_{0}^{3}\int_{0}^{x^{2}+y^{2}+1}dz  dydx

    A double integral may be better:

    \int_{0}^{3}\int_{0}^{3}[x^{2}+y^{2}+1]dydx
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  5. #5
    Moo
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    I'm not used to it, can a volume be obtained with a double integral ?
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  6. #6
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    Fornuately, for me the professor gave us a solution set

    a.) 62
    b.) 63
    c.) 64
    d.) None of these

    so if I'm in the ballpark, I'll know it. I might just try both
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  7. #7
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    Quote Originally Posted by Moo View Post
    I'm not used to it, can a volume be obtained with a double integral ?
    Well if you look at the change the only difference is he evaluated the innermost integral leaving you with two instead of three
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  8. #8
    Eater of Worlds
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    Yes, double integrals give volume. Look at choice b
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  9. #9
    Moo
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    Oh yeah indeed... I really have to learn my lessons
    Thanks !
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