What is the Correct Formula for the Circumference of an Ellipse?

[emc92]

all of my work so far is in the picture. I'm stuck on what i should do next.

 

[genericusrnme]

your handwriting is incredibly neat, good show!

 

[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.

 

[emc92]

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?

 

[emc92]

your handwriting is incredibly neat, good show!

thanks!

 

[Dick]

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.

 

[emc92]

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?

 

[Dick]

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.

 

[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))

 

[Dick]

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))??

 

[emc92]

right. k = 1-(b^2/a^2). did i cancel sin^2(θ) instead of cos(θ)?

 

[Dick]

right. k = 1-(b^2/a^2). did i cancel sin^2(θ) instead of cos(θ)?

Hard to say. What did you do?

 

[emc92]

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!

 

[Dick]

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.

 

[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?

 

[Dick]

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.