Find the volume of the regions enclosed by z = x 2 + y 2 − 2 and z = 30 − x 2 − y 2 I set up a triple integral with the bounds of the inmost as x 2 + y 2 − 2 to 30 − x 2 − y 2 . The two outer integrals both had the bounds from -4 to 4. When I solved it I got 1024 as the volume, but this isn't correct. Can someone please show me the steps to finding the bounds, and the correct answer?

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Find the volume of the regions enclosed by   z  =      x    2    +      y    2    −  2 and   z  =  30  −      x    2    −      y    2  I set up a triple integral with the bounds of the inmost as       x    2    +      y    2    −  2 to   30  −      x    2    −      y    2  . The two outer integrals both had the bounds from -4 to 4. When I solved it I got 1024 as the volume, but this isn&#039;t correct. Can someone please show me the steps to finding the bounds, and the correct answer?

wietselau · Accepted Answer

Step 1This is where it goes wrong:The two outer integrals both had the bounds from -4 to 4.If you let x and y run from -4 to 4, you are integrating over a square in the xy-plane, but you want to integrate over the projection of the given region onto the xy-plane. Observe that the two surfaces intersect at:      x    2    +      x    2    −  2  =  30  −      x    2    −      y    2      ⟺        x    2    +      y    2    =  16and this is a circle of radius 4, centered at the origin.Step 2You can keep   x  :  −  4  →  4 but then you need y as a function of x to integrate over this circle (or vice versa). It works, but the calculations can become a bit messy.

crazygbyo · Accepted Answer

Step 1Set   z  =  z in the two equations to find the circle of intersection:      x    2    +      y    2    −  2  =  30  −      x    2    −      y    2      2      x    2    +  2      y    2    =  32        x    2    +      y    2    =  16  ∭    d  z    d  y    d  x    ∬  z    d  y    d  x      ∫          −      4              4            ∫          −              16        −                  x          2                                    16        −                  x          2                      32  −  2      x    2    −  2      y    2      d  y    d  xStep 2Now we could keep cranking through that in Cartesian, but I think a conversion to cylindrical coordinates will make this easier.  x  =  r  cos  ⁡  θ    y  =  r  sin  ⁡  θ    d  y    d  x  =  r    d  r    d  θ      ∫    0          2      π            ∫          0              4        2  (  16  −      r    2    )  r    d  r    d  θ    u  =  (  16  −      r    2    )  ,  d  u  =  −  2  r    d  r        ∫    0          2      π            ∫          16              0        −  u    d  u    d  θ        ∫    0          2      π            16    2      d  θ    512  π

Find the volume of the regions enclosed by z=x^2+y^2-2 and z=30-x^2-y^2.

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