Finding the volume using triple integralsI am curious how I would describe the region y 2 = x bounded by the lines y = 1, x = 1, x = 0 and y = 0 revolved about the y axis in 3 dimensions in order to find the volume in terms of triple integrals. (The solid looks like a cone.) This can easily be done with 1 integral using the "disk" or "shell method, however suppose I don't know those methods and I wanted to find the volume from a very general idea (i.e. triple integrals). What would be the limits of integration in the z plane? Would the limits in the x and y planes simply be from 0 to 1?

Question

Finding the volume using triple integralsI am curious how I would describe the region       y    2    =  x bounded by the lines   y  =  1,   x  =  1,   x  =  0 and   y  =  0 revolved about the y axis in 3 dimensions in order to find the volume in terms of triple integrals. (The solid looks like a cone.) This can easily be done with 1 integral using the &quot;disk&quot; or &quot;shell method, however suppose I don&#039;t know those methods and I wanted to find the volume from a very general idea (i.e. triple integrals). What would be the limits of integration in the z plane? Would the limits in the x and y planes simply be from 0 to 1?

muilasqk · Accepted Answer

Step 1Let me clarify that the volume to be found is of the solid obtained by revolving around the y-axis the region bounded by   y  =  1,   x  =  0 and   y  =      x   (hence a cone-like shape).Cross-sections of the solid perpendicular to the y-axis are circles. If the cutting plane is   y  =  a then the circle is described by the equation       x    2    +      z    2    =  (      a    2        )    2  .The outermost layer of our triple integral in Cartesian coordinates will be with respect to y, which has range [0,1]:      ∫    0    1    …  d  yFor a given y, since the circular cross-section has radius       y    2   it makes sense to have the middle integral with respect to x and ranging over   [  −      y    2    ,      y    2    ], the circle&#039;s diameter:       ∫    0    1        ∫          −              y        2                            y        2              …  d  x     d  y.Step 2Given some x and y we must have   −            y      4        −          x      2        ≤  z  ≤            y      4        −          x      2       if the point (x,y,z) is to lie in the solid. The last integral with respect to z has these limits, and the integrand is just 1 since we are finding volume:  V  =      ∫    0    1        ∫          −              y        2                            y        2                  ∫          −                        y          4                −                  x          2                                              y          4                −                  x          2                      1     d  z     d  x     d  yEvaluating these integrals from the inside out we have  V  =      ∫    0    1        ∫          −              y        2                            y        2              2            y      4        −          x      2           d  x     d  y  V  =      ∫    0    1    π      y    4       d  y  =      π    5  This answer matches the volumes obtained from the shell method (  2  π      ∫    0    1    x  (  1  −      x    )     d  x) or the disc method (  π      ∫    0    1        y    4       d  y).Just for reference, here is a triple integral for the volume of the same solid in cylindrical coordinates:  V  =      ∭    R    r     d  r     d  φ     d  y    =      ∫    0    1        ∫    0          2      π            ∫    0                  y        2              r     d  r     d  φ     d  y  =      π    5

How I would describe the region y^2=x bounded by the lines y=1, x=1, x=0 and y=0 revolved about the y axis in 3 dimensions in order to find the volume in terms of triple integrals.

Open question

Answer & Explanation

New Questions in High school geometry