A quarterback needs to throw the ball quickly, and in his haste, the location it lands is Uniformly distributed within a $30×120$ sq. ft. area in the endzone. The receiver can only catch the ball the ball lands within $7$ feet of him, i.e., within a circle of radius $7$ feet, completely contained in the endzone. What is the probability that the receiver catches the ball? (Assume that the receiver is positioned so that a circle of $7$ feet around him is completely contained within the endzone.)
So the joint pdf for the rectangular endzone region is clearly $f(x,y) = 1/3600$ since it's just the area of the endzone region ($30 \cdot 120$), But I'm a little confused on where to go from here since we want to calculate the probability associated with the part of the endzone region that represents the receiver whose joint pdf is $1/(49\pi)$ as the area of the circle is $\pi r^2 = \pi7^2 = 49\pi$
I want to make sure that my logic is right.
Method 1: Using area. If the area of the rectangular region is $3600$ and the area of the circular region within the rectangular region is $49\pi$ then using proportionality, the probability is $\cfrac{49\pi}{3600}$
Method 2: Polar coordinates.
$P$(Reciever catching) $= \int_0^{2\pi}\int_0^7 \cfrac{r}{3600} drd\theta = \cfrac{49\pi}{3600}$ My logic is if $f(x,y) = 1/3600$ represents the rectangular region, then using the polar coordinates from $0 \leq r \leq 7$ and $0 \leq \theta \leq 2\pi$ for the rectangular joint pdf would represent the region of the circle within the rectangle giving us the probability.
Are both methods fine? Is one wrong and the other right? Is my answer way off? Please let me know! Thank You!