# Thread: Line integrals - how to determine limits

1. ## Line integrals - how to determine limits

Hy, i'm having problems with this example.

$\displaystyle \int_cxyds$

where c is one quarter of the ellipse

$\displaystyle \frac{x^2}{a^2}+\frac{y^2}{b^2}=1$

The line integral itself isn't the problem, the problem tis how to define the limits of integration.

I have the same problem with this example

$\displaystyle \int_c(2a-y)dx+xdy$

where c is an arc of the first bow of a trochoid
$\displaystyle x=a(t-\sin(t))$
$\displaystyle y=a(1-\cos(t))$

2. Originally Posted by Pinsky
Hy, i'm having problems with this example.

$\displaystyle \int_cxyds$

where c is one quarter of the ellipse

$\displaystyle \frac{x^2}{a^2}+\frac{y^2}{b^2}=1$

The line integral itself isn't the problem, the problem tis how to define the limits of integration.

I have the same problem with this example

$\displaystyle \int_c(2a-y)dx+xdy$

where c is an arc of the first bow of a trochoid
$\displaystyle x=a(t-\sin(t))$
$\displaystyle y=a(1-\cos(t))$
In the first, I'd switch to a parametric form: x = a cos(t), y = b sin(t). Then the simplest quarter of the ellipse would give limits t = 0 to t = pi/2.

In the second, have you tried sketching an arc of the first bow of the trochoid? It starts at t = 0. For what value of t does it end ......? By the way, did you notice that if a = 1 you have a cycloid? (This might help you answer my question.)

3. Let's say a have a line in 2D and i want to integrate over it in the first quadrant.
If the function of a line it given explicitly y=f(x) than i can easily see for which values of x is the line in the first quadrant. But how do i do that if the line is given in parametric equations x=f(t) y=f(t)? Is there a method for doing this?

Ask if you didn't understand my question.

4. Originally Posted by Pinsky
Let's say a have a line in 2D and i want to integrate over it in the first quadrant.
If the function of a line it given explicitly y=f(x) than i can easily see for which values of x is the line in the first quadrant. But how do i do that if the line is given in parametric equations x=f(t) y=f(t)? Is there a method for doing this?

Ask if you didn't understand my question.
Yes there is.

BUT it will be easier to show you by using an example. So ....... what line of the form y = f(x) do you have in mind?

5. Let's say we use a cykloide, and it is already given as a parametric equation x=f(t) y=f(t)

$\displaystyle x=a(t-\sin{t})$
$\displaystyle y=a(1-\cos{t})$

So now i have to calculate a line integral
$\displaystyle \int_c(2a-y)dx+xdy$
Where c is the first arc of the cycloide (the first example i gave, it turned to be a cycloide after all).

This is how i do it:
First
$\displaystyle dx=(a-a\cos{t})dt$
$\displaystyle dy=a\sin{t}dt$

So now i include all of that into the integral. But i don't know how to determine the borders. It can be seen from the picture that if i want to integrate until the first arc, i should use limits from 0 to 2*r*pi or in my case to 2*a*pi. But those would be the limits if i would integrate with dx.
So how do i find the limits when integrating with dt?

6. Originally Posted by Pinsky
Let's say we use a cykloide, and it is already given as a parametric equation x=f(t) y=f(t)

$\displaystyle x=a(t-\sin{t})$
$\displaystyle y=a(1-\cos{t})$

So now i have to calculate a line integral
$\displaystyle \int_c(2a-y)dx+xdy$
Where c is the first arc of the cycloide (the first example i gave, it turned to be a cycloide after all).

