Finding the bounds to solve for the volume using cylindrical coordinatesMy goal is to set up the triple integral that will solve for the volume inside the sphere x 2 + y 2 + z 2 = 2 z and above the paraboloid x 2 + y 2 = z using cylindrical coordinates.Upon solving, I know that the intersection of the sphere and the paraboloid is r = 1 and the bounds for θ is 0 ≤ θ ≤ 2 π.However, I am uncertain for the bounds of the z variable. I solve for the value of z in converted equation of the sphere r 2 + z 2 = 2 z. By quadratic formula, I got the value of z = 2 ± 4 − 4 r 2 2 .Should I only include z = 1 2 ( 2 + 4 − 4 r 2 ) as one of the bounds in the z variable? Or there will be a split integral? My yielding triple integral in this case if I only include z = 1 2 ( 2 + 4 − 4 r 2 ) would be ∫ 0 2 π ∫ 0 1 ∫ r 2 1 2 ( 2 + 4 − 4 r 2 ) r d z d r d θ

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Finding the bounds to solve for the volume using cylindrical coordinatesMy goal is to set up the triple integral that will solve for the volume inside the sphere       x    2    +      y    2    +      z    2    =  2  z and above the paraboloid       x    2    +      y    2    =  z using cylindrical coordinates.Upon solving, I know that the intersection of the sphere and the paraboloid is   r  =  1 and the bounds for   θ is   0  ≤  θ  ≤  2  π.However, I am uncertain for the bounds of the z variable. I solve for the value of z in converted equation of the sphere       r    2    +      z    2    =  2  z. By quadratic formula, I got the value of   z  =            2      ±              4        −        4                  r          2                      2  .Should I only include   z  =      1    2    (  2  +      4    −    4          r      2        ) as one of the bounds in the z variable? Or there will be a split integral? My yielding triple integral in this case if I only include   z  =      1    2    (  2  +      4    −    4          r      2        ) would be       ∫    0          2      π            ∫    0    1        ∫                  r        2                            1        2            (      2      +              4        −        4                  r          2                    )        r  d  z  d  r  d  θ

kliersel12g · Accepted Answer

Step 1Yes you are correct. Paraboloid   z  =      x    2    +      y    2    =      r    2   and Sphere       r    2    +      z    2    =  2  z.To find z at intersection, plug in       r    2    =  z from equation of paraboloid into the equation of sphere. We get       z    2    −  z  =  0    ⟹    z  =  0  ,  1. So at intersection above origin,   z  =  1. We also find that at intersection   r  =  1 which is the radius of the sphere too.Now as you mentioned, from the equation of the sphere,       r    2    +  (  z  −  1      )    2    =  1    ⟹    z  =  1  ±      1    −          r      2      Step 2As the sphere is centered at (0,0,1), for any   0  ≤  r  &amp;lt;  1, there are two z values on sphere - one below   z  =  1     ,  (  z  =  1  −      1    −          r      2        ) and one above   z  =  1     ,  (  z  =  1  +      1    −          r      2        ). The intersection of paraboloid and sphere is also at   z  =  1, z is bound below by paraboloid and above by sphere for   0  ≤  r  ≤  1. So the point that you should take on sphere as upper bound is   z  =  1  +      1    −          r      2      . Hence it should be,             ∫      0              2        π                    ∫      0      1              ∫                        r          2                            1        +                  1          −                      r            2                                r         d    z         d    r         d    θ  .

My goal is to set up the triple integral that will solve for the volume inside the sphere x^2+y^2+z^2=2z and above the paraboloid x^2+y^2=z using cylindrical coordinates.

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