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Fey
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Calculus: The region in the first quadrant bounded by the graphs of the equations y=(1/x), x=1, x=2, is revolved around the xaxis. Use the shell method to find the volume of the resulting solid.
 one year ago
 one year ago
Fey Group Title
Calculus: The region in the first quadrant bounded by the graphs of the equations y=(1/x), x=1, x=2, is revolved around the xaxis. Use the shell method to find the volume of the resulting solid.
 one year ago
 one year ago

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Fey Group TitleBest ResponseYou've already chosen the best response.0
I need help finding h(y),,,, (Height)
 one year ago

klimenkov Group TitleBest ResponseYou've already chosen the best response.0
dw:1360447641218:dw
 one year ago

Fey Group TitleBest ResponseYou've already chosen the best response.0
m,,, ok, I know the bounds are [1,2], the radius is y.... now, the height i want to say (1/y)1????
 one year ago

klimenkov Group TitleBest ResponseYou've already chosen the best response.0
\(V=\pi\int \limits_1^2y^2dx\)
 one year ago

Fey Group TitleBest ResponseYou've already chosen the best response.0
The shell method uses the 2(pi) at the beginning. Can u explain how u got that answere?
 one year ago

klimenkov Group TitleBest ResponseYou've already chosen the best response.0
Yes. You have to understand what integral means. This solid consists of many little cylinders like on the picdw:1360448107971:dw The volume of the cylinder is \(\pi R^2h\), on the pic \(R\) is \(y(x)\) and \(h\) is \(dx\). We want the whole solid, so we sum the volumes of this small cylinders and get the volume of the solid as an integral.
 one year ago

Fey Group TitleBest ResponseYou've already chosen the best response.0
well,,, im still not getting the right answere. its suppoased to be (pi)/2....
 one year ago

klimenkov Group TitleBest ResponseYou've already chosen the best response.0
You have to calculate the integral I wrote correctly.
 one year ago

Fey Group TitleBest ResponseYou've already chosen the best response.0
Oops. I made a mistake xD Ok, got it thanks.
 one year ago

klimenkov Group TitleBest ResponseYou've already chosen the best response.0
You are welcome.
 one year ago
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