A community for students.
Here's the question you clicked on:
 0 viewing
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 .

This Question is Closed

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

anonymous
 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

anonymous
 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

anonymous
 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 ?

anonymous
 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

anonymous
 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)

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

anonymous
 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

anonymous
 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

anonymous
 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

anonymous
 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 ?
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.