How do you use the disk or shell method to find the volume of the solid generated by revolving the regions bounded by the graphs of #y = x^(1/2)#, #y = 2#, and #x = 0# about the line #x = -1#?

1 Answer
Oct 19, 2015

See the explanation section, below.

Explanation:

Here is a picture of the region with a thin slice taken vertically.

So thickness is #dx# and we will use cylindrical shells.

enter image source here

The representative shell will have volume
#2pi "radius" * "height" * "thickness" = 2pi(x+1)(2-sqrtx)dx#.

We see that the values of #x# vary from #0# to #4#

So the volume of the solid is found by evaluating:

#int_0^4 2pi(x+1)(2-sqrtx)dx = 2pi int_0^4 (x+1)(2-sqrtx)dx #

Expand the product and integrate term by term.

To use disks/washers , we need to take our slices perpendicular to the axis of rotation as shown below.

enter image source here

The curve is now expressed as #x=y^2#

The representative washer has volume: #pi(R^2-r^2)dy# Where #r# is the greater radius (#x=-1# to #x=y^2#) and #r# is the lesser radius (#x=-1# to #x=0#) and the thickness is #dy#.

#y# goes from #0# to #2#, so the Volume is

#int_0^2 pi((y^2+1)^2-(1)^1) dy = pi int_0^2 ((y^2+1)^2-(1)^1) dy#

Again, expand and integrate term by term.