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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)
 one year ago
 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
 one year ago

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inkyvoydBest ResponseYou've already chosen the best response.1
I 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.
 one year ago

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

amistre64Best ResponseYou've already chosen the best response.2
dw:1365190324765:dw \[\int 2\pi ~r~h\]
 one year ago

amistre64Best ResponseYou've already chosen the best response.2
r = cx, h = sqrt(x), and dx from 0 to c
 one year ago

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

amistre64Best ResponseYou've already chosen the best response.2
that IS the shell method :)
 one year ago

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

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

amistre64Best 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)
 one year ago

amistre64Best ResponseYou've already chosen the best response.2
inverse of y = sqrt(x); x = y^2
 one year ago

amistre64Best ResponseYou've already chosen the best response.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
 one year ago

amistre64Best ResponseYou've already chosen the best response.2
the only issue that may present itself with that setup tho is that you get a negative summation
 one year ago

amistre64Best ResponseYou've already chosen the best response.2
why did i do 2pi r? thats circumference :/ .... area is \[\int pi r^2\]
 one year ago
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