Prove that equations $x^2+y^2=z^2$ and $y=z\sin(z)$ are bounded smooth curves on $D={(x,y,z) \in \mathbb{R}^3;x>0, \frac{\pi}{2}<z<\frac{3\pi}{2}}$

Question

I have to prove that equations $x^2+y^2=z^2$ and $y=z\sin(z)$ are bounded smooth curves on $D=\{(x,y,z) \in \mathbb{R}^3;x>0, \frac{\pi}{2}<z<\frac{3\pi}{2}\}$

So I put $y$ on other equation and I got $x^2=z^2\cos^2z$. If I consider $D$, where are $x>0$ and $\frac{\pi}{2}<z<\frac{3\pi}{2}$, then $x=-z\cos(z)$.

I came to this curve $\vec{r}=(-z\cos(z), z\sin(z), z)$ and I do not know how to prove that this is bounded smooth curve.

Any help?