circumference of an ellipse

emc92

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

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genericusrnme

your handwriting is incredibly neat, good show!

HallsofIvy

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.

emc92

I'm not sure I understand what you mean by the parametric equations.. how does that fit into what I already have?

emc92

thanks!

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.

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?

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Dick

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 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

Hard to say. What did you do?

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!

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Dick

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.

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

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.