Finding volume of hyperboloid bounded by two planesI want to find the volume bounded by hyperboloid x 2 a 2 + y 2 b 2 − z 2 c 2 = 1 and the planes z = − c , z = c.I do not know whether should I use the cylindirical coordinates or spherical coordinates.At first, I am thinking to set x = a u , y = b v , z = c w and now we have u 2 + v 2 − w 2 = 1. Jacobian of this transformation is abc. I guess that after this change of variables If I take a cross section then it will give me a circle on uv-plane but I do not know the reason.If we consider the part of the volume that is inside the first octant, my intuition says that the desired volume is ∫ 0 1 ∫ 0 1 − v 2 ∫ 0 u 2 + v 2 − 1 g ( u , v , w ) a b c × d w d u d v and I am not sure about g...To sum up, I appreciate if you could explain me what is going on exactly in a basic way.

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Finding volume of hyperboloid bounded by two planesI want to find the volume bounded by hyperboloid                                                               x            2                                                                                        a            2                                +                                                              y            2                                                                                        b            2                                −                                                              z            2                                                                                        c            2                                =  1 and the planes   z  =  −  c  ,  z  =  c.I do not know whether should I use the cylindirical coordinates or spherical coordinates.At first, I am thinking to set   x  =  a  u  ,  y  =  b  v  ,  z  =  c  w and now we have       u    2    +      v    2    −      w    2    =  1. Jacobian of this transformation is abc. I guess that after this change of variables If I take a cross section then it will give me a circle on uv-plane but I do not know the reason.If we consider the part of the volume that is inside the first octant, my intuition says that the desired volume is             ∫              0                    1                    ∫              0                    1        −                  v          2                            ∫              0                              u          2                +                  v          2                −        1              g    (    u    ,    v    ,    w    )    a    b    c    ×    d    w    d    u    d    v   and I am not sure about g...To sum up, I appreciate if you could explain me what is going on exactly in a basic way.

plomet6a · Accepted Answer

Step 1I like your change of coordinates step. So it&#039;s enough to show that the volume enclosed by the “standard” of one sheet       x    2    +      y    2    −      z    2    =  1  ,        −  1  ≤  z  ≤  1 is             8      π        3  .For this, use cylindrical coordinates. The curved surface has equation       r    2    −      z    2    =  1, and the enclosed solid can be represented as       {    (    r    ,    θ    ,    z    )    :    −    1    ≤    z    ≤    1    ,         0    ≤    θ    ≤    2    π    ,         0    ≤    r    ≤          1      +              z        2              }  .Therefore, the volume is                               ∫                      −            1                    1                          ∫          0                      2            π                                    ∫          0                      1            +                          z              2                                      r                d        r                d        θ                d        z        =                              8            π                    3                     as desired.Step 2Another user suggested, but unfortunately deleted, a “Calc II” solution where you realize this as a solid of revolution. It&#039;s the curve       z    2    −      x    2    =  1 revolved around the z-axis, between   z  =  −  1 and   z  =  1. A slice of the solid at a constant z-coordinate is a disk of radius       x    2  . Since       x    2    =  1  +      z    2  , the area of that disk is   π  (  1  +      z    2    ).Therefore the volume is       ∫          −      1        1    π  (  1  +      z    2    )    d  z  =            8      π        3  .In fact,       ∫    0          2      π            ∫    0                  1        +                  z          2                      r    d  r    d  θ  =  π  (  1  +      z    2    ), so this is no coincidence.

Find the volume bounded by hyperboloid x^2/a^2+y^2/b^2-z^2/c^2=1 and the planes z=-c, z=c.

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