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I have a task, to find a part of perimeter of an ellipse (on plane), I know it's density function. Both ellipse equation and density are given in cartesian coordinates. So I set up a line integral, but it seems to me quite complex to calculate: $${1\over4}\int_0^{2\pi}(2\cos t-7\sin t-\cos t\sin t+2)\sqrt{(7\cos t)^2+(-\sin t)^2}\,dt$$ or integral $$\int_{\frac{\pi}{2}}^{\pi}(2\cos t-7\sin t-\cos t\sin t+2)\sqrt{(7\cos t)^2+(-\sin t)^2}\,dt$$ both integrals should give the same result. How can I solve this? I thought about trying to rewrite all in polar coordinates, but got confused, because I'm not quite sure how.

Find a mass of a line arc

$$(x+1)^3 + {(y-2)^2\over49}=1, x>=-1, y<=2$$. Density: $$p(x,y)=2x-y-xy+2$$.

What was my logic: $$m=\int_{L}p(x,y)dl$$ where p - density, m - mass, L- is a quarter of ellipse perimeter length curve.

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  • $\begingroup$ "Elliptic curve" is a technical term in higher mathematics; it does not mean the same thing as "ellipse". The perimeter of an ellipse leads to what are known as "elliptic integrals", another advanced concept. $\endgroup$
    – Gerry Myerson
    Sep 17, 2013 at 13:25
  • $\begingroup$ edited. Ok, I'm reading now about elliptic integrals. $\endgroup$
    – user1956641
    Sep 17, 2013 at 13:30
  • $\begingroup$ Hm, I've read wiki about elliptic integrals, and I am quite sure I don't have to learn this concept to complete my task, because we don't have it in our program. There must be another way. $\endgroup$
    – user1956641
    Sep 17, 2013 at 13:41
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    $\begingroup$ Well, are you sure you want to find the perimeter of an ellipse? You write of "mass" and "density", neither of which are relevant to finding the perimeter. $\endgroup$
    – Gerry Myerson
    Sep 17, 2013 at 13:44
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    $\begingroup$ @GerryMyerson Oops, I have edited the second integral so that I can read it ;-p $\endgroup$
    – achille hui
    Sep 17, 2013 at 13:53

1 Answer 1

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Here are some remarks:

  • I think your integral should go from $\frac{3\pi}{2} \leq t \leq 2\pi$ if I'm not mistaken.

  • $p\left(x(t),y(t)\right)$ simplifies to

    $p\left(x(t),y(t)\right) = -7cos(t)sin(t) $

  • Now substitute:

    $u = sin^2(t) + 49 cos^2(t),$

    $\frac{du}{dt} = 2(1-49) * cos(t)sin(t)$

    leads to

    $m= \int -7 cos(t)sin(t) * \sqrt{u} \frac{du}{2(1-49) cos(t)sin(t)} $

    where the $cos(t)sin(t)$ cancel out. Is the rest clear?

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  • $\begingroup$ You're quite right about 3pi/2<=t<=2pi. But I'm not sure about simplification p(x(t),y(t))= -7cos(t)sin(t). Where did you lose/got rid of 2cost-7sint+2? $\endgroup$
    – user1956641
    Sep 17, 2013 at 16:25
  • $\begingroup$ Do you agree with $x(t)=cos(t)-1$, $y=7 sin(t) + 2$? From there it's just writing all the terms and deleting the ones that cancel out (I double checked and still get the same result) $\endgroup$
    – Lisa
    Sep 17, 2013 at 16:32
  • $\begingroup$ I added the next step (solving the integral using the substitution rule) $\endgroup$
    – Lisa
    Sep 17, 2013 at 16:40
  • $\begingroup$ oh. I see, we made different substitutions. I brought ellipse to canonical form, assuming that it wouldn't change a curve length. But we still have eleptical integral. i'm quite sure the good idea would be to use polar cordinates here. So that what I'm trying now. $\endgroup$
    – user1956641
    Sep 17, 2013 at 16:40
  • $\begingroup$ hm, I'll try your way. $\endgroup$
    – user1956641
    Sep 17, 2013 at 16:42

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