1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Surface area of solid of revolution (no calculus)

  1. Oct 26, 2012 #1
    1. The problem statement, all variables and given/known data

    Consider the region of the x y plane given by the inequality:

    x^2 + 4x + y^2 - 4x - 8 ≤ 0;

    If this region rotates an angle of π/6 radians around the line given by the equation x + y = 0, it will create a solid of revolution with surface area equal to

    (i) (128/3)π; (ii) (128/4)π; (iii) (128/5)π; (iv) (128/6)π; (v) (128/7)π

    2. Relevant equations

    area of a sphere = 4πr^2

    3. The attempt at a solution

    okay, so first I factored the inequality into:

    (x + 2)^2 + (y - 2)^2 ≤ 16;

    Which means it's a circle centered at (2, -2) and with r = 4. The line x + y = 0 goes through the center of the circle. Now I am stuck.. cause I though that if it rotated an angle of π radians around the line it would give the full external area of the sphere so if it rotated only π/6 it would give 1/6 of it. Which would equal to (4π4^2)/6 = (64/6)π. But that is not one of the answers..
     
  2. jcsd
  3. Oct 27, 2012 #2

    LCKurtz

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    I agree with everything you have said. My best guess is that either the problem meant for you to calculate its volume instead of surface area, or whoever wrote the answer choices accidentally did that instead of calculating the surface area. The volume does give one of those answers.
     
  4. Oct 27, 2012 #3

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    It's a solid, not a shell. Have you considered all surfaces of the solid?
     
  5. Oct 27, 2012 #4
    I haven't though about that! Of course it's not just a shell! It's an inequality that generates the solid of revolution! In that case I would have the (64/6)π + 2(π4^2) one area of the circle for where the solid is being generated from and one from where the solid is rotating to! And that equals (128/3)π!!

    Thanks!

    EDIT: Wait, is it a solid not a "shell" because they specified by saying so, or is it a solid not a shell because they gave the thing to generate the solid to be an inequality?
     
  6. Oct 27, 2012 #5

    LCKurtz

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Interestingly enough, that is also the answer to its volume.
     
  7. Oct 27, 2012 #6
    The inequality gives a filled in circle and rotated results in a solid no matter what they say :)
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Surface area of solid of revolution (no calculus)
  1. Solids of revolution (Replies: 3)

  2. Surface Area help (Replies: 3)

Loading...