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
Jun 17, 2015

I got area is 23.4 pi23.4π. The "How" has no one sentence answer -- see below.

Explanation:

The curves: y=x^2+1y=x2+1 and y=-x+3y=x+3 intersect at -22 and at 11.
On the interval [-2, 1}[2,1}, the value of -x+3x+3 is greater than that of x^x+1xx+1. (The line is above the parabola.)

We'll use washers and find int (pi R^2 - pi r^2) dx(πR2πr2)dx (Where the larger radius is RR and the smaller rr.

V = int_-2^1 (pi(-x+3)^2 - pi(x^2+1)^2) dxV=12(π(x+3)2π(x2+1)2)dx

= pi int_-2^1 ((x-3)^2 - (x^2+1)^2) dx=π12((x3)2(x2+1)2)dx

= pi int_-2^1 ((x^2-6x+9) - (x^4+2x^2+1)) dx=π12((x26x+9)(x4+2x2+1))dx

= pi int_-2^1 (-x^4-x^2-6x+8) dx=π12(x4x26x+8)dx

= pi[-x^5/5 -x^3/3-3x^2+8x]_-2^1=π[x55x333x2+8x]12

= pi[(-1/5-1/3-3+8)-(32/5+8/3-12-16)]=π[(15133+8)(325+831216)]

=pi[ -33/5-9/3+5+12+16]=π[33593+5+12+16]

= pi[-66/10 -3 +5+12+16]=π[66103+5+12+16]

= pi [-6.6 + 30]=π[6.6+30]

= 23.4 pi=23.4π