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anonymous

  • 5 years ago

the region enclosed by the graphs of y=e^(x/2), y=1, and x=ln3 is revolved about the x-axis. Find the volume of the solid generated??????? I NEED CALCULUS HELP PLEASE

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  1. amistre64
    • 5 years ago
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    yay!! an easy one

  2. anonymous
    • 5 years ago
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    easy? lol i would love your help then :)

  3. amistre64
    • 5 years ago
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    lets see where these curves meet; and what our boundaries are

  4. amistre64
    • 5 years ago
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    e^(x/2) = 1 when x=0 right?

  5. anonymous
    • 5 years ago
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    yes

  6. amistre64
    • 5 years ago
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    x = ln(3) is just a constant; we can find it on a calculator

  7. anonymous
    • 5 years ago
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    yea 1.098

  8. amistre64
    • 5 years ago
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    right; 1.09861...

  9. amistre64
    • 5 years ago
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    let me draw a pic of my interpretation of the bonds then

  10. anonymous
    • 5 years ago
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    i think they intersect at 0 and ln3 but i dont know how to integrate it

  11. amistre64
    • 5 years ago
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    like this?

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  12. amistre64
    • 5 years ago
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    theres only two things to integrate here; from 0 to ln(3) right?

  13. anonymous
    • 5 years ago
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    yes

  14. amistre64
    • 5 years ago
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    lets find the volume of the top curve; then we will cut out the section below y=1ok

  15. anonymous
    • 5 years ago
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    yes that would work but im suppose to do something with (pie)

  16. amistre64
    • 5 years ago
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    pi {S} [e^(1/2(x))]^2 dx ; [0,ln(3)]

  17. amistre64
    • 5 years ago
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    what is e^(x/2)^2 =? e^x right?

  18. anonymous
    • 5 years ago
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    yes because they would cancel

  19. amistre64
    • 5 years ago
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    the key here is to realize that we are adding up all the areas of circles that have been sliced; the area of a circle = pi r^r it just so happens the r = f(x)

  20. amistre64
    • 5 years ago
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    r^2 that is lol

  21. amistre64
    • 5 years ago
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    so; [f(x)]^2 = e^(x/2)^2 = e^x pi {S} e^x dx ; [0,ln(3)] whats the integral of e^x?

  22. anonymous
    • 5 years ago
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    the antiderivetive? its justs e^x

  23. amistre64
    • 5 years ago
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    e^x is its own derivative; so its its own integral :) yes

  24. amistre64
    • 5 years ago
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    pi [ e^(ln(3)) e^(0) ] = ?

  25. amistre64
    • 5 years ago
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    typoed that

  26. amistre64
    • 5 years ago
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    pi [ e^(ln(3)) - e^(0) ] = ?

  27. anonymous
    • 5 years ago
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    an example my teacher did shows it being {S} (pi (e^x)-1) and then taking the antiderivetive of that

  28. amistre64
    • 5 years ago
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    lets stay on one track here; well jump the track when we get to the end

  29. anonymous
    • 5 years ago
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    ok i understand that i just dont understand where the minus 1 came from?

  30. amistre64
    • 5 years ago
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    right now we are making a solid piece; we will then cut out the senter

  31. amistre64
    • 5 years ago
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    pi [e^(ln3) - e^0] = ??

  32. anonymous
    • 5 years ago
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    o so we r taking the whole volume minus the part we dont need which is the y=1

  33. amistre64
    • 5 years ago
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    exaclty :) we can do it in parts or all togther; makes no difference

  34. anonymous
    • 5 years ago
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    gotcha and i got 2 for that answer

  35. amistre64
    • 5 years ago
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    2 or 2pi?

  36. anonymous
    • 5 years ago
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    2 pi

  37. amistre64
    • 5 years ago
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    good :) now let integrate y = 1 from 0 to ln(3); or do you know a shortcut for this one?

  38. amistre64
    • 5 years ago
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    pi {S} 1 dx ; [0,ln3] becomes?

  39. anonymous
    • 5 years ago
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    pi (x)= pi (ln3)

  40. amistre64
    • 5 years ago
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    good :) 2pi - ln(3) pi is your answer then

  41. amistre64
    • 5 years ago
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    pi [ 2 - ln(3)] even lol

  42. anonymous
    • 5 years ago
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    and that is what the back of the book says...you are good lol

  43. amistre64
    • 5 years ago
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    toldja it was easy lol

  44. amistre64
    • 5 years ago
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    adding up areas of circles; thats all

  45. anonymous
    • 5 years ago
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    yea i just suck at it when my teacher tries to explain it..i may need more help in a few

  46. amistre64
    • 5 years ago
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    im sure well be around

  47. anonymous
    • 5 years ago
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    ok thank you :)

  48. anonymous
    • 5 years ago
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    u still there? i have a quick question?

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