- #1
joeblow
- 71
- 0
Suppose I have a region R whose boundary extremely complicated. While it would take me hundreds of years to approximate the boundary with formulae, I can easily estimate the area of R within a desired precision. I want to find the volume of the solid of revolution of R .
My intuition told me that I should be able to integrate some function solely dependent on the area with respect to θ from 0 to 2π .
However, using what knowledge I already have, this is not the case. For instance, the volume of the solid obtained by rotating a right triangle whose base is r and height is h about the axis coninciding with the leg of length h is given by the well-known formula [tex]\frac{1}{3}\pi r^{2}h=\pi r(\frac{1}{2}rh)\cdot\frac{2}{3}=\frac{2A\pi r}{3} [/tex]. The volume of a solid of a similar solid formed by a rectangle of base r and height 1/2 h is given by the well-known fromula [tex]\pi r^{2}(\frac{1}{2}h)=\pi r(\frac{1}{2}rh)=A\pi r [/tex]. The solids are both generated by regions of equal area, yet the volumes are different by a factor of 2/3 .
My question is what parameters are involved in this process? Clearly, I am missing at least one.
My intuition told me that I should be able to integrate some function solely dependent on the area with respect to θ from 0 to 2π .
However, using what knowledge I already have, this is not the case. For instance, the volume of the solid obtained by rotating a right triangle whose base is r and height is h about the axis coninciding with the leg of length h is given by the well-known formula [tex]\frac{1}{3}\pi r^{2}h=\pi r(\frac{1}{2}rh)\cdot\frac{2}{3}=\frac{2A\pi r}{3} [/tex]. The volume of a solid of a similar solid formed by a rectangle of base r and height 1/2 h is given by the well-known fromula [tex]\pi r^{2}(\frac{1}{2}h)=\pi r(\frac{1}{2}rh)=A\pi r [/tex]. The solids are both generated by regions of equal area, yet the volumes are different by a factor of 2/3 .
My question is what parameters are involved in this process? Clearly, I am missing at least one.