Parametrization of a parabola inside a cylinder

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So first of all, I would like to say that I wish to confirm whether the parametrization I am doing is correct. I've been told that $x = r\cos t, y = r\sin t, z = z$, which makes me hesitant on my answer.

Paraboloid: $y = x^2 + z^2 \text{ and } x^2 + z^2 = 16$

Which gives us a cylinder and parabola parallel to the $y$-axis with a radius of 4.

I THINK IT SHOULD BE THIS, SINCE THE CIRCLE IS ON THE X-Z PLANE:

$$x = r\cos t, y = y, z = r\sin t$$

Parametrization should be: $\langle r\cos t, y, r\sin t\rangle$

And then when I try to convert the scalar surface integral into the parametrized form in order to find surface area, I would use:

$$\int_{0}^{4}\int_{0}^{2\pi} \delta \text { } |r_t \text{ } x \text{ }r_y|\,\mathrm dt\mathrm dy$$

Can anyone tell me if I'm going on the correct route, or if I'm totally off track?

Thank you!

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I missed a bit of what you're doing. You're parametrizing the portion of the paraboloid inside the cylinder, so you need to put $y=r^2$ as your polar-coordinate expression in the parametrization. So the parametrization should be in terms of $r$ and $t$ (with $0\le r\le 4$ and $0\le t\le 2\pi$).