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Find the volume of the solid of revolution generated by revolving the equation \(y=\sqrt(x)\) about the line x=4 (The region is bounded by the x axis and x=4)

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I understand if I rewrite the equation as y=\(\sqrt{x-4}\) then integrate with appropriate bounds it will work, but there is some sort of alternate interpretation I am not understanding.
*rewrite the equation as y=\(\sqrt{x+4}\)
|dw:1365190324765:dw| \[\int 2\pi ~r~h\]

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Other answers:

r = c-x, h = sqrt(x), and dx from 0 to c
omg thank you - that applies for shell method as well correct?
that IS the shell method :)
oh shoot - is there any way to do thi with the disk method?
this problem is listed in the disk method section of my textbook, but i'm not sure how it's done...
\[\int~2\pi r\] \[r=g(y)\text{ ; which is the inverse of f(x)}\]and dy from 0 to f(c)
inverse of y = sqrt(x); x = y^2
inky suggests that we move the graph to something simpler to play with; shift if left by "c" units given us y = sqrt(x+c) r = x = sqrt(y) - c
r = x = y^2 - c :)
the only issue that may present itself with that setup tho is that you get a negative summation
why did i do 2pi r? thats circumference :/ .... area is \[\int pi r^2\]

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