Find the volume of a solid limited by 64 x 2 − 4 y 2 + 16 z 2 = 0 and y = 1.

Question

Find the volume of a solid limited by   64      x    2    −  4      y    2    +  16      z    2    =  0 and   y  =  1.

Bernard Scott · Accepted Answer

Explanation:I suppose that it&#039;s also limited by   y  =  0. We have elliptical cone, which volume is   V  =      1    3    π  a  b  h. Where the   h  =  1, because volume is bounded between   y  =  0 and   y  =  1. To find a and b we need to put   y  =  1 to equation:  64      x    2    +  16      z    2    =  4  16      x    2    +  4      z    2    =  1Hence we get   a  =  1      /    4 and   b  =  1      /    2. Therefore,   V  =      π    24  .

pramrok62 · Accepted Answer

Explanation:So your solid is limited by   y  =  1 and   y  =  16      x    2    +  4      z    2  . You can set this up as a triple integral (in cylindrical coordinates would be ideally easy), bounding   16      x    2    +  4      z    2    ≤  y  ≤  1 and integrate over the ellipse in the xz-plane.

Find the volume of a solid limited by 64x^2-4y^2+16z^2=0 and y=1.

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