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anonymous

  • 5 years ago

Consider the solid obtained by rotating the region bounded by the given curves about the line y = 5. y = 5x , y = 5 * squareroot of x Find volume.

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  1. amistre64
    • 5 years ago
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    considered....now what?

  2. anonymous
    • 5 years ago
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    find volume

  3. anonymous
    • 5 years ago
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    is it solid of revolution problem in calc

  4. amistre64
    • 5 years ago
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    you dont say..... ;)

  5. anonymous
    • 5 years ago
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    yup

  6. anonymous
    • 5 years ago
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    Ok, we first have to find intersections

  7. anonymous
    • 5 years ago
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    have you guys ever heard of the disk or washer method to finding volume?

  8. anonymous
    • 5 years ago
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    Yes

  9. amistre64
    • 5 years ago
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    and a shell

  10. amistre64
    • 5 years ago
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    first things first; whats our boundaries for integration

  11. anonymous
    • 5 years ago
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    thank you for that link :) how do you find the boundaries though?

  12. anonymous
    • 5 years ago
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    We look the the intersection point which are (1,5)(0,0)

  13. anonymous
    • 5 years ago
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    so would the bounaries be from 0 to 5?

  14. anonymous
    • 5 years ago
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    yes

  15. anonymous
    • 5 years ago
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    So what method would you use?

  16. amistre64
    • 5 years ago
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    the y=5 line seems spurious

  17. anonymous
    • 5 years ago
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    hmm, im not sure which method to choose

  18. anonymous
    • 5 years ago
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    because im looking through my notes right now, it looks like we could use the disk method

  19. anonymous
    • 5 years ago
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    Since you are going to be rotating around an horizontal point(y=5) you are going to use disk method( or washer)

  20. anonymous
    • 5 years ago
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    in Some cases you can use either but it is matter convienience

  21. amistre64
    • 5 years ago
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    the disk method is adding up all the areas of the circles; [S] 2pi [f(x)]^2 dx - [S] 2pi [g(x)]^2 dx

  22. amistre64
    • 5 years ago
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    ohhh....rotating AROUND y=5; haha :) helps to read the question lol

  23. anonymous
    • 5 years ago
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    I would advise against blindly memorizing the formula

  24. anonymous
    • 5 years ago
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    i have in my notes that the disk method formula is pi (R(x))^2 dx

  25. amistre64
    • 5 years ago
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    the formula is jsut the qnqlytic representation of what you are actually accomplishing;

  26. amistre64
    • 5 years ago
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    R(x) - 5 in this case

  27. amistre64
    • 5 years ago
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    analytic...

  28. anonymous
    • 5 years ago
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    So the bigger circle have radius of 5-5x the smaller circle have radius of 5-5Sqrt[x]

  29. amistre64
    • 5 years ago
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    i was combining the shell and the disk ....my mistake

  30. anonymous
    • 5 years ago
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    ohh i think i get it. so would it be (5-5sqrt)^2-(5-5x)^2 dx

  31. amistre64
    • 5 years ago
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    you simply find the volume of the outer function and cut out the volume of the inner function

  32. anonymous
    • 5 years ago
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    I think you have it reversed

  33. amistre64
    • 5 years ago
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    [5x-5] ^2 - [5sqrt(x)-5]^2

  34. anonymous
    • 5 years ago
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    how come its that way rather than the way i put it? i remember my professor i think tellung us that its the bottom function subtracted from the top one, and in this case, the top function is the 5sqrt of x

  35. anonymous
    • 5 years ago
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    i might be confusing it with something else x_X

  36. amistre64
    • 5 years ago
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    sine y = 5; you want to bring the function down to y = 0 therefore: f(x) - 5

  37. amistre64
    • 5 years ago
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    since y = 5...yadayada

  38. amistre64
    • 5 years ago
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    at y=5; the function is out of reach up on a shelf; you want to get it down to where you can work with it on the table.... bring it down to y=0 by subtracting 5 from it

  39. anonymous
    • 5 years ago
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    ohhh, i see what you mean now

  40. anonymous
    • 5 years ago
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    thank you guys so much for all the help but im running late for class haha so ill try to work this with what you guys helped me with so far. thank you guys times infinityy !

  41. amistre64
    • 5 years ago
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    youre welcome :)

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