volume of a frustrum of a pyramid

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mares09

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Use intergration techniques to find the formula for the volume of a frustrum of a pyramid with square base of side length b and square top with side length of a. the frustrum has height, h.

i cut the pyramid in 4 pieces and if i rotate it against the x-axis then the limits will be from 0 to h. then i foud the slope m=(b-a)/2h. I used the slope to find the line equation y=(b-a/2h)x+(a/2). there fore i suposed that what i have to intergrate is Pi((b-a/2h)x+(a/2))^2.

im stuck since i dont know how to get the intergral of that.
 
mares09 said:
Use intergration techniques to find the formula for the volume of a frustrum of a pyramid with square base of side length b and square top with side length of a. the frustrum has height, h.

i cut the pyramid in 4 pieces and if i rotate it against the x-axis then the limits will be from 0 to h. then i foud the slope m=(b-a)/2h. I used the slope to find the line equation y=(b-a/2h)x+(a/2). there fore i suposed that what i have to intergrate is Pi((b-a/2h)x+(a/2))^2.

im stuck since i dont know how to get the intergral of that.
Click to expand...

It is very similar to the cap of the sphere problem - that Galactus solved for you.

Use disk method. Put the axis of the cone on y-axis (or x-axis).

Then use Riemannian sum as before.
 
You will find the volume of a cone frustum, but not for a square pyramid, by using that method. For a square pyramid, use similar triangles. I assume you meant a square pyramid?.
 
galactus said:
You will find the volume of a cone frustum, but not for a square pyramid, by using that method. For a square pyramid, use similar triangles. I assume you meant a square pyramid?.
Click to expand...

You could do it that way - only in this case the disks would be square parallelopepid of square base and height of 'dz'.
 
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