# Homework Help: Equation of the circumference of an ellipse parametric equations

1. Jul 30, 2010

### mickellowery

1. The problem statement, all variables and given/known data
Consider the ellipse given by the parametric equation x=3cos(t) y=sin(t) 0$$\leq$$t$$\leq$$2$$\Pi$$. Set up an integral that gives the circumference of the ellipse. Also find the area enclosed by the ellipse.

2. Relevant equations
$$\int$$$$\sqrt{1+(dy/dx)^2}$$dt

3. The attempt at a solution
$$\int$$$$\sqrt{1+(-2/3 cot(t))^2}$$dt It should also be the integral from 0 to 2$$\Pi$$ I'm not sure what I did wrong but I know that -2/3 cot(t) is not right.
area: A=2$$\int$$1/2 (2/3 tan(t))dt
=$$\int2/3 tan(t)dt$$
=-2/3ln(lcos(t)l) evaluated from $$\Pi$$ to 0
I know that I got something wrong here too, and I assume it is the 2/3 tan(t) but I'm not sure what I did wrong again.

Last edited: Jul 30, 2010
2. Jul 30, 2010

[STRIKE]Your formula for arc length is wrong. It's

$$\int \sqrt{1 + (\frac{dy}{dx})^2} \ dx$$

$$= \int \sqrt{1 + (\frac{\frac{dy}{dt}}{\frac{dx}{dt}})^2} \ \frac{dx}{dt}dt$$[/STRIKE]

EDIT: Pathetic lapse of judgement on my part >_>...ignore this post.

Last edited: Jul 30, 2010
3. Jul 30, 2010

### Staff: Mentor

Arc length for a parametrized curve can also be written this way:
$$\int_a^b \sqrt{(\frac{dx}{dt})^2 + (\frac{dy}{dt})^2}dt$$

4. Jul 30, 2010

### gomunkul51

$$= \int \sqrt{1 + (\frac{\frac{dy}{dt}}{\frac{dx}{dt}})^2} \ \frac{dx}{dt}dt$$

how did you get that and how do you plan to integrate it? :)

But as far as I know the ds element for parametrization: f(x(t),y(t)) = x(t) + y(t) is:

$$\int f(x,y) ds = \int f(x(t),y(t)) \sqrt{\left ( \frac{dx}{dt} \right )^2 + \left ( \frac{dy}{dt} \right )^2}dt$$

5. Jul 30, 2010

### mickellowery

OK the way I did it I actually used $$\int$$$$\sqrt{1+dy/dt/dx/dt}$$ and that's how I came up with the -$$\frac{2}{3}$$cot(t)2 I had -$$\frac{2}{3}$$$$\frac{cos(t)}{sin(t)}$$ and I simplified it to cot(t). Would this even be the right formula to use for the circumference? I thought that arc length would be the right choice as long as I evaluated it from 0 to 2$$\Pi$$.

6. Jul 30, 2010

### Staff: Mentor

This isn't the right formula for arc length. If you simplify Raskolnikov's formula in post 2, you get the one I showed in the next post.

7. Jul 30, 2010

### mickellowery

Oh geez I just noticed a typo in the original problem. It should be x=3cos(t) y=2sin(t) not y=sin(t) sorry about that.

8. Jul 30, 2010

### mickellowery

Alright so with the correct equations would the proper integral for the circumference be:

$$\int$$$$\sqrt{(-3sin(t))^2 +(2cos(t))^2}$$dt

And then for the area enclosed by the ellipse would I use $$\int$$(3cos(t)-2sin(t))2 evaluated from 0 to $$\Pi$$?

9. Jul 30, 2010

### Staff: Mentor

You should simplify the integrand. Also, the limits are from 0 to $2\pi$.

Tip: Put all your LaTeX code inside one pair of tex tags.
$$\int$$$$\sqrt{(-3sin(t))^2 +(2cos(t))^2}$$
$$\int \sqrt{(-3sin(t))^2 +(2cos(t))^2}dt$$