• January 13th 2013, 03:44 PM
skinsdomination09
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

(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)

Thanks.
• January 13th 2013, 04:29 PM
topsquark
Just for the record $(3 - \sqrt{x - 1})^2 = ( \sqrt{x - 1} - 3)^2$

-Dan
• January 13th 2013, 04:47 PM
skinsdomination09
Quote:

Originally Posted by topsquark
Just for the record $(3 - \sqrt{x - 1})^2 = ( \sqrt{x - 1} - 3)^2$

-Dan

well that's embarassing.

Thanks.
• January 14th 2013, 06:00 AM
topsquark