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The question asks to find the following line integral $$\int |y|ds$$ along the curve $$C: (x^2+y^2)^2=8^2(x^2-y^2)$$ I used the parameterization $$x=\sec m; y=\tan m$$ and solved it further to get $$\sec^2m+\tan^2m=8$$, which is $$x^2+y^2=8$$. Then I parameterize once again using $$x=\cos t, y=\sin t$$ and the final answer I get is 16. However, I don't have the answer to this question so I am not sure if what I have done is correct.

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  • $\begingroup$ Your parametrization imposes $x^2-y^2=1$ out of nowhere. You can't just choose such thibgs at random! $\endgroup$ Nov 18, 2013 at 22:29
  • $\begingroup$ @TedShifrin: I followed your method to the dot and got the answer as $0$. However, the system says it's incorrect. I will paste the comment given below : //First, you need to parameterize the curve. You can use polar coordinates to do it. Second, you need to find the period, which is to be done in Cartesian coordinates.// We parameterized the curve and the period is from $-3\pi/4$ to $5\pi/4$. Why is $0$ not the right answer? $\endgroup$
    – Artemisia
    Nov 21, 2013 at 16:02
  • $\begingroup$ Haha yikes. I completely forgot about the $\sqrt{\cos 2t}$ term. Thanks a ton :D $\endgroup$
    – Artemisia
    Nov 21, 2013 at 17:43

1 Answer 1

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HINT: Here's a correct way to parametrize the curve. Start with polar coordinates: $x=r\cos t$, $y=r\sin t$. Substitute to obtain $$r^2 = 8^2 \cos 2t\,.$$ Note that this limits $t$ to the intervals where $\cos 2t\ge 0$, i.e., $|t|\le \pi/4$ and $-3\pi/4\le t\le 5\pi/4$. So we have the parametrization \begin{align*} x(t) &= 8\sqrt{\cos 2t}\cos t \\ y(t) &= 8\sqrt{\cos 2t}\sin t \end{align*} on these intervals.

enter image description here

EDIT: By symmetry, the integral is $4$ times what we get integrating from $0$ to $\pi/4$, so we get (you should check that $ds/dt = 8/\sqrt{\cos 2t}$) $$4\cdot 64\int_0^{\pi/4} \sqrt{\cos 2t}\sin t\frac1{\sqrt{\cos 2t}}\,dt = 256\big(1-\frac1{\sqrt2}\big)\,.$$

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  • $\begingroup$ But is the previous answer that I got incorrect? Can I parameterize twice? $\endgroup$
    – Artemisia
    Nov 18, 2013 at 17:36
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    $\begingroup$ The previous approach is totally wrong. This curve is not a circle. See the figure above. $\endgroup$ Nov 18, 2013 at 19:37

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