Curves and Surfaces

Question

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!

Answer 1

Here are 2 pictures of the intersection:

Answer 2

You are correct with the $z=1/5,1$. Notice that taking $z$ to be constant in either $S$ or $T$ results in an equation of the form $$x^2+y^2=\text{const}. $$ Can you recognize what this looks like in the $xy$-plane?