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# tobypuff asked a qn regarding volume from integration

• Repost from tobypuff : Would appreciate some help with part (iii) in the link http://imgur.com/a/Imf3U

How do I go about finding the volume generated by R?

π∫(e^√y+1)^2 dy from 0 to -1 minus π∫(e^-√y+1)^2 dy from 0 to -1

As from the pic, volume generated by R when rotated 2π about y axis, can be obtained by vol generated from outer curve minus vol generated by inner curve.

Welcome Further discussions.

• Hi, thanks very much for the reply! You have the correct integral set up but  I just can't seem to figure out why the outer radius is e^√y+1 and the inner radius is e^-√y+1 since i don't know what these functions look like graphically, unless its possible to plot functions of the form x = g(y) into my graphing calculator?

• One way to see is that the outer radius is of a value larger than the inner radius.

So e^√y+1 being larger than e^-√y+1 for all values of y between -1 and 0, e^√y+1 is the outer radius.

• Thanks eagle! That clarifies it for me :)