Find volume under given contraints on the Cartesian plane.The constrains are given as x 2 + y 2 + z 2 ⩽ 64 , x 2 + y 2 ⩽ 16 , x 2 + y 2 ⩽ z 2 , z ⩾ 0 .With the goal of finding the Volume. Personally, I have trouble interpreting the constrains in terms of integrals to find the Volume. However, logically z can be maximum of 8 and x or y no larger than 4.

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Find volume under given contraints on the Cartesian plane.The constrains are given as      x    2    +      y    2    +      z    2    ⩽  64  ,        x    2    +      y    2    ⩽  16  ,        x    2    +      y    2    ⩽      z    2    ,    z  ⩾  0    .With the goal of finding the Volume. Personally, I have trouble interpreting the constrains in terms of integrals to find the Volume. However, logically z can be maximum of 8 and x or y no larger than 4.

Rebecca Benitez · Accepted Answer

Step 1The problem is simpler if you convert to cylindrical coordinates. In this coordinate system, we have   ρ  =            x      2        +          y      2      ,   z  =  z, and   ϕ  =      arctan        (          y      x        )  . An important point is that   ρ is taken to be positive, while   ϕ dictates direction completely.Under this system, we have the bounds       ρ    2    +      z    2    ≤  64  ,      ρ    2    ≤  16  ,      ρ    2    ≤      z    2    ,  z  ≥  0. From these inequalities, we have that z may range anywhere from 0 to 8, and there is no restriction on   ϕ (so it ranges from 0 to   2  π). Combining the inequalities involving   ρ, we get that as z ranges from 0 to 4,   ρ ranges from 0 to z. Then, as z ranges from 4 to       48  ,   ρ ranges from 0 to 4. Finally, as z ranges from       48   to 8, ρ ranges from 0 to       64    −          z      2      .Step 2We can write each of these three volumes as triple integrals. The volume element in cylindrical coordinates, dV, is   ρ    d  ρ    d  z    d  ϕ. So the first will be      ∫    0          2      π            ∫    0    4        ∫    0    z    1  ×    ρ    d  ρ    d  z    d  ϕStep 3Similarly the second will be      ∫    0          2      π            ∫    4                  48                  ∫    0    4    1  ×    ρ    d  ρ    d  z    d  ϕ and the third will be       ∫    0          2      π            ∫                  48              8        ∫    0                  64        −                  z          2                      1  ×    ρ    d  ρ    d  z    d  ϕ.Solve each triple integral from the inside out, add the three together, and you&#039;ll have your answer.

The constrains are given as x^2+y^2+z^2 leq 64, x^2+y^2 leq 16, x^2+y^2 leq z^2, z geq 0. With the goal of finding the Volume.

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