# circumference of an ellipse

by emc92
Tags: circumference, ellipse
 P: 33 all of my work so far is in the picture. i'm stuck on what i should do next. Attached Thumbnails
 P: 615 your handwriting is incredibly neat, good show!
 Math Emeritus Sci Advisor Thanks PF Gold P: 38,895 You are making the substitution $x= a sin\theta$ but then your integral has both x and $\theta$. That's not right. However, I would advise using the parametric equations $x= a sin(\theta)$, $y= b cos(\theta)$ rather than that complicated equation.
P: 33

## circumference of an ellipse

 Quote by HallsofIvy You are making the substitution $x= a sin\theta$ but then your integral has both x and $\theta$. That's not right. However, I would advise using the parametric equations $x= a sin(\theta)$, $y= b cos(\theta)$ rather than that complicated equation.
I'm not sure I understand what you mean by the parametric equations.. how does that fit into what I already have?
P: 33
 Quote by genericusrnme your handwriting is incredibly neat, good show!
thanks!
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 Quote by emc92 I'm not sure I understand what you mean by the parametric equations.. how does that fit into what I already have?
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.
P: 33
 Quote by Dick 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.
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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P: 25,174
 Quote by emc92 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?
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.
 P: 33 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))
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 Quote by emc92 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))
I thought you were supposed to get the integral of sqrt(1-k*sin^2(θ))??
 P: 33 right. k = 1-(b^2/a^2). did i cancel sin^2(θ) instead of cos(θ)?
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P: 25,174
 Quote by emc92 right. k = 1-(b^2/a^2). did i cancel sin^2(θ) instead of cos(θ)?
Hard to say. What did you do?
P: 33
 Quote by Dick Hard to say. What did you do?
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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P: 25,174
 Quote by emc92 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!
You are going in circles. Look you've got $\int \sqrt{1+\frac{b^2 \sin^2{\theta}}{a^2 \cos^2{\theta}}} a \cos{\theta} d\theta$. Bring the cos into the square root, so you've got $\int \sqrt{(1+\frac{b^2 \sin^2{\theta}}{a^2 \cos^2{\theta}}) \cos^2{\theta}} a d\theta$. Simplify inside the radical. Then use your trig identity and change the x limits to theta limits.
 P: 33 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?
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P: 25,174
 Quote by emc92 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?
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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