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Evaluate the line integral,

$$\int_C |y|\, {\rm d}s,$$

Where the curve $C$ is the is given by the equation $(x^2+y^2)^2=29^2(x^2-y^2)$

What I tried.

I recognize this as a scalar valued function, hence I know that I must first parameterize the curve C. What I did was to change it to polar coordinates and getting $r^4=29^2(r^2\cos(t)-r^2\sin(t))$ and then simplifying it to get $r=29(\cos2t)^{0.5}$ and hence getting a parameterization of $r(t)=(29(\cos2t)^{0.5}\cos(t),29(\cos2t)^{0.5}\sin(t))$ but I'm unsure whether my parametrisation is correct , could anyone please explain. Thanks

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1 Answer 1

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Hint: Your curve looks like this:

enter image description here

And it has the following solutions:

$$y = -\sqrt{-2 x^2-29 \sqrt{8 x^2+841}-841}/\sqrt{2}$$ $$y = \sqrt{-2 x^2-29 \sqrt{8 x^2+841}-841}/\sqrt{2}$$ $$y = -\sqrt{-x^2+29/2 \sqrt{8 x^2+841}-841/2}$$ $$y = \sqrt{-x^2+29/2 \sqrt{8 x^2+841}-841/2}$$

This should allow you to split the integral into a form like this: $\int_a^b|f|\,dx$

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  • $\begingroup$ I still dont get it. Could u explain, Thanks $\endgroup$
    – ys wong
    Nov 11, 2014 at 2:58

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