Finding the volume of a region below a plane and above a paraboloid (Triple integrals)I have to find the volume below the plane z = 3 − 2 y and above the paraboloid z = x 2 + y 2 .Integrating by z first, it looks like the "arrow" I draw parallel to z-axis enters the region at z = x 2 + y 2 and exits the region at z = 3 − 2 y. So ∫ x 2 + y 2 3 − 2 y 1 d z.How I am supposed to find the other two integrals?

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Finding the volume of a region below a plane and above a paraboloid (Triple integrals)I have to find the volume below the plane   z  =  3  −  2  y and above the paraboloid   z  =      x    2    +      y    2  .Integrating by z first, it looks like the &quot;arrow&quot; I draw parallel to z-axis enters the region at   z  =      x    2    +      y    2   and exits the region at   z  =  3  −  2  y. So       ∫                  x        2            +              y        2                    3      −      2      y        1  d  z.How I am supposed to find the other two integrals?

Malierb6 · Accepted Answer

Explanation:Set       x    2    +      y    2    =  3  −  2  y to find the intersection in the xy-plane. You can then get the limits for x by solving for it in the above equation, or you can complete the square to get       x    2    +  (  y  +  1      )    2    =  4 and solve for y.

Find the volume below the plane z=3-2y and above the paraboloid z=x^2+y^2.

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