Probability hit “bullseye”

I have to finish en example, but don't know how to do it.

We hahve a shooting target. Let's say that distribution in this target is bivariate normal distribution (x,y). . We have this formula for probability density:

The center of the target is . The points (x,y) will have polar coordinate:

Probability density f(x,y) is:

I need to calculate the (integral) probability hit on "bullseye" if we have r=1 for the last formula. Can anyone know how to do it?

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@Walls You beat me by 5 seconds! Ah! :P – Aneesh Dogra Mar 21 '13 at 16:34
Ok. I will move there. Thanks – Like Mike Mar 21 '13 at 16:54

You're trying to integrate

$$P = \int\limits_0^R {\frac {rdr}{2π \sigma^2}} e^{\frac{-(r/\sigma)^2}{2}}$$

The derivative of

$$e^{\frac{-(r/\sigma)^2}{2}}$$ is

$$-\frac{r}{\sigma^2} \cdot e^{\frac{-(r/\sigma)^2}{2}}$$

So the integral of

$$\frac{r}{2π \sigma^2} e^{\frac{-(r/\sigma)^2}{2}}$$

is

$$-\frac{e^{\frac{-(r/\sigma)^2}{2}}} {2\pi}$$

which is easy enough to evaluate at $R$ and $0$.

EDIT:

To review, if

$$f(r)=\frac{d}{dr}g(r)$$

then

$$\int\limits_a^b {f(r)}dr = g(b)-g(a)$$

In this case

$$\frac{r}{2π \sigma^2} e^{\frac{-(r/\sigma)^2}{2}} = \frac{d}{dr}\left(-\frac{e^{\frac{-(r/\sigma)^2}{2}}} {2\pi}\right) = \frac{-1}{2\pi}\frac{d}{dr}\left({e^{\frac{-(r/\sigma)^2}{2}}}\right)$$

so

$$P = \int\limits_0^1\frac{r dr}{2π \sigma^2} e^{\frac{-(r/\sigma)^2}{2}} = \frac{-1}{2\pi}\left({e^{\frac{-(1/\sigma)^2}{2}}}-{e^{\frac{-(0/\sigma)^2}{2}}}\right)$$

$$= \frac{-1}{2\pi}\left({e^{\frac{-(1/\sigma)^2}{2}}}-1\right)$$

$$= \frac{1}{2\pi}\left(1-{e^{\frac{-(1/\sigma)^2}{2}}}\right)$$

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@Harold: Thanks! – Beta Mar 21 '13 at 19:38
Thanks. But still have problems. I did learn 10 years ago (just basic integrals and derivates) and i forgot how to calculate this. – Like Mike Mar 22 '13 at 12:36
@LikeMike: which step is giving you trouble? – Beta Mar 22 '13 at 12:44
Is there just enough to put value for r=1 in the $$-\frac{e^{\frac{-(r/\sigma)^2}{2}}} {2\pi}$$ and get result? – Like Mike Mar 22 '13 at 12:58
if sigma is 0 does that mean the chance is zero? – ldgorman Mar 22 '13 at 17:08