Q:

Curves and Surfaces

Suppose $$S=\{(x,y,z) \in \mathbb{R}^3 : x^2+y^2+(z-1)^2=1\}$$ and                $$T=\left\{(x,y,z) \in \mathbb{R}^3 : \frac{(z+1)^2}{4}=x^2+y^2, z \geq-1 \right\}.$$
(1) Find the $z$-coordinates of the points of intersection of $S$ and $T$ and sketch the projection into the $xy$-plane of the curves of intersection.
I tried to solve simultaneously and found that $z=1/5, 1$. Not sure if this is correct. Also am unsure about the sketch.
Thanks!

A:

Here are 2 pictures of the intersection:

