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

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

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

$= 178/15 pi$

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