<s>[INST] Finding a line integral along the curve of intersection of two surfaces Find \begin{align*} \int_C \sqrt{1+4x^2 z^2} ds, \end{align*} where $C$ is the curve of intersection of the surfaces $x^2 + z^2 = 1$ and $y = x^2$. 
Attempt at solution: So first I need a parametrization of this curve. I let $x = t$. Then we have $y = t^2$ and $z = +- \sqrt{1-t^2}$. But I'm not sure what sign I should pick here, and what my integration bounds are?
Any help would be appreciated.
 [/INST] Parameterize a curve by letting $x=\cos u$, $z = \sin u$, and $y=\cos^2 u$, where $-\pi \le u \le \pi$.
Then, 
$$\begin{align} 
ds &=\sqrt{x'(u)^2+y'(u)^2+z'(u)^2}\, du\\\\
&=\sqrt{\sin^2u+4\cos^2u\sin^2u+\cos^2u}\, du\\\\
&=\sqrt{1+4\cos^2u\sin^2u}\, du
\end{align}$$
The integral becomes
$$\int_{-\pi}^{\pi} \left(1+4\cos^2 u \sin^2 u\right) du$$
</s>