revolve the region bounded by y = x^2 , y = 1, about the y axis .

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- anonymous

revolve the region bounded by y = x^2 , y = 1, about the y axis .

- katieb

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- anonymous

before you say anything, its not possible to do. x = sqrt y, and x = -sqrt y , but i have

- dumbcow

\[V = \pi \int\limits_{0}^{1}(\sqrt{y})^{2} dy\]

- dumbcow

what do you mean?

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## More answers

- anonymous

but what about -1 to 0

- dumbcow

function is not defined for y<0

- anonymous

but the region is defined

- anonymous

ok here is another one like it, revolve y = x^2 and y = x+2 bounded, about y axis

- anonymous

it exists in the math universe, not in the real world ( you can only revolve one revolution at a time about an axis )

- anonymous

dumbcow, do you see y = x^2 , y = x+2

- dumbcow

oh yes i see, sorry i was also assuming it was bounded by x=0

- anonymous

so does it make sense , you get extra volume, correct ?

- dumbcow

wait no it would still be same volume if rotating around y-axis
the whole region is from y=0 and y=1

- anonymous

but you have extra volume in the second quadrant

- dumbcow

hmm ok its double the volume then

- anonymous

not quite double though

- anonymous

its a real feather of a problem, lets use shell method , see what we get

- anonymous

easier than using disc method here

- anonymous

integral 2pi * x ( x+2 - x^2)

- dumbcow

ok
\[=2 \pi \int\limits_{-1}^{1}x(x ^{2})dx\]

- dumbcow

oh we are doing the other one

- anonymous

as i suspected, we get negative volume from -1 to 0

- anonymous

i think in the shell method, its important that the radius remain positive

- anonymous

here 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

yes but volume can't be neg in the practical sense

- anonymous

right

- anonymous

so there are limitations to the revolving region problems

- dumbcow

you just switch the bounds from neg to positive, because when revolving around axis there is symmetry

- dumbcow

not in every case though

- anonymous

what about y = x^2 - 4 and bounded by y = 0, revolve it about x axis

- anonymous

i think that works, we only get problems with the shell method because the radius is assumed to be positive

- anonymous

and 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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