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!

Deriving the first moment of area of semicircle

  1. Apr 25, 2011 #1
    1. The problem statement, all variables and given/known data
    Derive via integration the first moment of area Q of a semicircle with radius r.

    2. Relevant equations
    [tex]Q = \int_{A} y dA[/tex]

    [tex] A_{semicircle} = \frac{\pi r^{2} }{2}[/tex]

    3. The attempt at a solution
    [tex] A = \frac{\pi r^{2} }{2}[/tex]
    [tex] A(y) = \frac{\pi y^{2} }{2}[/tex]
    [tex] dA = \pi y dy[/tex]

    [tex]Q = \int^{y=r}_{y=0} y dA[/tex]
    [tex] = \int^{r}_{0} \pi y^{2} dy[/tex]
    [tex] = \frac{\pi}{3} [y^{3}]^{r}_{0}[/tex]

    [tex] Q = \frac{\pi r^{3}}{3}[/tex]


    But the answer is [tex]\frac{2 r^{3} }{3}[/tex], which my textbook derived from the equation [tex]Q = (area) \times (centroidal height) [/tex]. I want to know how to derive the Q for any shape without knowing its centroidal height beforehand. Can someone help me out with why I got a different and wrong answer?
     
  2. jcsd
  3. Apr 26, 2011 #2

    nvn

    User Avatar
    Science Advisor
    Homework Helper

    Elbobo: dA is not pi*y*dy. Hint: Shouldn't dA instead be, dA = 2[(r^2 - y^2)^0.5]*dy? Try again.
     
  4. Apr 26, 2011 #3
    Sorry, I really don't understand why dA equals that. My A(y) must be wrong then? What should it be and why?
     
  5. Apr 26, 2011 #4

    nvn

    User Avatar
    Science Advisor
    Homework Helper

    Elbobo: A(y) = integral(dA), integrated from y = y1 to y = r. In your particular case, y1 = 0.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Similar Discussions: Deriving the first moment of area of semicircle
  1. Area moment (Replies: 1)

Loading...