given
$A:= \{ (x_1,x_2,x_3) \in \mathbb{R} : x_1^2 + x_2^2 + x_3^2 = 18, x_3 \leq 2\}$
show that $A$ is a manifold with boundary and calculate $ \delta A$ where $\delta A$ is the boundary of A.
I know that $B_{18}:=\{ (x_1,x_2,x_3) \in \mathbb{R} : x_1^2 + x_2^2 + x_3^2 = 18 \}$ is a Manifold since for
$f(x_1,x_2,x_3):= x_1^2 + x_2^2 + x_3^2 - 18 $
$B_{18}= \{ \{ (x_1,x_2,x_3) \in \mathbb{R} : f(x_1,x_2,x_3) = 0 \}$ and $df|_x \neq 0 \forall x \in B_{18}$
So for every $p \in B_{18} $ there are open sets $U,V$ and a diffeomorphism $\phi :U \to V$ with $\phi(U \cap B_{18}) = V \cap \mathbb{R}^d_0$
$A$ is a subset of $B_{18}$ therefore the same is true for $A$ which means $A$ is also a manifold. Is this correct?
To calculate the boundary of $A$
we have $\mathbb{R}^3_+:=\{ (x_1,x_2,x_3) \in \mathbb{R}^3 : x_3 = 0 \} $
$\delta A=\{ (x_1,x_2,x_3) \in \mathbb{R} : x_1^2 + x_2^2 + x_3^2 = 18, x_3 \leq 2\} \cap \mathbb{R}^3_+= \{ (x_1,x_2,u) \in \mathbb{R} : x_1^2 + x_2^2 + (u+2)^2 = 18, u = 0 \} = \{ (x_1,x_2,0) \in \mathbb{R} : x_1^2 + x_2^2 = 14 \} $
is this correct?
The tangent space and the normal space would be the same as for $B_{18}$ right? $\forall p \in A$, $ T_pA = T_pB_{18}$ the same for the normal space.