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
Jim H
Jun 17, 2015

I got area is #23.4 pi#. The "How" has no one sentence answer -- see below.

Explanation:

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

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

#V = int_-2^1 (pi(-x+3)^2 - pi(x^2+1)^2) dx#

#= 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#

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