The cylinder $x^2+(y-\frac{1}{2})^2=\frac{1}{4}$ crosses the ball $x^2+y^2+z^2=1$ along a curve $\gamma$.
Calculate the angle at which $\gamma$ crosses itself.
First, I tried to find $\gamma$:
$x^2=1-y^2-z^2 \;\Rightarrow \; 1-y^2-z^2+y^2-y+\frac{1}{4}=\frac{1}{4} \Rightarrow z^2=1-y$
How can I parametrize what I got? and How can I then find the desired angle?
You'll need to convert $x=r\cos\theta$, $y=r\sin\theta$. By doing so, the cylinder has the equation $r=\sin\theta$, and the equation of the ball is $z^2=1-r^2$. The curve $\gamma(\theta)$ is parameterized as: $$\begin{align*} x&=\sin\theta\cos\theta\\ y&=\sin^2\theta\\ z&=\pm\sqrt{1-\sin^2\theta}=\pm\cos\theta. \end{align*}$$
To find such $\theta\in[0,2\pi]$: observe that $z$ is either plus or minus. For the same $\theta$, $z$ could be in two positions. You can conclude that $\cos\theta$ has to be zero, otherwise that point would not be intersecting point. There are two solutions to $\cos\theta=0$, namely $\theta=\pi/2$ and $\theta=3\pi/2$. You can check $\gamma(\frac{\pi}{2})=\gamma(\frac{3\pi}{2})=(0,1,0)$, the crossing point.