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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.

user1956641
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  • "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. – Gerry Myerson Sep 17 '13 at 13:25
  • edited. Ok, I'm reading now about elliptic integrals. – user1956641 Sep 17 '13 at 13:30
  • 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. – user1956641 Sep 17 '13 at 13:41
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    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. – Gerry Myerson Sep 17 '13 at 13:44
  • Are we discussing an ellipse-shaped wire, or an ellipse-shaped plate? – bubba Sep 17 '13 at 13:46
  • I have edited one of your integrals to make it look nicer. You can look at the edit to see how I did it, and then you can do it for your second integral. – Gerry Myerson Sep 17 '13 at 13:48
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    @GerryMyerson Oops, I have edited the second integral so that I can read it ;-p – achille hui Sep 17 '13 at 13:53
  • it is a wire. I'll edit my question to provide you with a whole task. – user1956641 Sep 17 '13 at 15:42
  • updated post. PS: Starting to understand this strange notation)) – user1956641 Sep 17 '13 at 15:52

1 Answers1

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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?

Lisa
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  • 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? – user1956641 Sep 17 '13 at 16:25
  • 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) – Lisa Sep 17 '13 at 16:32
  • I added the next step (solving the integral using the substitution rule) – Lisa Sep 17 '13 at 16:40
  • 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. – user1956641 Sep 17 '13 at 16:40
  • hm, I'll try your way. – user1956641 Sep 17 '13 at 16:42
  • This is a helpful link for solving line integrals which you might be interested in. And here they applied the same substitution to solve a similar problem! – Lisa Sep 17 '13 at 16:43