# Solve this number equation

1. Jul 19, 2010

### vorcil

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

$$n(r) = \frac{N_{0}}{\pi} (R - r)$$

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
calculus

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
$$2/pi r * dr$$ in length

multiplying that by my equation I get the equation

$$\int dn(r) = \int \frac{No}{\pi} (R-r) 2\pi r dr$$

moving the constant outside the equation

$$\int dn(r) = \frac{No}{\pi} \int (R-r) 2\pi r dr$$

expanding the right side of the equation I get

$$R * (2\pi r dr) + -r * (2\pi r dr)$$

integrating I get
$$n = \frac{No}{\pi} (\frac{R 2\pi r^2}{2} - \frac{2\pi r^3}{3})$$

simplifying it giving me the final equation

$$n = \frac{No}{\pi} ({R\pi r^2} - \frac{2}{3}\pi r^3 )$$

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. Jul 19, 2010

### 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
$$\int_?^? \frac{N_0}{\pi} (R-r) 2\pi r dr$$

Before integrating, you can bring all your constants out of the integral. You can substitute for N0.

3. Jul 19, 2010

### vorcil

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

4. Jul 19, 2010

### Staff: Mentor

I get 100 sheep, too.
The limits in your integral are backwards - it should be
$$\int_0^R \frac{N_0}{\pi} (R-r) 2\pi r dr$$
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.

5. Jul 19, 2010

thanks