Hi,

I have done the question on the picture attached

I don't believe it could be the correct answer, can someone help me please

Thanks

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- Feb 26th 2010, 09:14 AMwolfhoundV of solid rev..........
Hi,

I have done the question on the picture attached

I don't believe it could be the correct answer, can someone help me please

Thanks - Feb 26th 2010, 09:28 AMwolfhound
actually should my answer be 1114?

- Feb 26th 2010, 10:19 AMione
- Feb 26th 2010, 10:37 AMwolfhound
Silly me,

Hows it looking now please?

still wrong I think - Feb 26th 2010, 02:49 PMione
- Feb 26th 2010, 03:12 PMwolfhound
but is it a rule to make it into a fraction and keep the pi(sign) or is it ok to do like I did

- Feb 27th 2010, 08:15 AMskeeter

- Feb 27th 2010, 08:15 AMArchie Meade
Hi wolfhound.

It always helps to understand how the situation looks on a diagram.

The volume of revolution is the sum of the areas of a series of "wafer-thin" discs (an "infinite" sum of them, evaluated as their widths "deta-x" goes to zero).

The radius of the discs in this case is f(x) because the curve is being rotated around the x-axis.

the area of a disc is

All of these disc areas are integrated from x=0 to x=3.

This gives the volume of revolution.

You need to get used to how to do this notationally,

as your written work needs to be redone.

Therefore, the volume of revolution is

from x=0 to 3

I've shown the disc from an angle (hence it looks like an ellipse!) - Feb 27th 2010, 11:09 AMwolfhound
Thanks for the help, I must make more sense when I write it out..