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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

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  1. anonymous
    • 5 years ago
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    Ok which function is on top?

  2. anonymous
    • 5 years ago
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    y=sqrt2, bounded on left by y -axis and below by x=y^2

  3. anonymous
    • 5 years ago
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    just confirming it is y=\[y=\sqrt(2)\] x= x^2 right functions?

  4. anonymous
    • 5 years ago
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    -- correction x=y^2

  5. anonymous
    • 5 years ago
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    yes

  6. anonymous
    • 5 years ago
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    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
    • 5 years ago
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    y=sqrt2 right?

  8. anonymous
    • 5 years ago
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    The intersection point will have y=sqrt(2)

  9. anonymous
    • 5 years ago
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    when x = 2

  10. anonymous
    • 5 years ago
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    what is x intersection is 2

  11. anonymous
    • 5 years ago
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    so (2,sqrt(2)

  12. anonymous
    • 5 years ago
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    good so far?

  13. anonymous
    • 5 years ago
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    yes

  14. anonymous
    • 5 years ago
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    So the region we are rotating has these points(0,0)-- origin (0,sqrt(2)top left corner, (2,sqrt(2) intersection

  15. anonymous
    • 5 years ago
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    any questions?

  16. anonymous
    • 5 years ago
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    so far good, i have a drawing set up

  17. anonymous
    • 5 years ago
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    So we are going to use shell method which is just add up bunch off cylinder

  18. anonymous
    • 5 years ago
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    Volume of cylinder is pi*radius^2*h

  19. anonymous
    • 5 years ago
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    got ya

  20. anonymous
    • 5 years ago
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    So if you look your drawing(may be draw horizonal rectangle) you will see that the y value of function represent radius

  21. anonymous
    • 5 years ago
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    and the function x=y^2 represent h

  22. anonymous
    • 5 years ago
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    yea from the integral of 0 to sqrt2 right?

  23. anonymous
    • 5 years ago
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    Yes

  24. anonymous
    • 5 years ago
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    \[2\pi\int\limits(y*y^2 dy(\]

  25. anonymous
    • 5 years ago
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    you got the limit of the integration right

  26. anonymous
    • 5 years ago
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    so just integrate that? with the height as y^2?

  27. anonymous
    • 5 years ago
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    Yes

  28. anonymous
    • 5 years ago
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    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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