How do you find the volume of the region bounded by the graph of #y = x^2+1# for x is [1,2] rotated around the x axis?

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Answer 1 · 2016-11-12T17:40:24

See below.

The region is in blue. The rotation is shown by the red circular arrow.

A representative slice has been taken perpendicular to the axis or rotation to use the method of discs. The slice has black borders.

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The volume of a representative slice is

#pir^2* "thickness" #

In general, the radius #r# is the longer distance minus the shorter. In this case the shorter distance to the axis of rotation is #0# so we have

#r = x^2+1#

The thickness is a differential. It is the thin side of the slice. IN this case #"thickness" = dx#

The values of #x# go from #1# to #2#.

The volume we seek is

#V = int_1^2 pi (x^2+1)^2 dx#

# = pi int_1^2 (x^4+2x^2+1) dx#

# = 178/15 pi#