If #f (x)= x^2+ 1# , how do you find the volume of the solid generated by revolving the region under graph of f from x=-1 to x=1 about the x- axis?
1 Answer
Oct 21, 2017
11.73 (2. d.p.)
Explanation:
First we need to integrate
So we have:
We have:
Plugging in 1 and -1 for our upper and lower bounds:
11.73 (2. d.p.)