Finding the Total Number of Sheep in a Circular Pen

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Homework Help Overview

The problem involves calculating the total number of sheep in a circular pen, where the density of sheep per unit area varies with the radius. The density function is given, and the task is to integrate this function over the area of the pen.

Discussion Character

  • Mathematical reasoning, Problem interpretation, Assumption checking

Approaches and Questions Raised

  • Participants discuss setting up the integral for the total number of sheep, considering the limits of integration and the correct interpretation of the density function. There is a focus on ensuring the integral is definite and correctly oriented.

Discussion Status

Some participants have provided feedback on the setup of the integral and the interpretation of the density variable. There is an acknowledgment of potential errors in the limits of integration and the units of the density constant. Multiple interpretations of the problem are being explored, particularly regarding the integration limits.

Contextual Notes

There is a noted confusion regarding the units of the density constant, with a suggestion that it may have been misinterpreted as cubic meters instead of square meters. This could affect the understanding of the problem setup.

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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?
 
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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.
 
thanks
 

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