Finding the Total Number of Sheep in a Circular Pen

  • Thread starter Thread starter vorcil
  • Start date Start date
Join the discussion
Registration is free. Start your own thread to ask a follow-up.
4 replies · 2K views
vorcil
Messages
395
Reaction score
0

Homework Statement



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)


Homework Equations


calculus


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?
 
Physics news on Phys.org
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.
 
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

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

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

(assuming I did the integral right)

I got

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

I got 100 sheep
 
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.