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

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

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

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

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

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

YUNoWorkOS
 one year 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
 one year 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
 one year ago
Best ResponseYou've already chosen the best response.2inverse of y = sqrt(x); x = y^2

amistre64
 one year 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
 one year 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
 one year 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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