A community for students.

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

anonymous

  • 5 years ago

how do you find the volume of the region defined by y = sqrt(x), y = 0 and x = 6 when the region is revolving around the line x = 6?

  • This Question is Closed
  1. amistre64
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    you can do areas of circles then.... pi {S} [sqrt(x)]^2 dx ; [0,6]

  2. amistre64
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    pi x^2 ------ at 6 = 18pi 2

  3. amistre64
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    sooo close... i forgot to read the hwhole thing

  4. amistre64
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    since you are at the line x=6; subtract 6 from the function to bring it down to x = 0

  5. amistre64
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    over down..someplace; same idea

  6. amistre64
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    sqrt(x+6) is the function you use to move everything to the x = 0 line

  7. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    okay....so my 18 pi answer was wrong; where do you subtract 6 from? I did and equation where it was just sqrt(x) and then one where it was sqrt(x)-6, and for the first one I found it to be 18pi, and the second came out to be a weird decimal

  8. amistre64
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    1 Attachment
  9. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    that's the graph I have, yep

  10. amistre64
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    let me redo what I said lol

  11. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    okay

  12. amistre64
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    pi {S} [sqrt(x+6)]^2 dx ; [0,6] pi {S} x+6 dx ; [0,6] pi(x^2/2 + 6x) at x = 6

  13. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    you added 6 under the sqrt sign. How do you know to do that?

  14. amistre64
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    thats how we move things left and right.... see for yourself; when you put in x = -6 you want a zero right? to match the original

  15. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    OKAY. wow, brain lapse.

  16. amistre64
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    we dont want to cahng ethe value of the function; just move it :)

  17. amistre64
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    and I steered wrong again i think; I sppun it around the x axis in that..

  18. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    so you used the disk method to solve the problem

  19. amistre64
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    you can use disk; but youd end up doing it with respect to y and would have to change the formulas again.. shell is just as well

  20. amistre64
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    when I just did it I spun it around the x axis like an idiot lol

  21. amistre64
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    2pi {S} x(sqrt(x-6)) dx is good too

  22. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    using shell method would leave me with \[2\pi \int\limits_{0}^{6}(6-x)(\sqrt(x+6)dx ?\]

  23. amistre64
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    maybe harder; so lets reinterpret the graph for y instead of x... does that make sense?

  24. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    let's try

  25. amistre64
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    not 6-x; just x; the radius is x as x moves from 0 to 6, but that integral looks scary

  26. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    yes it does

  27. amistre64
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    sqrt(x^2(x+6)) = sqrt(x^3 + 6x^2) and integrating that looks painful lol

  28. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    \[\pi \int\limits_{0}^{6}y(y^2+6)dy ?\]

  29. amistre64
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    x = y^2-6 is a good tranlation dont you think?

  30. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    oh, y^2 minus 6. But the rest of the integration looks okay?

  31. amistre64
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    hmm.... if we shell it with respect to y, that doesnt get us out spin around the x=0 part

  32. amistre64
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    yes; but we need a new interval; one that works along the y line

  33. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    now i am confused...can we try again with x? That makes the most sense to me

  34. amistre64
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    when x = 6 to begin with; y = sqrt(6) so our interval is along the y axis from 0 to sqrt(6)

  35. amistre64
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    if we do it with respect to x; we run into an awful integral.....

  36. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    okay, now I understand. The limits were wrong

  37. amistre64
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    if we do it with respect to y; its easier, we just have to know how far to go onthe y to match

  38. amistre64
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    at x= 6, our original; y = sqrt(6) so our boundary is from 0 to sqrt(6) of y^2-6

  39. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    and it's minus 6 because of "top-bottom"?

  40. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    \[\pi \int\limits_{0}^{\sqrt(6)}y(y^2-6)dy\]

  41. amistre64
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    1 Attachment
  42. amistre64
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    now we do the disc method of areas of circles..... to spin around x=0

  43. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    yeah! now I get it

  44. amistre64
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    pi {S} [y^2-6]^2 dy; from [0,sqrt(6)]

  45. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    wait, so my integral is wrong again, since we revert to disk method?

  46. amistre64
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    correct; remember shell method gave us a nightmare integral

  47. amistre64
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    we might get a negative out of this; but that just means we went the wrong direction on the interval and just ignore the sign

  48. amistre64
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    pi {S} (y^4 -12y^2 +36) dy ints to: pi(y^5/5 -12y^3/3 + 36y) at y = sqrt(6) right?

  49. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    yeah; okay so now my answer is -9.797958971, the same answer I had, but didn't get right. Is there a fractional way of writing this?

  50. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    my bad. recalculating...

  51. amistre64
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    ((sqrt(6)^5) / 5) - ((12 * (sqrt(6)^3)) / 3) + (36 * sqrt(6)) = 47.0302031 according to the google

  52. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    hmmm

  53. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    I put that into my calculator and got -940898/7275....

  54. amistre64
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    http://www.wolframalpha.com/input/?i=int%28pi%28y^2-6%29^2%29+dy+from+0+to+sqrt%286%29

  55. amistre64
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    i mighta typoed it in google lol

  56. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    no, you have yours right I think. What the fractional form?

  57. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    it's right!!! thanks a bunch!

  58. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    I have another problem just like that so I will use this method

  59. amistre64
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    :)

  60. Not the answer you are looking for?
    Search for more explanations.

    • Attachments:

Ask your own question

Sign Up
Find more explanations on OpenStudy
Privacy Policy

Your question is ready. Sign up for free to start getting answers.

spraguer (Moderator)
5 → View Detailed Profile

is replying to Can someone tell me what button the professor is hitting...

23

  • Teamwork 19 Teammate
  • Problem Solving 19 Hero
  • You have blocked this person.
  • ✔ You're a fan Checking fan status...

Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.

This is the testimonial you wrote.
You haven't written a testimonial for Owlfred.