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What is the volume of the structure bounded by the regions x^2+2y^2+z≤16, 0≤x≤2, 0≤y≤2 and z≥0?

  • Math -

    The volume to be integrated is bounded by the planes
    0≤x≤2
    0≤y≤2
    z=0
    and the surface
    z=16-x²-2y².

    Thus the volume is given by the triple integral:

    V=∫∫∫z dz dx dy

    Where z is evaluated from 0 to 16-x²-2y²,
    x and y are both evaluated from 0 to 2.

    Thus
    V=∫∫(16-x²-2y²)dx dy
    =∫(88-12y²)/3 dy
    =48

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