Just for the record
-Dan
Let R be the region enclosed by the graph of y=√x-1 and x=10 and the x axis
b) find the volume of the solid generated when R is revolved about the horizontal line y=3
I thought it would be
3.14 X the integral from 1 to 10 of (3)² - (3-√x-1)²
according to my answer key, it is instead
(3)² - (√x-1 -3)²
I really don't get this. 3 is above √x-1 so why wouldn't you do 3 - √x-1 instead of √x-1 -3 (which would give a negative number)
Please help guys, maybe my answer key is wrong?
Thanks.