0
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ Your parametrization imposes $x^2-y^2=1$ out of nowhere. You can't just choose such thibgs at random! $\endgroup$ – Ted Shifrin Nov 18 '13 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 '13 at 16:02
  • $\begingroup$ Haha yikes. I completely forgot about the $\sqrt{\cos 2t}$ term. Thanks a ton :D $\endgroup$ – Artemisia Nov 21 '13 at 17:43
0
$\begingroup$

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 $ds/dt = 8/\sqrt{\cos 2t}$) $$4\cdot 64\int_0^{\pi/4} \sqrt{\cos 2t}\sin t\frac1{\sqrt{\cos 2t}}\,dt = 256\big(1-\frac1{\sqrt2}\big)\,.$$

$\endgroup$
  • $\begingroup$ But is the previous answer that I got incorrect? Can I parameterize twice? $\endgroup$ – Artemisia Nov 18 '13 at 17:36
  • 1
    $\begingroup$ The previous approach is totally wrong. This curve is not a circle. See the figure above. $\endgroup$ – Ted Shifrin Nov 18 '13 at 19:37

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.