This is how i do it:
First
$\displaystyle dx=(a-a\cos{t})dt$
$\displaystyle dy=a\sin{t}dt$

So now i include all of that into the integral. But i don't know how to determine the borders. It can be seen from the picture that if i want to integrate until the first arc, i should use limits from 0 to 2*r*pi or in my case to 2*a*pi. But those would be the limits if i would integrate with dx.
So how do i find the limits when integrating with dt?
$\displaystyle \int_c (2a-y) \, dx + x \, dy = \int_{t = t_1}^{t = t_2} \left( (2a-y) \, \frac{dx}{dt} + x \, \frac{dy}{dt} \right) \, dt = .......$

where I leave it to you to sub in $\displaystyle x=a(t-\sin{t})$, $\displaystyle y=a(1-\cos{t})$, $\displaystyle \frac{dx}{dt} = a(1 - \cos{t})$, $\displaystyle \frac{dy}{dt} = a\sin{t}$, $\displaystyle t_1 = 0$ and $\displaystyle t_2 = 2\pi$ (NOT $\displaystyle 2 a \pi$) and then evaluate the resulting definite integral.

7. In parametric, one arch of the cycloid.

$\displaystyle x = a(t-sint), \;\ y=a(1-cost)$

$\displaystyle (\frac{dx}{dt})^{2}+(\frac{dy}{dt})^{2}=[a(1-cost)]^{2}+[asint]^{2}=2a^{2}(1-cost)=4a^{2}sin^{2}(\frac{t}{2})$

$\displaystyle L=\int_{0}^{2\pi}2asin(\frac{t}{2})dt$

I included a diagram. Hope it helps. Perhaps you already know it, though.

8. Originally Posted by galactus
In parametric, one arch of the cycloid.

$\displaystyle x = a(t-sint), \;\ y=a(1-cost)$

$\displaystyle (\frac{dx}{dt})^{2}+(\frac{dy}{dt})^{2}=[a(1-cost)]^{2}+[asint]^{2}=2a^{2}(1-cost)=4a^{2}sin^{2}(\frac{t}{2})$

$\displaystyle L=\int_{0}^{2\pi}2asin(\frac{t}{2})dt$

I included a diagram. Hope it helps. Perhaps you already know it, though.
But just to make it clear, you've set up the integral for calculating the length of the arch (an interesting and useful result btw ), not the line integral that Pinsky posted. Right?

9. DUH, Thanks MrF. I'll leave it anyway. May help.

BTW, isn't the cycloid also known as the 'brachistochrone'?. Wasn't this the problem one of the Bernoulli's posed to Newton and he solved it rather quickly, according to the story?. The problem was to determine the best shape for an object, such as a bearing might roll from a point P to a point Q, not directly below it, the fastest. This is the shape Newton came up with.

10. Originally Posted by galactus
DUH, Thanks MrF. I'll leave it anyway. May help.

BTW, isn't the cycloid also known as the 'brachistochrone'?. Wasn't this the problem one of the Bernoulli's [Mr F adds: Johann, in fact. You need a score card to keep a track of that family ] posed to Newton and he solved it rather quickly, according to the story?. The problem was to determine the best shape for an object, such as a bearing might roll from a point P to a point Q, not directly below it, the fastest. This is the shape Newton came up with.
Indeed. But I think Bernoulli's own solution is the one that's mostly mentioned. The problem has a very interesting history, as does the cycloid (which had been known for at least 100 years prior to the problem Bernoulli posed).

11. But why did you take the limits of integration from 0 to $\displaystyle 2\pi$.
How do you figure that out from the parametric equation, which is is my case a cycloide?

Like in this example.
$\displaystyle \int_cxyds$

where c is one quarter of the ellipse

$\displaystyle \frac{x^2}{a^2}+\frac{y^2}{b^2}=1$
Mr.fantastic said to take the limits of integration from 0 to $\displaystyle \frac{\pi}{2}$

Why?

12. Found an almost analog problem, that deals whit double integrals and changeint the variables to polar cordinates.

The task is to calculate an integral

$\displaystyle \int_D\int\sqrt{1-x^2-y^2}dxdy$

Where D is a half circle bounded by a circle $\displaystyle (x-\frac{1}{2})^2+y^2=\frac{1}{4}$
and the x axis.

