anonymous
  • anonymous
revolve the region bounded by y = x^2 , y = 1, about the y axis .
Mathematics
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get our expert's

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this
and thousands of other questions.

anonymous
  • anonymous
revolve the region bounded by y = x^2 , y = 1, about the y axis .
Mathematics
katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this
and thousands of other questions

anonymous
  • anonymous
before you say anything, its not possible to do. x = sqrt y, and x = -sqrt y , but i have
dumbcow
  • dumbcow
\[V = \pi \int\limits_{0}^{1}(\sqrt{y})^{2} dy\]
dumbcow
  • dumbcow
what do you mean?

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
but what about -1 to 0
dumbcow
  • dumbcow
function is not defined for y<0
anonymous
  • anonymous
but the region is defined
anonymous
  • anonymous
ok here is another one like it, revolve y = x^2 and y = x+2 bounded, about y axis
anonymous
  • 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
  • anonymous
dumbcow, do you see y = x^2 , y = x+2
dumbcow
  • dumbcow
oh yes i see, sorry i was also assuming it was bounded by x=0
anonymous
  • anonymous
so does it make sense , you get extra volume, correct ?
dumbcow
  • 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
  • anonymous
but you have extra volume in the second quadrant
dumbcow
  • dumbcow
hmm ok its double the volume then
anonymous
  • anonymous
not quite double though
anonymous
  • anonymous
its a real feather of a problem, lets use shell method , see what we get
anonymous
  • anonymous
easier than using disc method here
anonymous
  • anonymous
integral 2pi * x ( x+2 - x^2)
dumbcow
  • dumbcow
ok \[=2 \pi \int\limits_{-1}^{1}x(x ^{2})dx\]
dumbcow
  • dumbcow
oh we are doing the other one
anonymous
  • anonymous
as i suspected, we get negative volume from -1 to 0
anonymous
  • anonymous
i think in the shell method, its important that the radius remain positive
anonymous
  • 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
  • dumbcow
yes but volume can't be neg in the practical sense
anonymous
  • anonymous
right
anonymous
  • anonymous
so there are limitations to the revolving region problems
dumbcow
  • dumbcow
you just switch the bounds from neg to positive, because when revolving around axis there is symmetry
dumbcow
  • dumbcow
not in every case though
anonymous
  • anonymous
what about y = x^2 - 4 and bounded by y = 0, revolve it about x axis
anonymous
  • anonymous
i think that works, we only get problems with the shell method because the radius is assumed to be positive
anonymous
  • 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 ?

Looking for something else?

Not the answer you are looking for? Search for more explanations.