I need to calculate the volume of a shape which is kind of like a truncated cone but has an elliptical base
eg. Minor axis a: 5 Major axis b: 15
and a second more round elliptical top
eg. Minor axis c: 1 Major axis d: 2
The 'vertex' ellipse is more round than the base which is more elliptical.
The height of the shape would be
h = 10
I've been looking for a formula for this but can't seem to find it ...
Can You help ?
Edit. The shape is simple enough 2 cocentric ellipses on parallel planes form the top and bottom of the shape. The walls of the shape resemble those of a truncated cone. Basically I need the formula for a truncated elliptical cone but the ratios of minor to major axis on the ellipses describing the top and bottom are different.
Please help :)
Thanks a million :)
I assume that each cross section is an ellipse and that the length of the major axis and the length of the minor axis vary linearly with the height of the cross section above the bottom.
Let $a_0$ be the length of semimajor axis at the bottom, $b_0$ the length of the semiminor axis at the bottom, $a_1$ be the length of semimajor axis at the top, $b_1$ the length of the semiminor axis at the top. Let $a(z)$ be the length of the semimajor axis at the height $z$ above the bottom and $b(z)$ the length of the semiminor axis at the height $z$ above the bottom. Then $$ a(z) = \frac{z}{h}(a_1-a_0)+a_0 \\ b(z) = \frac{z}{h}(b_1-b_0)+b_0 $$ And the volume is $$ V=\int\limits_0^h \pi\,a(z)\,b(z)\,dz $$ After some lengthy calculations, I got $$ V = \pi\,h\,\frac{2a_0b_0 + 2a_1b_1 +a_0b_1+a_1b_0}{6} $$ If you use the lengths of the complete axes instead of the semi axes, the formula is $$ V = \pi\,h\,\frac{2A_0B_0 + 2A_1B_1 +A_0B_1+A_1B_0}{24} $$