Evaluating the line integral of $\sqrt{1 + 4x^2z^2}$ over the curve of intersection of $x^2 + z^2 = 1$ and $y = x^2$

Question

I am trying to evaluate the line integral of the function $f(x, y, z) = \sqrt{1 + 4x^2z^2}$ over the curve of intersection of $x^2 + z^2 = 1$ and $y = x^2$. Since at the curve $x^2 + z^2 = 1 \Longleftrightarrow y = 1 - z^2$, $f(x, y, z) = \sqrt{1 + 4yz^2} = \sqrt{1 + 4(1 - z^2)z^2} = \sqrt{1 + 4z^2 - 4z^4}$. Thus I figured that a parametrization of $z \in [-1 ,1], y = 1 - z^2$ is sufficient, but then as $r(z) = (z, 1 - z^2)$, $||r'(z)|| = \sqrt{1 + 4z^2}$ so that $f(r(z))||r'(z)|| = \sqrt{1 + 8z^2 + 12z^4 - 16z^6}$. Is this really the way of evaluating the line integral, or is there an easier way I haven't figured out?

Answer 1

Parametrize the path $(x,y,z) = (\cos t, \cos^2 t, \sin t)$ and the integral is simply

$$\int_0^{2\pi} 1 + \sin^2 2t \:dt = 3\pi$$