Question

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

Answer 1

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

Answer 2

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

Answer 3

what do you mean?

Answer 4

but what about -1 to 0

Answer 5

function is not defined for y<0

Answer 6

but the region is defined

Answer 7

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

Answer 8

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

Answer 9

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

Answer 10

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

Answer 11

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

Answer 12

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

Answer 13

but you have extra volume in the second quadrant

Answer 14

hmm ok its double the volume then

Answer 15

not quite double though

Answer 16

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

Answer 17

easier than using disc method here

Answer 18

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

Answer 19

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

Answer 20

oh we are doing the other one

Answer 21

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

Answer 22

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

Answer 23

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

Answer 24

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

Answer 25

right

Answer 26

so there are limitations to the revolving region problems

Answer 27

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

Answer 28

not in every case though

Answer 29

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

Answer 30

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

Answer 31

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 ?