This is a solved example from a book.
So the first thing they do is transform the formula of the circuit into polar coordinates.
So

$\displaystyle (x-\frac{1}{2})^2+y^2=\frac{1}{4}\Rightarrow r=\cos{\varphi}$

So in the end they get:

$\displaystyle \int_{0}^{\frac{\pi}{2}}\int_{0}^{\cos{\varphi}}\s qrt{1-r^2}drd\varphi$

I've drawn the half circle in the first quadrant. An then again my question, how where the limits 0 to $\displaystyle \frac{\pi}{2}$ defined?
Why aren't the limits from 0 to $\displaystyle \pi$? That's a half circle then.

13. Originally Posted by Pinsky
But why did you take the limits of integration from 0 to $\displaystyle 2\pi$.
How do you figure that out from the parametric equation, which is is my case a cycloide?

Like in this example.

Mr.fantastic said to take the limits of integration from 0 to $\displaystyle \frac{\pi}{2}$

Why?
The easiest way for you to see this is to actually substitute values of t into the parametric equations defining each curve and then plot the points .....

Eg. What do you get if you substitute values of t from 0 to pi/2 into the parametric equations defining the ellipse ..... What happens as you go past t = pi/2 .....

(If you have access to a graphics calculator you can draw curves in parametric mode and use the trace feature)

It looks to me like you don't have a lot of experience with curves defined in parametric form. It will be well worth your academic while to jump onto that particular learning curve.

14. Originally Posted by Pinsky
Found an almost analog problem, that deals whit double integrals and changeint the variables to polar cordinates.

The task is to calculate an integral

$\displaystyle \int_D\int\sqrt{1-x^2-y^2}dxdy$

Where D is a half circle bounded by a circle $\displaystyle (x-\frac{1}{2})^2+y^2=\frac{1}{4}$
and the x axis.

This is a solved example from a book.
So the first thing they do is transform the formula of the circuit into polar coordinates.
So

$\displaystyle (x-\frac{1}{2})^2+y^2=\frac{1}{4}\Rightarrow r=\cos{\varphi}$

So in the end they get:

$\displaystyle \int_{0}^{\frac{\pi}{2}}\int_{0}^{\cos{\varphi}}\s qrt{1-r^2}drd\varphi$

I've drawn the half circle in the first quadrant. An then again my question, how where the limits 0 to $\displaystyle \frac{\pi}{2}$ defined?
Why aren't the limits from 0 to $\displaystyle \pi$? That's a half circle then.
On the circle $\displaystyle (x-\frac{1}{2})^2+y^2=\frac{1}{4}$ defining the region of integration, the point (1, 0) corresponds to the polar coords [1, 0] (capisce?). The point (0.5, 0.5) corresponds to $\displaystyle \left[\frac{1}{\sqrt{2}}, \frac{\pi}{4}\right]$ (capisce?). The point (0, 0) corresponds to [0, pi/2] (capisce?).

I hope you can see that as you move around the circle from (1, 0) to (0.5, 0.5), r goes from 1 to $\displaystyle \frac{1}{\sqrt{2}}$ and $\displaystyle \varphi$ goes from 0 to $\displaystyle \frac{\pi}{4}$.

And I hope you can see that as you move around the circle from (0.5, 0.5) to (0, 0), r goes from $\displaystyle \frac{1}{\sqrt{2}}$ to 0 and $\displaystyle \varphi$ goes from $\displaystyle \frac{\pi}{4}$ to $\displaystyle \frac{\pi}{2}$.

Indeed, if you express the circle $\displaystyle (x-\frac{1}{2})^2+y^2=\frac{1}{4}\Rightarrow x^2 - x + y^2 = 0$ in polar coordinates, you get $\displaystyle r = \cos \varphi$ ...... If you draw a graph of r versus $\displaystyle \varphi$ you will clearly see r changing from 1 to 0 as $\displaystyle \varphi$ goes from 0 to $\displaystyle \frac{\pi}{2}$. And obviously r = 1 corresponds to the point (1, 0) and r = 0 corresponds to the origin (0, 0).

What happens as $\displaystyle \varphi$ goes from $\displaystyle \frac{\pi}{2}$ to $\displaystyle \pi$ .....? What happens as $\displaystyle \varphi$ increases from $\displaystyle \pi$ ....?

Again, you need to jump onto the bus going to parametric city. Make sure it's an express.