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anonymous
 5 years ago
revolve the region bounded by y = x^2 , y = 1, about the y axis .
anonymous
 5 years ago
revolve the region bounded by y = x^2 , y = 1, about the y axis .

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0before you say anything, its not possible to do. x = sqrt y, and x = sqrt y , but i have

dumbcow
 5 years ago
Best ResponseYou've already chosen the best response.0\[V = \pi \int\limits_{0}^{1}(\sqrt{y})^{2} dy\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0but what about 1 to 0

dumbcow
 5 years ago
Best ResponseYou've already chosen the best response.0function is not defined for y<0

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0but the region is defined

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ok here is another one like it, revolve y = x^2 and y = x+2 bounded, about y axis

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0it exists in the math universe, not in the real world ( you can only revolve one revolution at a time about an axis )

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0dumbcow, do you see y = x^2 , y = x+2

dumbcow
 5 years ago
Best ResponseYou've already chosen the best response.0oh yes i see, sorry i was also assuming it was bounded by x=0

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so does it make sense , you get extra volume, correct ?

dumbcow
 5 years ago
Best ResponseYou've already chosen the best response.0wait no it would still be same volume if rotating around yaxis the whole region is from y=0 and y=1

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0but you have extra volume in the second quadrant

dumbcow
 5 years ago
Best ResponseYou've already chosen the best response.0hmm ok its double the volume then

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0not quite double though

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0its a real feather of a problem, lets use shell method , see what we get

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0easier than using disc method here

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0integral 2pi * x ( x+2  x^2)

dumbcow
 5 years ago
Best ResponseYou've already chosen the best response.0ok \[=2 \pi \int\limits_{1}^{1}x(x ^{2})dx\]

dumbcow
 5 years ago
Best ResponseYou've already chosen the best response.0oh we are doing the other one

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0as i suspected, we get negative volume from 1 to 0

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i think in the shell method, its important that the radius remain positive

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0here is a similiar question, revolve region y = (x+2)^2 and y = 1 about y axis, using shell method. you get a negative answer

dumbcow
 5 years ago
Best ResponseYou've already chosen the best response.0yes but volume can't be neg in the practical sense

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so there are limitations to the revolving region problems

dumbcow
 5 years ago
Best ResponseYou've already chosen the best response.0you just switch the bounds from neg to positive, because when revolving around axis there is symmetry

dumbcow
 5 years ago
Best ResponseYou've already chosen the best response.0not in every case though

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0what about y = x^2  4 and bounded by y = 0, revolve it about x axis

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i think that works, we only get problems with the shell method because the radius is assumed to be positive

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0and in the disc method we have radius squared, so its more flexible, but why cant we integrate y = x^2 and y=1 about y axis, using horizontal discs ?
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