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anonymous
 one year ago
Using shell method find the volume of the solid bounded by y=1+x^1/2 ,y=2,y=1,x=1 when it revolve around x=1?
anonymous
 one year ago
Using shell method find the volume of the solid bounded by y=1+x^1/2 ,y=2,y=1,x=1 when it revolve around x=1?

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triciaal
 one year ago
Best ResponseYou've already chosen the best response.0dw:1443682301707:dw

FireKat97
 one year ago
Best ResponseYou've already chosen the best response.3dw:1443682610450:dw This is the volume you want to find

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Ok,then what shall I do?

FireKat97
 one year ago
Best ResponseYou've already chosen the best response.3so now, imagine taking a shell, dw:1443683814127:dw

FireKat97
 one year ago
Best ResponseYou've already chosen the best response.3dw:1443683890033:dw

FireKat97
 one year ago
Best ResponseYou've already chosen the best response.3dw:1443683966585:dw so lets take the thickness of the shell to be dx, the radius, to be x and the height to be f(x), so height is 1 + x^1/2

FireKat97
 one year ago
Best ResponseYou've already chosen the best response.3dw:1443684519400:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I have just started understanding this:D

FireKat97
 one year ago
Best ResponseYou've already chosen the best response.3:D I'm glad. Now that we have "opened" the shell up it is easier to write up the equation for the volume, so try coming up with the integral

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0So I think the radious will =1x And the height=y1 Am I right?
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