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!

Homework Help: Trouble visualizing what is going on. Volume of object

  1. Sep 3, 2013 #1
    1. The problem statement, all variables and given/known data
    Here is a link to a pdf .http://www.mrskeller.net/documents/hwsols14.pdf
    I'm having issues with number 56. The solution makes no sense to me.

    2. Relevant equations

    3. The attempt at a solution

    So, I did graph the equation y = sqrt(r^2 - x^2) and I understand that height of the circle. I understand that you need to integrate from -r to r. But I do not understand the integral by any means.

    They have ##\int 4(r^2-x^2)) dx ## from -r to r
    I thought that maybe there were doing something of the form ∏∫y^2 because that is what this chapter uses. But I guess not. They say that the length of a side is 2y = 2sqrt(r^2 - x^2). And the only way that I can see how to get from that equation to the integral they suggest is ly squaring 2y = 2sqrt(r^2 - x^2). So I don't understand that even because if you squared it and used the integral of the form ∏∫y^2 dx then your 4 dissapears. I just don't understand what is going on with this problem. Thanks
  2. jcsd
  3. Sep 3, 2013 #2


    User Avatar
    Homework Helper

    The volume of a solid with a known cross section is basically
    [tex]{\int_a}^b A(x) dx[/tex],
    where A(x) represents the area of the cross section.

    I believe the attached graphic illustrates the problem. Imagine a square lying on its side, the side being the thin rectangle in the diagram. Now imagine a bunch of thin vertical rectangles going from left to right (from -r to r). The distance from the x-axis to the top of this particular rectangle is y. Then you know the length of this thin rectangle (= the length of the square cross section), which is 2y. The area of the square cross section is
    [tex]A = s^2[/tex]
    (s is the length of the side),
    so plug in 2y for s. Now, what does y equal in terms of x and r? Plug that in for y, and the resulting expression serves as the integrand for this problem.

    Attached Files:

  4. Sep 3, 2013 #3
    OMG! Thank you that makes a lot more sense. A lot more!
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted