How do you find the volume of the solid obtained by rotating the region bounded by the curves #y=x^2+1# and #y=-x+3# rotated around the x-axis?
1 Answer
Jun 17, 2015
I got area is
Explanation:
The curves:
On the interval
We'll use washers and find
#= pi int_-2^1 ((x-3)^2 - (x^2+1)^2) dx#
#= pi int_-2^1 ((x^2-6x+9) - (x^4+2x^2+1)) dx#
#= pi int_-2^1 (-x^4-x^2-6x+8) dx#
#= pi[-x^5/5 -x^3/3-3x^2+8x]_-2^1#
#= pi[(-1/5-1/3-3+8)-(32/5+8/3-12-16)]#
#=pi[ -33/5-9/3+5+12+16]#
#= pi[-66/10 -3 +5+12+16]#
#= pi [-6.6 + 30]#
#= 23.4 pi#