How do you find the volume of the solid obtained by rotating the region bounded by the curves y=x^2+1y=x2+1 and y=-x+3y=−x+3 rotated around the x-axis?
1 Answer
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=π∫1−2((x−3)2−(x2+1)2)dx
= pi int_-2^1 ((x^2-6x+9) - (x^4+2x^2+1)) dx=π∫1−2((x2−6x+9)−(x4+2x2+1))dx
= pi int_-2^1 (-x^4-x^2-6x+8) dx=π∫1−2(−x4−x2−6x+8)dx
= pi[-x^5/5 -x^3/3-3x^2+8x]_-2^1=π[−x55−x33−3x2+8x]1−2
= pi[(-1/5-1/3-3+8)-(32/5+8/3-12-16)]=π[(−15−13−3+8)−(325+83−12−16)]
=pi[ -33/5-9/3+5+12+16]=π[−335−93+5+12+16]
= pi[-66/10 -3 +5+12+16]=π[−6610−3+5+12+16]
= pi [-6.6 + 30]=π[−6.6+30]
= 23.4 pi=23.4π