Someone plays darts that has a join density function based on the surface z = 6 – 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.
$$f(r,\theta)=
\begin{cases}
C_{1}(6-r), & \text{if} (r,\theta)∈ R\\
0, & \text{otherwise}
\end{cases}$$
Where R is the region inside of r = 2+4cosθ (which includes the little loop) and S is just the region inside the inner loop.
Right now I am thinking this might be how to solve for C if that's correct how would I actually find the probability?
Here is your integral
Best Answer
Your calculation of $C_1$ is correct, although most people would have chosen the limits $-2\pi/3 \le \theta \le 2\pi/3$. Yours is simply $2\pi$ added to this. The exact value is $$C_1 = \left(12 \sqrt{3} + \frac{208\pi}{9}\right)^{-1}.$$
To get the desired probability, we have to integrate for the inner lobe, i.e. $$2 \int_{\theta = 2\pi/3}^\pi \int_{r=0}^{|2 + 4 \cos \theta|} C_1 (6-r)r \, dr \, d\theta.$$ Note the use of $2$ (to exploit symmetry) and the absolute value on the upper limit of $r$ because for $2\pi/3 < \theta \le \pi$, we have $r(\theta) < 0$. I get $$\frac{82\pi - 135\sqrt{3}}{52\pi + 27\sqrt{3}}.$$
Since there is interest in the actual computation of the integral for the inner lobe, I will provide an outline. You should be able to fill in the gaps using various elementary trigonometric identities. Note that on $\theta \in [2\pi/3, \pi]$, $r \le 0$, thus $(6-|r|)r = (6+r)r$, and
$$\begin{align} 2C_1 \int_{\theta = 2\pi/3}^\pi \int_{r=0}^{2 + 4 \cos \theta} (6+r)r \, dr \, d\theta &= 2C_1 \int_{\theta = 2\pi/3}^\pi \left[3r^2 + \frac{r^3}{3}\right]_{r=0}^{2 + 4 \cos \theta} \, d\theta \\ &= \frac{8C_1}{3} \int_{\theta = 2\pi/3}^\pi (1 + 2 \cos \theta)^2 (11 + 4 \cos \theta) \, d\theta \\ &= \frac{8C_1}{3} \int_{\theta = 2\pi/3}^\pi 41 + 60 \cos \theta + 30 \cos 2\theta + 4 \cos 3\theta \, d\theta \\ &= \frac{8C_1}{3} \left[ 41\theta + 60 \sin \theta + 15 \sin 2\theta + \frac{4}{3} \sin 3\theta \right]_{\theta=2\pi/3}^\pi. \end{align}$$ The rest is obvious.