Find the parametrization of the surface delimited by the revolution of the parabole $$z=x^2$$, for $$z$$ between $$1$$ and $$3$$.

### Development:

Since it is a revolution body, we can use the last example of surfaces, this way $$$\varphi(x,\theta) =(x\cdot\cos\theta,x\cdot\sin\theta,x^2)$$$

The domain will be $$\theta\in[0,2\pi]$$ and for $$x$$, it is necessary to bear in mind that $$x=\sqrt{z}$$, therefore if $$z\in[1,3], \ x\in[\sqrt{1},\sqrt{3}]=[1,\sqrt{3}]$$.

Another form to parametrize would be to take $$z$$ as a variable, then $$$\varphi(z,\theta) =(\sqrt{z}\cdot\cos\theta,\sqrt{z}\cdot\sin\theta,z), \ \ z\in[1,3]$$$

### Solution:

The parametrization of the surface is $$\varphi(x,\theta) =(x\cdot\cos\theta,x\cdot\sin\theta,x^2), \ \ x\in[1,\sqrt{3}]$$ or $$\varphi(z,\theta) =(\sqrt{z}\cdot\cos\theta,\sqrt{z}\cdot\sin\theta,z), \ \ z\in[1,3]$$