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Math Help - Triple Integral

  1. #1
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    Triple Integral

    Find the volume of the region.

    The region between the cylinder z=y^2 and the xy-plane that is bounded by the planes x=0,x=1,y=-1,y=1

    When I graph this it looks like a halfpipe(skateboarding) extending in the positive x direction.

    Could someone please help me determine the limits of integration.

    my attempt:

    int 0..1 int x^2+1.. x^2-1 int 0.. y^2 dz dy dx
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  2. #2
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    That's just a double integral. You're integrating from the x-y plane up to the surface that is z=f(x,y)=y^2. That's the z-direction. And we're integrating that over the foot-print in the x-y plane that's just the two unit squares underneath the surface which because it's symmetric, just multiply by two, the volume underneath the unit square as x goes from 0 to 1 and y goes from 0 to one or:

    V=2\int_0^1\int_0^1 y^2 dydx

    or if you wanna' be a purist, a triple integral:

    V=2\int_0^1\int_0^1\int_0^{y^2} dzdydx

    but same dif.
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  3. #3
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    Quote Originally Posted by shawsend View Post
    That's just a double integral. You're integrating from the x-y plane up to the surface that is z=f(x,y)=y^2. That's the z-direction. And we're integrating that over the foot-print in the x-y plane that's just the two unit squares underneath the surface which because it's symmetric, just multiply by two, the volume underneath the unit square as x goes from 0 to 1 and y goes from 0 to one or:

    V=2\int_0^1\int_0^1 y^2 dydx

    or if you wanna' be a purist, a triple integral:

    V=2\int_0^1\int_0^1\int_0^{y^2} dzdydx

    but same dif.
    I will take the purist direction. The only reason is that it is under the triple integral section. Thanks for the help.
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