How do you find the volume of the region bounded by #y=sqrt x,# and the lines #y=2# and# x=0# and it is revolved about the line #y=2#?

First, try and imagine how it would look. This could be described as a "cone" where the tracing of the surface starting from every point on the base's circumference (at the same time) grows like #x^2# until you reach the tip.

Thankfully, this is a closed surface with a finite volume (not Gabriel's horn), due to the revolution occurring around #y = 2#; the functions #y = sqrtx# and #y = 4 - sqrtx# are stopped at #x = 2^2 = 4#. This "cone" can be shifted down to #y = 0# for simplicity without changing its shape.

So the interval you would have to integrate over is #[0, 4]#. You can imagine this as the infinite layering of circles with continuously varying radii from the base of the "cone" to the tip of the "cone". The circles' radii, defined as #r(x)#, vary according to either the curve #y = sqrtx - 2# or #y = 2-sqrtx#.

The volume is defined as:
#V = pi(r(x))^2*x#

#=> int_0^4 pi[r(x)]^2dx#

where #x# is now represented by #int_0^4dx#.

#y = 2-sqrtx#
#y = sqrtx - 2#

Just pick one of them and use it, since we've redefined the revolution to occur using one of these symmetric curves around the x-axis. I picked the second one.

#V = int_0^4 pi[sqrtx - 2]^2dx#

#= piint_0^4 x-4x^(1/2)+4dx#

#= pi [x^2/2 - 4*2/3x^(3/2) + 4x]_(x = 0->4)#

#= pi [(4)^2/2 - 4*2/3(4)^(3/2) + 4(4)]#

#= pi [8 - 64/3 + 16]#

#= pi (8/3)#

#= (8pi)/3 "u"^2 ~~ 2.bar66pi "u"^2 ~~ 8.378 "u"^2#

Comparing with a regular right cylindrical cone, we would expect:

#V_(sqrtx"curve") < V_"cone" => V_"cone" = pir^2h/3 = pi*2^2*4/3 = 5.bar33pi ~~ 16.755 "u"^2#

#8.378 < 16.755#

So our answer is reasonable. You can also see here that our answer is identical to the result we would have gotten if we didn't shift the "cone":
http://www.wolframalpha.com/input/?i=volume+of+y+%3D+sqrtx+from+x+%3D+0+to+4+revolved+around+y+%3D+2