Finding volume inside a paraboloid and inside a sphereI want to find the volume inside the paraboloid z = x 2 + y 2 and inside the sphere x 2 + y 2 + z 2 = 12.So, I wrote x = r cos ⁡ t , y = r sin ⁡ t, find out 0 ≤ r ≤ 3 and 0 ≤ 2 π. We also have z = r 2 the paraboloid and z = 12 − r 2 . Now, why is the desired volume is given as: ∫ 0 2 π ∫ 0 3 ( 12 − r 2 − r 2 ) r d r d t ? Why do we subtract two z values in this way and not use just one of them?

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Finding volume inside a paraboloid and inside a sphereI want to find the volume inside the paraboloid   z  =      x    2    +      y    2   and inside the sphere       x    2    +      y    2    +      z    2    =  12.So, I wrote   x  =  r  cos  ⁡  t  ,  y  =  r  sin  ⁡  t, find out   0  ≤  r  ≤      3   and   0  ≤  2  π. We also have   z  =      r    2   the paraboloid and   z  =      12    −          r      2      . Now, why is the desired volume is given as:            ∫              0                    2        π                    ∫              0                              3                      (          12      −              r        2              −          r      2        )    r            d    r    d    t    ?  Why do we subtract two z values in this way and not use just one of them?

inmholtau5 · Accepted Answer

Step 1The region is bounded above by       z    1    =      12    −          r      2       and bounded below by       z    2    =      r    2  . That is why we subtract them.Step 2For example, you might want to consider a simpler problem. Suppose you want to compute the area between   y  =      x    2   and   y  =  x where x is in between 0 and 1. We would evaluate this as       ∫    0    1    (  x  −      x    2    )    d  x.

Find the volume inside the paraboloid z=x^{2}+y^{2} and inside the sphere x^{2}+y^{2}+z^{2}=12.

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