How do you find the volume of the solid obtained by rotating the region bounded by the curves #y=x^2# and #y=2-x^2# and #x=0# about the line #x=1#?

1 Answer
Oct 24, 2015

I would use shells (to avoid doing two integrals).

Explanation:

Here is a picture of the region. I have included the axis of rotation (#x=1#) as a dashed black line, a representative slice taken parallel to the axis of rotation (solid black segments inside the region) and the radius of the rotation (red line segment taken at about #y=1.7#).

enter image source here

When we rotate the slice, we'll get a cylindrical shell.

The representative cylindrical shell will have volume:

#2 pixx"radius"xx"height"xx"thickness"#.

Here we have:
#"radius"=r=(1-x)#.

The height will be the greater #y# value minus the lesser (the top y minus the bottom y).
#"height" = y_"top"-y_"bottom" = (2-x^2)-(x^2) = (2-2x^2)#

#"thickness" = dx#

We also not that in the region, #x# varies from #0# to #1#

Using shell, the volume of the solid is found by evaluating:

#V - int_a^b 2 pixx"radius"xx"height"xx"thickness"#

So we need

#V - int_0^1 2 pi (1-x)(2-2x^2) dx#

# = 2piint_0^1 (2-2x-2x^2+2x^3) dx#

# = 5/3 pi#