# 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#?

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.

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

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": https://tutor.hix.ai

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To find the volume of the region bounded by ( y = \sqrt{x} ), ( y = 2 ), ( x = 0 ), and revolving it about the line ( y = 2 ), you can use the method of cylindrical shells.

The integral setup for this problem is:

[ V = 2\pi \int_{0}^{4} (2 - y)(y^2) , dy ]

Solving this integral will give you the volume of the solid formed by revolving the region bounded by ( y = \sqrt{x} ), ( y = 2 ), and ( x = 0 ) about the line ( y = 2 ).

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