Finding a line integral along the curve of intersection of two surfaces

Question

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.

Answer 1

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

$$\int_{-\pi}^{\pi} \left(1+4\cos^2 u \sin^2 u\right) du$$

Answer 2

And just completing Dr. MV answer, $$ (x,z,y) = \left(\cos\theta,\sin\theta,\cos^2\theta\right) $$ gives: $$ ds = \sqrt{1+4\sin^2\theta \cos^2\theta}\,d\theta $$ so: $$\int_C\sqrt{1+4 x^2 z^2}\,ds = \int_{-\pi}^{\pi}\left(1+\sin^2(2\theta)\right)\,d\theta=\color{red}{3\pi}.$$