# 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.

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

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

• How do I know how to parametrize such a curve? Is there some 'trick' you using? Mar 31, 2015 at 19:35
• Well, I used the fact that a unit circle is traced by $(\cos \theta,\sin \theta)$ for $-\pi \le \theta \le \pi$. Mar 31, 2015 at 19:46

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: $$\int_C\sqrt{1+4 x^2 z^2}\,ds = \int_{-\pi}^{\pi}\left(1+\sin^2(2\theta)\right)\,d\theta=\color{red}{3\pi}.$$

• How did you find the integration bounds though? Mar 31, 2015 at 19:39
• @Kamil: The curve is given by the intersection between a cylinder and a parabolic surface, so it just does one turn around the cylinder axis, i.e. $\theta\in[-\pi,\pi]$. Mar 31, 2015 at 20:04