Find the probability density function of a dart.

Question

Someone plays darts that has a join density function based on the surface z = 3 - r where z is the likelihood of the dart landing at $(r,\theta)$. How can I find the value of C so that f is a joint density function and then find the probability that the thrown dart lands in S?

$$g(r,\theta)= \begin{cases} C_{1}(3-r), & \text{if} (r,\theta)∈ R\\ 0, & \text{otherwise} \end{cases}$$

R is the region inside of r = 1 + tan $(\theta /4)$ (which includes the little loop) and S is just the region inside the inner loop.

So far this is how I am solving for C but how can I actually get the probability? Also is this correct so far?

Answer 1

Limits of your integral do not seem correct.

The outer loop of the region $R$ forms for $\theta \in \displaystyle \small \bigg(\big (0, \frac{4\pi}{3} \big), \big(\frac{7\pi}{3}, \frac{8\pi}{3} \big) , \big(\frac{11\pi}{3}, 4\pi\big)\bigg)$

And the inner loop $S$ forms for $ \ \theta \in \displaystyle \small \big(\frac{8\pi}{3}, \frac{11\pi}{3}\big)$.

So the integral to find $\small C$ should be,

$\displaystyle \small \int_{0}^{4 \pi/3} \int_{0}^{|1 + \tan (\theta/4)|} C(3 - r) \ r \ dr \ d\theta + \int_{7 \pi /3}^{8 \pi/3} \int_{0}^{|1 + \tan (\theta/4)|} C(3 - r) \ r \ dr \ d\theta$

$+ \displaystyle \small \int_{11 \pi /3}^{4 \pi} \int_{0}^{|1 + \tan (\theta/4)|} C(3 - r) \ r \ dr \ d\theta$

$ \small \approx 14.043 C \implies \small C \approx 0.0712 $

Then the desired probability is,

$\displaystyle \small \int_{8 \pi /3}^{11 \pi / 3} \int_{0}^{|1 + \tan (\theta/4)|} C(3 - r) \ r \ dr \ d\theta \approx 0.0551 $

Answer 2

To help you understand just how complicated the integral setup is, here is the polar plot for the curve $r(\theta) = 1 + \tan \frac{\theta}{4}$ and the Cartesian plot of the same curve, with $\theta$ on the horizontal axis and $r$ on the vertical axis:

I have limited the domain of $\theta$ to the area of interest, namely $\theta \in [-5\pi/3, 4\pi/3]$. I leave it as an exercise to show that that this corresponds to the point of intersection in Quadrant III. The other intersection point, in Quadrant IV, corresponds to $\theta = -4\pi/3$ and $\theta = -\pi/3$. I have color coded the segments. Note the red and green segments correspond to $r < 0$; the blue and orange correspond to $r > 0$. The red and orange segments correspond to the larger (outer) lobe, and the green and blue to the smaller (inner) lobe. In this way, we can see that the region enclosed by the outer lobe is the sum of an integral on $\theta \in \color{red}{[-5\pi/3, -4\pi/3]}$ and on $\theta \in \color{orange}{[-\pi/3, 4\pi/3]}$, but because the red interval has $r < 0$, we need to integrate $(3 - |r|)r = (3 + r)r$, so $$\frac{1}{C} = \int_{\theta = -5\pi/3}^{-4\pi/3} \int_{r=0}^{1 + \tan \theta/4} (3+r)r \, dr \, d\theta + \int_{\theta = -\pi/3}^{4\pi/3} \int_{r = 0}^{1 + \tan \theta/4} (3-r)r \, dr \, d\theta.$$ Then the inner lobe is handled similarly, with $\theta \in \color{green}{[-4\pi/3, -\pi]}$ and $\theta \in \color{blue}{[-\pi, -\pi/3]}$: $$\int_{\theta = -4\pi/3}^{-\pi} \int_{r = 0}^{1 + \tan \theta/4} (3+r)r \, dr \, d\theta + \int_{\theta = -\pi}^{-\pi/3} \int_{r = 0}^{1 + \tan \theta/4} (3-r)r \, dr \, d\theta.$$ This expression is divided by $C$ to get the desired probability.