The proof of the volume of the truncated cone

[Jasty]

Please I need a respectable proof how to get the volume of the truncated cone. I need it really quick. So please could you help me. No numbers just "the method" how to get that formula. Thanks.

 

[HallsofIvy]

Starting from where? If you can use the formula for volume of a cone, its trivial.

 

[mathman]

Please I need a respectable proof how to get the volume of the truncated cone. I need it really quick. So please could you help me. No numbers just "the method" how to get that formula. Thanks.

If the truncation is parallel to the base, Halls of Ivy is correct. Otherwise it is some what more complicated.

 

[HallsofIvy]

Yeah, I'll leave that to mathman!

 

[Jasty]

I mean the most basic conditions. I can imagine the whole cone and than remove the top. Is there any other way how to prove it?

 

[HallsofIvy]

Let C be a truncated, right circular cone with height H, upper radius R₁ and lower radius R₂. Set it up on a coordinate system with the center of the base at (0,0,0), and center of the top at (0,0,H). Looking at it from the side, so that you see the xz-plane, you see a "trapezoid" with one side starting at (R2,0,0) and ending at (R1,0,H). Since any nonvertical line in the xz-plane can be written in the form z= Ax+ B. you must have 0= R2A+ B and H= R1A+ B. Subtracting the first from the second, H= (R1-R2)A so A= H/(R1-R2), B= -R2A so B= -R2H/(R1-R2). The equation of the line is z= H(x-R2)/(R1-R2) or you can write it x= (R1-R2)z/H+ R2.

Now imagine the entire cone, divided into thin disks: each has thickness "dz" and radius, x= (R1-R2)z/H+ R2 so area $\pi [(R1-R2)z/H+ R2)^2$ and volume $\pi [(R1-R2)z/H+ R2)^2dx$. To find the entire volume integrate that from z= 0 to z= H.

 

[Jasty]

ok thanks all.

 

[trambolin]

Can we do this by using a deformation matrix and then using the regular formula if it is not parallel to the base? Just out of curiosity, couldn't see it right away

 

[socalmusic]

Now imagine the entire cone, divided into thin disks: each has thickness "dz" and radius, x= (R1-R2)z/H+ R2 so area $\pi [(R1-R2)z/H+ R2)^2$ and volume $\pi [(R1-R2)z/H+ R2)^2dx$. To find the entire volume integrate that from z= 0 to z= H.

dz instead of dx?

 

[HallsofIvy]

Yes, of course. Thanks.