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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.

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2 Answers 2

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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

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

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  • $\begingroup$ How do I know how to parametrize such a curve? Is there some 'trick' you using? $\endgroup$
    – Kamil
    Mar 31, 2015 at 19:35
  • $\begingroup$ Well, I used the fact that a unit circle is traced by $(\cos \theta,\sin \theta)$ for $-\pi \le \theta \le \pi$. $\endgroup$
    – Mark Viola
    Mar 31, 2015 at 19:46
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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: $$\int_C\sqrt{1+4 x^2 z^2}\,ds = \int_{-\pi}^{\pi}\left(1+\sin^2(2\theta)\right)\,d\theta=\color{red}{3\pi}.$$

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  • $\begingroup$ How did you find the integration bounds though? $\endgroup$
    – Kamil
    Mar 31, 2015 at 19:39
  • $\begingroup$ @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. $\theta\in[-\pi,\pi]$. $\endgroup$ Mar 31, 2015 at 20:04

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