A community for students.

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing


  • 5 years ago

powerseries Lifesaver (1 fan)Become a fan You're a fanNot a fan Surface area of a torus which has radius r which is generated by the circle centered at (R,0) with radius r. And assume R>r. So the equation of the circle is (x-R)^2 +y^2 = r^2. FOr some reason my integral does not work. i have integral 2pi y dS, where y = sqrt [r^2 - (x-R)^2], and ds = sqrt ( 1 + dy/dx ^2). dy/dx = -2(x-R)/ (2sqrt(r^2 - (x-R)^2)). anyways, integration limits , the lower limit is R-r, and upper R+r

  • This Question is Closed

    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...


  • 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.