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?
Flyinggrenade's suggested solution :
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 :)