## anonymous 5 years ago find the volume of the solid using shell method, region of y=sqrt2, x=y^2 , x=0. about the x-axis

1. anonymous

Ok which function is on top?

2. anonymous

y=sqrt2, bounded on left by y -axis and below by x=y^2

3. anonymous

just confirming it is y=$y=\sqrt(2)$ x= x^2 right functions?

4. anonymous

-- correction x=y^2

5. anonymous

yes

6. anonymous

Ok so x=y^2 is a side way parabola facing in the positive x direction(right) y=sqrt(2) is just a horizonal striaght line We need to find intersection of of lines

7. anonymous

y=sqrt2 right?

8. anonymous

The intersection point will have y=sqrt(2)

9. anonymous

when x = 2

10. anonymous

what is x intersection is 2

11. anonymous

so (2,sqrt(2)

12. anonymous

good so far?

13. anonymous

yes

14. anonymous

So the region we are rotating has these points(0,0)-- origin (0,sqrt(2)top left corner, (2,sqrt(2) intersection

15. anonymous

any questions?

16. anonymous

so far good, i have a drawing set up

17. anonymous

So we are going to use shell method which is just add up bunch off cylinder

18. anonymous

19. anonymous

got ya

20. anonymous

So if you look your drawing(may be draw horizonal rectangle) you will see that the y value of function represent radius

21. anonymous

and the function x=y^2 represent h

22. anonymous

yea from the integral of 0 to sqrt2 right?

23. anonymous

Yes

24. anonymous

$2\pi\int\limits(y*y^2 dy($

25. anonymous

you got the limit of the integration right

26. anonymous

so just integrate that? with the height as y^2?

27. anonymous

Yes

28. anonymous

oooh thank you, i kept thinking of subtracting sqrt2 from the height wasn't really paying attention that this was respect to y. Thank you for all your help:)

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