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Circumference of an ellipse

  1. Feb 18, 2012 #1
    all of my work so far is in the picture. i'm stuck on what i should do next.
     

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  2. jcsd
  3. Feb 18, 2012 #2
    your handwriting is incredibly neat, good show!
     
  4. Feb 18, 2012 #3

    HallsofIvy

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    You are making the substitution [itex]x= a sin\theta[/itex] but then your integral has both x and [itex]\theta[/itex]. That's not right.

    However, I would advise using the parametric equations [itex]x= a sin(\theta)[/itex], [itex]y= b cos(\theta)[/itex] rather than that complicated equation.
     
  5. Feb 19, 2012 #4
    I'm not sure I understand what you mean by the parametric equations.. how does that fit into what I already have?
     
  6. Feb 19, 2012 #5
    thanks!
     
  7. Feb 19, 2012 #6

    Dick

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    You were doing alright until after you drew the triangle. But you want to get rid of all of the x's in the thing you are integrating. And you never used dx=a cos(θ) dθ, probably because you weren't writing the dx in the integration. You need that.
     
  8. Feb 21, 2012 #7
    oh darn! i did forget. okie well, now that I have dθ in there and i changed x^2, i still have a really ugly equation.. what should i do next?
     

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  9. Feb 21, 2012 #8

    Dick

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    Bring the cos(θ) inside the square root where it becomes cos^2(θ). And i) replacing the a^2 with x^2/sin^2(θ) doesn't do you any good and ii) somewhere you missed cancelling an a^2.
     
    Last edited: Feb 21, 2012
  10. Feb 21, 2012 #9
    i completely reworked it, and it looks so much better now! lol.
    now i have integral from 0 to a of sqrt(1- (b^2/a^2)cos(θ))
     
  11. Feb 21, 2012 #10

    Dick

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    I thought you were supposed to get the integral of sqrt(1-k*sin^2(θ))??
     
  12. Feb 21, 2012 #11
    right. k = 1-(b^2/a^2). did i cancel sin^2(θ) instead of cos(θ)?
     
  13. Feb 21, 2012 #12

    Dick

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    Hard to say. What did you do?
     
  14. Feb 21, 2012 #13
    i've completely messed up. i don't know what to use for substitutions.. i always end up in the same place. and i don't know where dx = a cos(θ) dθ fits into all of this.

    sorry i'm a lot confused!
     

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  15. Feb 21, 2012 #14

    Dick

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    You are going in circles. Look you've got [itex]\int \sqrt{1+\frac{b^2 \sin^2{\theta}}{a^2 \cos^2{\theta}}} a \cos{\theta} d\theta[/itex]. Bring the cos into the square root, so you've got [itex]\int \sqrt{(1+\frac{b^2 \sin^2{\theta}}{a^2 \cos^2{\theta}}) \cos^2{\theta}} a d\theta[/itex]. Simplify inside the radical. Then use your trig identity and change the x limits to theta limits.
     
  16. Feb 21, 2012 #15
    but if k = 1 - (b^2/a^2) and under the radical says 1-ksin^2(θ), shouldn't the end result under the radical, when expanded, be 1 - (sin(θ))^2- (b^2/a^2)(sin(θ))^2?

    and where does the 4 outside the integral come from?
     
  17. Feb 21, 2012 #16

    Dick

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    No! It should be 1 - (sin(θ))^2 + (b^2/a^2)(sin(θ))^2! You are subtracting k. That's just being sloppy. And if you are integrating x from 0 to a you are only integrating over 1/4 of the ellipse. You are just covering the first quadrant.
     
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