I have a three dimensional solid defined with respect to an orthonormal frame $Oxyz$ by

$$3 z^2 < ax + by < z$$

I would like to find its volume using Cavalieri's principle where the horizontal cross sections are normal to $z$ axis.

The integration tool I would like to use is one dimensional integration rather than double integrals.

Is there a numerical answer to this problem 3?

What is  the differential volume that gets multiplied by dx or dz?

There could be a number of ways to do this.

Any help is greatly appreciated.

  • $\begingroup$ numerical solution? $\endgroup$ – Narasimham Nov 4 '16 at 19:13
  • $\begingroup$ @Narasimham, What do you mean by numerical solution? Could I integrate by the z axis? $\endgroup$ – Frank Nov 4 '16 at 19:29
  • $\begingroup$ seems that you are confusing up some different concepts $\endgroup$ – G Cab Nov 4 '16 at 19:38
  • $\begingroup$ Do you agree with the new presentation I have given to your question ? $\endgroup$ – Jean Marie Nov 4 '16 at 20:46
  • $\begingroup$ Cavalieri's principle is interesting only if the areas of the cross sections are easy to compute. What is their shapes in this case ? $\endgroup$ – Jean Marie Nov 4 '16 at 20:47

If $0<c<{1\over3}$ then $3c^2<c$. It follows that for such a $c$ the plane $z=c$ intersects your solid in the infinite parallel strip $$3c^2<ax+by<c\tag{1}$$ of positive width, hence infinite area. Using Cavalieri's principle (or Fubini's theorem) it follows that this solid has infinite volume.

Concerning volumes of bodies as integrals: If you are given a finite "body" $B\subset{\mathbb R}^3$ you can project it to the $z$-axis and obtain an interval $[p,q]$, and you can intersect it with planes $z={\rm const.}$ and obtain plane slices $$B_z:=\{(x,y)\>|\>(x,y,z)\in B\}$$ of area ${\rm area}(B_z)\geq 0$. Now by Fubini's theorem one has $${\rm vol}(B)=\int_B 1\>{\rm d}(x,y,z)=\int_p^q\int_{B_z}1\> {\rm d}(x,y)\>dz=\int_p^q {\rm area}(B_z)\>dz\ .$$ In the case at hand $B_z$ is empty unless $3z^2<z$, and this is the case iff $0<z<{1\over3}$. We therefore have $p=0$, $q={1\over3}$. But ${\rm area}(B_z)=\infty$ $(p<z<q)$, since $(1)$ defines an infinite parallel strip of positive width, whatever $(a,b)\ne(0,0)$ for these $z$. It follows that ${\rm vol}(B)=\infty$ not only per inspection, but also per integral.

  • $\begingroup$ Your answer is what I was hoping to see. Could I ask what was your integral and the limits of integration? Thank you. $\endgroup$ – Frank Nov 5 '16 at 0:21
  • $\begingroup$ Blattter, Suppose a = 1 and b = -1 , would we still have infinite area? How would I calculate area as a function of z where a = 1 and b = -1? Why is p = 0 and q = 1/3? Thank you. $\endgroup$ – Frank Nov 5 '16 at 12:55
  • $\begingroup$ Suppose a = 1 and b = -1 , would we still have infinite area? How would I calculate area as a function of z where a = 1 and b = -1? Why is p = 0 and q = 1/3? Thank you. $\endgroup$ – Frank Nov 5 '16 at 13:01
  • $\begingroup$ @Frank: Given that you are still trying to understand Cavalieri's principle and all that it is not very helpful to provide a degenerate example as working piece. $\endgroup$ – Christian Blatter Nov 5 '16 at 16:56
  • $\begingroup$ Why is this a degenerate case? Thank you. $\endgroup$ – Frank Nov 6 '16 at 2:48

Assuming Monge form of surface and using Green's theorem find area enclosed from closed curve boundary. Then integrate functionally with respect to z.


Monge form is like

$$ z= f(x,y) $$

e.g., like

$$ z= (x/a)^2 + (y/b)^2 $$

Rather than repeat it, referring to anonymous's answer here among others in a particular example to compute cross-section $A$ of an ellipse using line integral. A suitable $single $ parameter is needed to link $(x,y)$.

$$ A =\frac{1}{2} \int\limits_{\partial{D}} xdy - ydx = ... $$

and then the usual

$$ V = \int A\cdot dz $$

Suggested this as you wanted a separate evalution for each level cross-section.

An added advantage is the possibility of using any curvilinear coordintes, not just Cartesian, here we are using variables $(x,y)$ in place of more general curvilinear parameters $(u,v). $

  • $\begingroup$ I think your answer is great. Please show us how you used Green's theorem and integrated with respect to z. What is the Monge form of surface ? Thank you. $\endgroup$ – Frank Nov 5 '16 at 0:14

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