Volume between under parabaloidSuppose I want to find the volume between z = 2 x 2 + 3 y 2 and z = 4. Is there a way finding that with a double integral? I tried to use 4 − 2 x 2 − 3 y 2 inside the integral and then convert it to polar coordinates...

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Volume between under parabaloidSuppose I want to find the volume between   z  =  2      x    2    +  3      y    2   and   z  =  4. Is there a way finding that with a double integral? I tried to use   4  −  2      x    2    −  3      y    2   inside the integral and then convert it to polar coordinates...

Dominic Paul · Accepted Answer

Step 1The volume of the solid of interest E is,      ∬          D        (  4  −  2      x    2    −  3      y    2    )  d  AWhere D is the projection of the solid E onto the xy plane. That is, the region   2      x    2    +  3      y    2    ≤  4 in the xy plane. Or,             (              x                  2                    )        2    +            (              y                              4            3                              )        2    ≤  1.To evaluate the double integral, use the modified polar transformation:  x  =      2      r  cos  ⁡  (  θ  )  y  =            4      3          r  sin  ⁡  (  θ  )Whose Jacobian is             8      3          r. Hence the volume is,   V  =      ∫          0              2      π            ∫          0              1                  8      3        r  (  4  −  4      r    2    )  d  r  d  θ  =  4  π            2      3        =      4    3    π      6  The above transformation can be thought of as combining two transformations.  X  =      x          2        Y  =      y                  4        3            Then   X  =  r  cos  ⁡  (  θ  )  Y  =  r  sin  ⁡  (  θ  )Step 2At a height of   z  ≥  0, a slice of our solid of interest (parallel to the xy plane) has elliptical area   A  (  z  )  =  π            z      2                  z      3        =      π          6        z.So the volume of the solid is,   V  =      ∫          0              4        A  (  z  )  d  z  =            8      π              6        =      4    3    π      6  Which matches the result from the double integral.

Find the volume between z=2x^2+3y^2 and z=4. Is there a way finding that with a double integral?

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