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In summary, the truncated cone can be found using the formula for volume of a cone if the truncation is parallel to the base. Otherwise it is a little more complicated.f

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

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

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Yeah, I'll leave **that** to mathman!

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Let C be a truncated, right circular cone with height H, upper radius R_{1} and lower radius R_{2}. 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 [itex]\pi [(R1-R2)z/H+ R2)^2[/itex] and volume [itex]\pi [(R1-R2)z/H+ R2)^2dx[/itex]. To find the entire volume integrate that from z= 0 to z= H.

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

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ok thanks all.

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

dz instead of dx?

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Yes, of course. Thanks.

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