Csdrhrt

It would help me if someone can verify the following two proofs I made for the statement below (it's regarding only the second proof).

Let us have a unit ball centered at $(0,1,0)$. And let $(X,Y,Z)$ be a multivariate random variable uniformly distributed in that ball. Then I want to prove/disprove the statement that the distribution of $(X,Y,Z)/sqrt{X^2+Y^2+Z^2}$ is cosine distributed over the unit hemisphere, that is: $p(theta) = frac{costhetasintheta}{pi}$.

First let's consider the $2D$ case. For each point on the unit hemisphere $(x,y) : x = rsintheta, y = rcostheta, theta in [-pi/2,pi/2], r = 1$, I find the intersection of the ray starting at $(0,0)$ with direction $(x,y)$ with the unit circle centered at $(0,1)$, with canonical equation $x^2+(y-1)^2=1$. Plugging in x and y I solve for r: $r^2sintheta+r^2costheta - 2rcostheta + 1 - 1 = 0$, $r(r-2costheta)=0$. Obviously one of the roots is $0$ and the other is $r = 2costheta$. Then the first intersection point is $(0,0)$ and the second is $(2costhetasintheta,2cos^2theta)$. Considering that, I argue that the probability density function induced by the normalization transformation is $p(theta)=frac{costheta}{2}$ over the unit hemicircle. Here's the part I am not so sure about: since we have uniformly distributed points in the unit ball, then the probability for picking a specific direction $theta$ is $p(theta) = C|(2costhetasintheta,2cos^2theta)|$. I am not certain that I can do this - precisely using the distance as the corresponding probability, since I do not provide justification about this except for the fact that it 'follows' from the fact that the distribution inside the ball is uniform, so I can integrate it along each ray to get the corresponding probability for picking a point onto the unit hemisphere in that direction $theta$. Expanding $p(theta) = Csqrt{4cos^2thetasin^2theta + 4cos^4theta} = 2C|costheta|$. After integrating: $int_{-pi/2}^{pi/2}{2C|costheta|dtheta} = 1$, one gets $C = 1/4$, and $p(theta) = frac{costheta}{2}$.

The $3D$ case is similar. Once again we generate the ray: $x=rsinthetacosphi, y=rcostheta, z=rsinthetasinphi$, and intersect it with $x^2 + (y-1)^2 + z^2 = 1$. Plugging in the ray equation into the canonical equation once again I get the solutions: $r(r-2costheta)=0$. Then $p(theta) = C|(2costhetasinthetacosphi, 2cos^2theta, 2costhetasinthetasinphi)|sintheta$. After a few transformations: $p(theta) = 2Ccosthetasintheta$. Integrating $int_{0}^{2pi}{dphi}int_{-pi/2}^{pi/2}{2Ccosthetasintheta dtheta}$ yields $C = 1/2pi$. Ultimately I get $p(theta) = frac{costhetasintheta}{pi}$.

My proof seems to be incorrect as I didn't take into account the $r$ and $r^2$ factors when integrating, yielding $C_1cos^2theta$ and $C_2cos^3theta$ for the pdfs.