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Math Help - Triple integral, doubt

  1. #1
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    Triple integral, doubt

    Hi there, I have this iterated integral \displaystyle\int_{0}^{2}\displaystyle\int_{0}^{2-y}\displaystyle\int_{0}^{4-y^2}dxdzdy, and the thing is, does it define a solid? because I think that as it is given it doesn't, but I'm not sure. I think that the cylindric paraboloid never cuts the x axis, and thats why I think this integral doesn't define any solid.
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  2. #2
    MHF Contributor FernandoRevilla's Avatar
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    The integral has sense in the region:

    T\equiv{}\begin{Bmatrix}0\leq{y}\leq{2}\\ 0\leq{z}\leq{2-y} \\ 0\leq{x}\leq{4-y^2}\end{matrix}

    Besides:

    \displaystyle\iiint_{T}dV=\displaystyle\int_{0}^{2  }\displaystyle\int_{0}^{2-y}\displaystyle\int_{0}^{4-y^2}dxdzdy=\displaystyle\int_{0}^{2}dy\displaystyl  e\int_{0}^{2-y}dz\displaystyle\int_{0}^{4-y^2}dx

    Regards.
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  3. #3
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    Quote Originally Posted by Ulysses View Post
    Hi there, I have this iterated integral \displaystyle\int_{0}^{2}\displaystyle\int_{0}^{2-y}\displaystyle\int_{0}^{4-y^2}dxdzdy, and the thing is, does it define a solid? because I think that as it is given it doesn't, but I'm not sure. I think that the cylindric paraboloid never cuts the x axis, and thats why I think this integral doesn't define any solid.
    What? x= 4- y^2 cuts the x-axis when x= 4.
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  4. #4
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    I'm starting to figure it out now. As the concavity is pointing at the x axis I thought it wouldn't cut ever the other integrands, but now I think I'm seeing it, the base would be at the yz plane I think. I'll try again.

    Thank you both.
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