A community for students.

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

liizzyliizz

  • 5 years ago

Can someone help me with Cross sections please? -Region bounded by y=e^x +1 , y=8 and the y-axis cross sections are semi circles.

  • This Question is Closed
  1. amistre64
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    |dw:1327356271299:dw|

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

    Amistre64, can you hep me once you're done here please?

  3. amistre64
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    i can try

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

    o.O lol oh lord.

  5. amistre64
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    im assuming by cross section you are trying to find the area when the cross sections are semicircles?

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

    Yes sir.

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

    for the y=8 would hat be considered another line o.o its just that my teacher gave us a piece of paper and the information listed above is all that is written. So i've been afraid to do anything crazy because it is a project lol

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

    that***

  9. amistre64
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    y=8 is a horizontal line thru y=8 yes

  10. amistre64
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    do you kow how the semicircles are crosssectioned in there? say flat part up hopefully?

  11. amistre64
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    |dw:1327359342185:dw|

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

    That last part has me a bit thrown off, can you explain?

  13. amistre64
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    im trying to determine how the semicircles are orientated in the shaded area

  14. amistre64
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    with any luck the flat end is up and the rounded end down to make life simpler

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

    ooh i see.

  16. amistre64
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    we can also move the graph and not change the area; lower it down so it begine at 0,0 - in effect we subtract 2 from our heights

  17. amistre64
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    y = e^x - 2 y = 6 would be the new way it would look but it would retain the same content

  18. amistre64
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    it might be good to know where these lines cross at so that we can set up boundaries to integrate by

  19. amistre64
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    e^x - 2 = 6 e^x = 8 x = ln(8) looks about right to me

  20. amistre64
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    |dw:1327360185033:dw|

  21. amistre64
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    the radius of each semicircle is then the distance from y=6 to y=e^x-2 or simply: R = 6-(e^x -2) = 6 -e^x +2 = 8 - e^x

  22. amistre64
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    got some numbers off in me head; e^x is normal ar y=1 +1 means we are at y=2 drop 2 means we are at e^x -1, not e^x -2 it bites getting old

  23. amistre64
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    R = 7 - e^x is our radiuses then

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

    oh my lol. hmm.

  25. amistre64
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    :) its easier than it looks, i hope

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

    This is like going over my head, lol I'll try it out, it doesn't look so hard, I guess I just get intimidated by graphs /

  27. amistre64
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    the graph is more of a picture to help focus your thought on; gives you something more concrete to play with than an abstract notion of integrations and such

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

    It's just that I have to find someway to make 10 semicircles on my graph. so that has me nervous. lol I think the actually solving the integral is much easier.

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

    But thank you anyway :D your help is always greatly appreciated !

  30. amistre64
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    10? ugh yeah, wish i could be more helpful :) good luck with it tho

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

    Thank you very much. and lol yesss -__- cutting 10 little semi circles is not fun :c

  32. 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.