# Line Integral and swept area

1. Jan 10, 2012

### bugatti79

1. The problem statement, all variables and given/known data

Find the area swpet out by the line from the origin to the ellipse x=cos t and y=sin t as t varies from 0 to t_0 where t_0 is constant between 0 and 2 pi

2. Relevant equations

Use $A= \frac{1}{2} \oint_{C} y dx -xdy$

3. The attempt at a solution

How does one determine how to graph this in order to aid calculation?

Thanks

2. Jan 10, 2012

### LCKurtz

You could eliminate the parameter t if you want the xy equation, but you don't really need to for this problem. You have x and y given in terms of t; just plug x, y, dx, and dy in the integral and integrate with respect to t.

3. Jan 11, 2012

### bugatti79

Ok,

1) I calculate

$\displaystyle A= \frac{1}{2} \int_{0}^{2 \pi} a sin t (-a sin t) dt - a cos t (a cos t)dt = \frac{-a^2}{2} \int_{0}^{ 2 \pi} (sin^2 t +cos^2 t) dt = -a^2 \pi$..? A negative area is not right..

2)I am interested in how you would eliminate the parameter t as another method...can you advise?

Thanks

4. Jan 11, 2012

### LCKurtz

That is because your formula I have highlighted in red is wrong. It should have a minus sign in front of it. And where did the $a$ come from in your answer. There is no $a$ in your problem.
What do you get if you calculate $x^2+y^2$ in terms of your parameter $t$?

5. Jan 11, 2012

### bugatti79

Well spotted, thanks. I left that out by mistake.The question has x= a cos t and y = a sin t
I guess there isn't enough info to calculate the area numerically, just in terms of 'a' like I have seen in similar example elsewhere.

Well, equation of ellipse is x^2/a^2 +y^2/b^2 = 1. But based on this, shouldn't the above y=a sin t be y = b sin t?

I think we get a more difficult integral using the 'eliminate t' method........?

6. Jan 11, 2012

### LCKurtz

There isn't any $b$ in your problem so why are you bringing that up? What do you get if you calculate $x^2+y^2$ in your example?

7. Jan 11, 2012

### bugatti79

I calculate x^2 +y^2 =a^2. So I am guessing we could rearrange for x and y and put back into original 'corrected' integral in post 1...?

8. Jan 11, 2012

### LCKurtz

We are now just going around in circles. You asked in your original post how you could graph the curve. I said one way was to eliminate the parameter but I told you that was unnecessary to work the problem. But if you want to see the curve, graph $x^2+y^2=a^2$ and look at it. I presume you do know what that graph is and how to graph it, right?

9. Jan 11, 2012

### bugatti79

Sorry, I mis-interpreted your first quote above. I thought you meant 'evaluating the integral' by eliminating the variable t instead of 'plotting it'.

Yes I do know how to plot it. In the original question, I was thrown off by the word 'line' I thought it was more than just an ellipse,ie some combination of a revolved line with the ellipse...?

Other than that, I'm pretty happy :-)