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anonymous
 3 years ago
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)
anonymous
 3 years ago
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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inkyvoyd
 3 years ago
Best ResponseYou've already chosen the best response.1I understand if I rewrite the equation as y=\(\sqrt{x4}\) then integrate with appropriate bounds it will work, but there is some sort of alternate interpretation I am not understanding.

inkyvoyd
 3 years ago
Best ResponseYou've already chosen the best response.1*rewrite the equation as y=\(\sqrt{x+4}\)

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.2dw:1365190324765:dw \[\int 2\pi ~r~h\]

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.2r = cx, h = sqrt(x), and dx from 0 to c

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0omg thank you  that applies for shell method as well correct?

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.2that IS the shell method :)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0oh shoot  is there any way to do thi with the disk method?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0this problem is listed in the disk method section of my textbook, but i'm not sure how it's done...

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.2\[\int~2\pi r\] \[r=g(y)\text{ ; which is the inverse of f(x)}\]and dy from 0 to f(c)

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.2inverse of y = sqrt(x); x = y^2

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.2inky 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

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.2the only issue that may present itself with that setup tho is that you get a negative summation

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.2why did i do 2pi r? thats circumference :/ .... area is \[\int pi r^2\]
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