Question

How do I revolve y=sqrt(x) around x= 6 around x is pi int( (sqrt(x))^2 dx

Answer 1

what? Is that two questions?

Answer 2

Or do you mean how do you revolve the region bounded by y=sqrt(x) and y=0 and x=6
about x=pi?

Answer 3

no i am just saying i know how to do it for the x axis.

Answer 4

around the x axis

Answer 5

what is the region that we are trying to rotate about the x-axis?

Answer 6

Find the volumes of the solids generated by revolving the regions bounded by the graphs of the equations about the given lines.
y = sqrt( x)
y = 0
x = 6