Line Integral and swept area

[bugatti79]

Homework Statement

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

Homework Equations

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

The Attempt at a Solution

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

Thanks

 

[LCKurtz]

Homework Statement

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

Homework Equations

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

The Attempt at a Solution

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

Thanks

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.

 

[bugatti79]

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.

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

 

[LCKurtz]

Homework Statement

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

Homework Equations

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

The Attempt at a Solution

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

Thanks

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.

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..

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.

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

Thanks

What do you get if you calculate $x^2+y^2$ in terms of your parameter $t$?

 

[bugatti79]

And where did the $a$ come from in your answer. There is no $a$ in your problem.

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.

What do you get if you calculate $x^2+y^2$ in terms of your parameter $t$?

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...?

 

[LCKurtz]

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...?

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?

 

[bugatti79]

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?

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...?

 

[LCKurtz]

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...?

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?

 

[bugatti79]

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.

Homework Statement

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

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?

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 :-)