anonymous
  • anonymous
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)
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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inkyvoyd
  • inkyvoyd
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.
inkyvoyd
  • inkyvoyd
*rewrite the equation as y=\(\sqrt{x+4}\)
amistre64
  • amistre64
|dw:1365190324765:dw| \[\int 2\pi ~r~h\]

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amistre64
  • amistre64
r = c-x, h = sqrt(x), and dx from 0 to c
anonymous
  • anonymous
omg thank you - that applies for shell method as well correct?
amistre64
  • amistre64
that IS the shell method :)
anonymous
  • anonymous
oh shoot - is there any way to do thi with the disk method?
anonymous
  • anonymous
this problem is listed in the disk method section of my textbook, but i'm not sure how it's done...
amistre64
  • amistre64
\[\int~2\pi r\] \[r=g(y)\text{ ; which is the inverse of f(x)}\]and dy from 0 to f(c)
amistre64
  • amistre64
inverse of y = sqrt(x); x = y^2
amistre64
  • amistre64
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
amistre64
  • amistre64
r = x = y^2 - c :)
amistre64
  • amistre64
the only issue that may present itself with that setup tho is that you get a negative summation
amistre64
  • amistre64
why did i do 2pi r? thats circumference :/ .... area is \[\int pi r^2\]

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