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

1. anonymous

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

2. anonymous

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

3. anonymous

what do you mean?

4. anonymous

but what about -1 to 0

5. anonymous

function is not defined for y<0

6. anonymous

but the region is defined

7. anonymous

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

8. anonymous

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

9. anonymous

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

10. anonymous

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

11. anonymous

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

12. anonymous

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

13. anonymous

but you have extra volume in the second quadrant

14. anonymous

hmm ok its double the volume then

15. anonymous

not quite double though

16. anonymous

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

17. anonymous

easier than using disc method here

18. anonymous

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

19. anonymous

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

20. anonymous

oh we are doing the other one

21. anonymous

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

22. anonymous

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

23. 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

24. anonymous

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

25. anonymous

right

26. anonymous

so there are limitations to the revolving region problems

27. anonymous

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

28. anonymous

not in every case though

29. anonymous

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

30. anonymous

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

31. 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 ?