1. Limited time only! Sign up for a free 30min personal 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!

Volumes of revolution

  1. Mar 19, 2014 #1
    1. The problem statement, all variables and given/known data

    Calculate the volume obtained by rotating the triangle bounded by y = 0, y = x, and y = 2 - x, about the line x = -2. You may use either horizontal or vertical rectangles.

    3. The attempt at a solution

    So since this is a triangle, I tried to split up the volume down to two integrals, one from 0 - 1, and the other from 1-2. I used vertical rectangles by the way. So I set up my first integral as

    [itex]2 \cdot \pi \int_{0}^{1} (x+2)(x) \,dx[/itex]
    since the shell radius would be a distance x + another 2, and then the shell height would be the x.
    My second integral is

    [itex]2 \cdot \pi \int_{1}^{2} (x+2)(2-x) \,dx[/itex]

    then I added these together, but I did not get the correct answer, could anyone show me what the error in my steps are?
     
  2. jcsd
  3. Mar 19, 2014 #2

    BvU

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    I am issing 2). relevant equations. Where did you get the idea to set up the integrals as you did ?
    Did you make a drawing to see if you want to integrate over dx or over dy ?

    [edit]Ok, you are using vertical shells. What is their height ? x seems strange: in my drawing height decreases with x.

    [edit][edit]vertical shells require bounds in x, not in y.
     
  4. Mar 19, 2014 #3
    y=x is the first height,no? from 0-1 if not, then what is it
     
  5. Mar 19, 2014 #4

    BvU

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Ah, I see. You want to integrate over x (i.e. horizontal rings), the bounds are x=0 at the lower en x=x at the upper end , so the height is x for the first integral. And also for the second integral.

    Or do you want to use vertical rings and integrate over x ? x does not exceed 1, right ?

    Can you show your drawing ?
     
  6. Mar 19, 2014 #5
    I vertical cylinders, I don't think you can have vertical rings. You can have horizontal rings. Here is the graph 105v0b8.jpg
     
    Last edited: Mar 19, 2014
  7. Mar 19, 2014 #6

    LCKurtz

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Your integrals are set up correctly. You must have some mistake in your evaluating them.
     
  8. Mar 19, 2014 #7
    Yea, I think I did miss a factor of (1/3) somewhere when I solved my integral. Thanks for the help!
     
  9. Mar 19, 2014 #8

    BvU

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Oh boy, my mistake: I revolved around x= -2 and had line x=0 as boundary instead of y=0. Scuse me! Eating humble pie. At least it explains why communicating was difficult.

    And yes, integrals are set up correctly -- from the start :redface:.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Volumes of revolution
  1. Volume by revolution (Replies: 2)

  2. Volume of Revolution (Replies: 2)

  3. Volume of revolution (Replies: 2)

  4. Volume by revolution (Replies: 4)

  5. Volume of revolution (Replies: 10)

Loading...