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Homework Help: Solve this number equation

  1. Jul 19, 2010 #1
    1. The problem statement, all variables and given/known data

    Sheep are collected in a circular pen in such a way that the number per unit area at radius r is given by

    [tex] n(r) = \frac{N_{0}}{\pi} (R - r)[/tex]

    Where R = 10, and N0 = 0.3m^-3

    Find the Total number of sheep in the pen (round your answer to the nearest integer)

    2. Relevant equations

    3. The attempt at a solution

    Drawing it out, if you take the segments of the circular pen to be small rings,

    then you get a strip, that is
    [tex] 2/pi r * dr [/tex] in length

    multiplying that by my equation I get the equation

    [tex] \int dn(r) = \int \frac{No}{\pi} (R-r) 2\pi r dr [/tex]

    moving the constant outside the equation

    [tex] \int dn(r) = \frac{No}{\pi} \int (R-r) 2\pi r dr [/tex]

    expanding the right side of the equation I get

    [tex] R * (2\pi r dr) + -r * (2\pi r dr) [/tex]

    integrating I get
    [tex] n = \frac{No}{\pi} (\frac{R 2\pi r^2}{2} - \frac{2\pi r^3}{3}) [/tex]

    simplifying it giving me the final equation

    [tex] n = \frac{No}{\pi} ({R\pi r^2} - \frac{2}{3}\pi r^3 ) [/tex]

    I need to figure out how to calculate a number!

    pls help

    i probably got it wrong, can I just substitute in 10, for the values R and also r?
  2. jcsd
  3. Jul 19, 2010 #2


    Staff: Mentor

    Your integral looks fine, but you should be working with a definite integral. What are the possible values of r (as an interval)? IOW, what are the minimum and maximum values of r?
    This makes your integral
    [tex] \int_?^? \frac{N_0}{\pi} (R-r) 2\pi r dr [/tex]

    Before integrating, you can bring all your constants out of the integral. You can substitute for N0.
  4. Jul 19, 2010 #3
    Oh, so it's from the centre of the circle, where r=0 out to the edge, where I'm guessing the edge is R=10m

    \int_R^0 \frac{N_0}{\pi} (R-r) 2\pi r dr

    n = \left \frac{No}{\pi} ({R\pi r^2} - \frac{2}{3}\pi r^3 ) \right|_R^0

    (assuming I did the integral right)

    I got

    n = \frac{No}{\pi} ({\pi r^3} - \frac{2}{3}\pi r^3 )
    substituting in 10 for r,

    I got 100 sheep
  5. Jul 19, 2010 #4


    Staff: Mentor

    I get 100 sheep, too.
    The limits in your integral are backwards - it should be
    [tex]\int_0^R \frac{N_0}{\pi} (R-r) 2\pi r dr [/tex]
    If you had evaluated the integral you wrote correctly, you would have gotten -100 sheep.

    Another thing - you said that N0 = 0.3m^-3
    That should be N0 = 0.3m^(-2), or .3 sheep per square meter, not cubic meter. It might be per cubic meter if the sheep were stacked up in the pen.
  6. Jul 19, 2010 #5
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