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?

Question

Charles Anctil · Accepted Answer

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.

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#

Evelyn Agard · Answer

undefined To find the volume of the region bounded by the graph of $y = x^2 + 1$ for $x$ in $[1,2]$ rotated around the x-axis, you use the method of cylindrical shells. The formula for the volume of a solid obtained by rotating a region bounded by a curve around the x-axis is:

$$V = \int_{a}^{b} 2\pi x \cdot f(x) \, dx$$

Where:
$a$ and $b$ are the limits of integration (in this case, 1 and 2).
$f(x)$ is the function representing the curve ($x^2 + 1$ in this case).

So, substituting the given values, the volume can be calculated as follows:

$$V = \int_{1}^{2} 2\pi x \cdot (x^2 + 1) \, dx$$