A community for students.

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

anonymous

  • 4 years ago

**Calc2 Help Needed** Find the volume of the solid generated by revolving the described region about the given axis: The region bounded above by the line y=6 , below by the curve y=sqrt(x), and to the left by the y-axis, rotated along the following lines: x=36

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

    another one huh :)

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

    Yep.. I'm going to start tutoring next week. I am having the worst time with these..

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

    try drawing out the region to get the bounds

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

    yep, i've got that

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

    |dw:1327644073706:dw|

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

    so the bounds should be in y? so 0 to 6

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

    I was thinking of using the cylindrical shell method

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

    2pi integral (radius)height

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

    radius = y height = y-36?

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

    i'm not as familiar with the shell method, mostly because i don't like using it anyway, wouldn't the radius be in terms of x, and the height be 6-y

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

    oh yes..

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

    so we would use the disc method?

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

    so radius = y^2 and height is 6-y?

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

    you could use either i believe for this one. if you used the disc method: outer radius = 36 inner radius = 36-y^2 ...yes

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

    so we have the integral of pi(36-36+y^2)?

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

    think area of circle: A = pi*r^2

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

    forgot the ^2

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

    what should you integrate from, or what are the bounds

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

    0 to 36?

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

    the outer radius is 36 right, that is in x-direction now imagine the figure being sliced horizontally so we integrate over y what are the bounds of y

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

    oh 0 to 6 in terms of y

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

    yep

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

    ok so from 0 to 6 integral (36 - 36 + x^2)^2?

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

    = pi(y^5/5)

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

    not quite its R^2 - r^2 Not (R-r)^2 -> 36^2 - (36-y^2)^2

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

    because you're not subtracting the radius, you are subtracting the areas

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

    ohh.. do I have two separate integrals?

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

    no just keep them separate...you can simplify it

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

    nvm.. i see now

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

    I got it right!! yay.. haha.. =3628.8pi. Thank you so much!

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

    great...welcome

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

    Now it's time to study biology... fun :/

  33. 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.