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Math Help - volume of a region bounded by two surfaces

  1. #1
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    volume of a region bounded by two surfaces

    I have this problem

    find the voulme of the solid bounded above by surface z=50-4(y^2)
    and below by z=9(y^2)+14

    I only know how to calculate voules of solid when the side parameters are given(eg. bounded on sides by cylinder x^2+y^2=1)
    but i dont know how to apply the double integral for solid bounded by two surfaces.

    I would be grateful if someone can answer how to do it(or point to some website where it is clearly shown)

    thanks in advance
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  2. #2
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    Quote Originally Posted by Dili View Post
    I have this problem

    find the voulme of the solid bounded above by surface z=50-4(y^2)
    and below by z=9(y^2)+14

    I only know how to calculate voules of solid when the side parameters are given(eg. bounded on sides by cylinder x^2+y^2=1)
    but i dont know how to apply the double integral for solid bounded by two surfaces.

    I would be grateful if someone can answer how to do it(or point to some website where it is clearly shown)

    thanks in advance
    Try drawing a rough sketch of things.

    The volume is given by:

    V = \int \int_{R_{xy}} (50 - 4y^2) - (9y^2 + 14) \, dx \, dy \, where \, R_{xy}\, is the region of the xy-plane defined by the infinite rectangular strip -\frac{6}{\sqrt{13}} \leq y \leq \frac{6}{\sqrt{13}} and -\infty < x < \infty.

    Note: the bounding values of y are found by solving 50 - 4y^2 = 9y^2 + 14.

    Unsurprisingly the volume is infinite.
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