# Homework Help: Evaluate the Following Line Integral Part 1

1. Nov 1, 2011

### bugatti79

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

$\int_{c}cos (x)dx+sin(y)dy$ where c consist of the top half of the circle x^2+y^2=1 from (1,0) to (-1,0)

3. The attempt at a solution

Do I parameterise x=t and then y becomes $y= (1-t^2)^{1/2}$....? Replace the corresponding dx and dy and then integrate between the limits?

2. Nov 1, 2011

### I like Serena

Hi bugatti79!

Well, yes, you have to parametrize to calculate the result.
But you can integrate before you parametrize.

3. Nov 1, 2011

### HallsofIvy

That's one way to do it. Another is to think of this in polar coordinates with r= 1. In polar coordinates, $x= r cos(\theta)$ and $y= r sin(\theta)$ so on the unit circle, $x= cos(\theta)$ and $y= sin(\theta)$. These are the "standard" parameterizations of the unit circle.

Of course, going form (1, 0) to (-1, 0), $\theta$ goes from 0 to $\pi$. You will need to determine dx and dy in terms of $d\theta$.

4. Nov 1, 2011

### bugatti79

Thanks guys,
I think the notes only show parameterisation etc. I will take note of the other option. I suspect integrating first before parameterising is easier. Will try it later...

5. Nov 1, 2011

### bugatti79

Ok, I arrive at

$\displaystyle \int_{0}^{\pi }cos(-r sin(\theta) )d\theta+ sin(r cos(\theta))d\theta$

...this doesnt look right...?

6. Nov 1, 2011

### I like Serena

I'm afraid you didn't substitute properly.
It should be:

$\displaystyle \int_{0}^{\pi }cos(r \cos(\theta) ) \cdot -r \sin(\theta) d\theta+ \sin(r \sin(\theta)) \cdot r \cos(\theta) d\theta$

Furthermore, you should substitute r=1.

But rather than doing this, I recommend integrating first, then substituting the parameterisation (one or the other), and then calculate the result.

How would you integrate $\int \cos(x)dx$?

7. Nov 1, 2011

### bugatti79

That was an error I made but I continue on and use substitution for the integral

$\displaystyle \int_{0}^{\pi }-cos^2(\theta) sin(\theta) d\theta + cos (\theta) sin^2(\theta) d\theta$

I let $u=cos \theta$ for the first integral and $u=sin \theta$ for second integral. I get an answer of -2/3....?

If I integrate first then I get sin(x) -cos(y) to be evaluated to some limits...not sure how to proceed..

In addition to this, in my notes it states not to "calculate"

$\int f_1(x,y,z)dx+f_2(x,y,z)dy+f_3(x,y,z)dz$ etc with respect to its variable while keeping the other 2 constant.....I suspect it is ok to do in this example because each integral is only a function of one variable...?

8. Nov 1, 2011

### I like Serena

Uhhm... did you know that cos(cos(θ))≠cos2(θ)?

With your original parametrization you would have:
$$\left.sin(x(t)) - cos(y(t))\right|_{t=1}^{t=-1}$$
or with HoI's parametrization
$$\left.sin(x(θ)) - cos(y(θ))\right|_{θ=0}^{θ=\pi}$$

If you keep one variable constant, say x, then you'd have $y=\sqrt{1-x^2}$.
$$\int_{x=1}^{x=-1} \cos(x)dx + \sin(\sqrt{1-x^2}) \cdot d(\sqrt{1-x^2})$$
$$\int_{x=1}^{x=-1} \cos(x)dx + \sin(\sqrt{1-x^2}) \cdot {1 \over 2\sqrt{1-x^2}} \cdot -2x \cdot dx$$

I recommend solving the second part of the integral with the substitution $y=\sqrt{1-x^2}$.

9. Nov 1, 2011

### bugatti79

Oh, I over looked that substitution mistake!! Thanks!

How is yours evaluated? The sine or cos of 1 is not correct...?

So based on my comment on the way NOT to calculate...is it okay to do here because we are only dealing with one variable per integral?

I get the answer 2 using HoI's method.
Attempt on difficult integral will follow.

10. Nov 1, 2011

### I like Serena

You have:
$$\left.\sin(x(t)) - \cos(y(t))\right|_{t=1}^{t=-1}$$
with $x(t)=t$ and $y(t)=\sqrt{1-t^2}$

Substituting:
$$\left.\sin(t) - \cos(\sqrt{1-t^2})\right|_{t=1}^{t=-1}$$

So yes, you would get sin(1) in there.

You would get:
$$(\sin(-1) - \cos(\sqrt{1-(-1)^2})) - (\sin(1) - \cos(\sqrt{1-1^2})) = -2 \sin(1)$$

I'm not sure what it means NOT to calculate.
Does it mean you should not actually integrate or something?

11. Nov 1, 2011

### bugatti79

How would -2sin(1) evaluate to 2?

Yes, it says

"you should never calculate

$\int f_1(x,y,z)dx+f_2(x,y,z)dy+f_3(x,y,z)dz$ by integrating f_1(x,y,z) wrt x (treating y+z like constants), same for f_2(x,y,z) etc. THis will give wrong answer....."

12. Nov 1, 2011

### I like Serena

It doesn't. Why should it?

Ah, now I get it.
Indeed you can not do the integration wrt to 1 variable and expect it to work out.

And yes, in this case it was possible, because the integral could be separated into separate integrals with only 1 variable.

13. Nov 1, 2011

### bugatti79

But arent the 3 different methods suppose to yield the saem answer...ie your method as in post #2 should yield the same as HoI..right? :-)

14. Nov 1, 2011

### I like Serena

Of course.
Any reason to think that it doesn't?

15. Nov 1, 2011

### LCKurtz

There is a much easier way to do this problem. Call C1 the given curve, C2 the straight line segment from [-1,0] to [1,0], and let C be $C_1\cup C_2$. Since you have the hypotheses for Green's Theorem you have:

$$0 = \int_C \cos x\, dx + \sin y\, dy = \left(\int_{C_1}+\int_{C_2}\right)( \cos x\, dx + \sin x\, dy)$$

$$\int_{C_1} \cos x\, dx + \sin y\, dy = - \int_{C_2} \cos x\, dx + \sin y\, dy = -\int_{-1}^1\cos x\, dx$$

16. Nov 1, 2011

### I like Serena

@LCKurtz:

Uhh... :uhh: Aren't you glossing over the fact the the curl of the function over the surface enclosed by C is zero in this case?

17. Nov 1, 2011

### bugatti79

Im afraid I dont know. In your method, we have integrated wrt x and used the x coordinates...so it seems right is it not?

For the difficult integral I end up with

$sin(x)|_1^{-1} + (-cos(1-x^2)^{1/2})|_1^{-1}$

18. Nov 1, 2011

### I like Serena

Yes.....

19. Nov 1, 2011

### bugatti79

This doesnt give HoI answer of 2 either. I give up. I least I get nearly full marks :-) Thanks.

BTW..when I say I nearly get full marks...'if' it was an exam. It is just exercises :-)

20. Nov 1, 2011

### I like Serena

Ah well, with HoI's parametrization you get:
$$\left.\sin(x(θ)) - \cos(y(θ))\right|_{θ=0}^{θ=\pi}$$
$$\left.\sin(\cos(θ)) - \cos(\sin(θ))\right|_{θ=0}^{θ=\pi}$$
$$(\sin(\cos(\pi)) - \cos(\sin(\pi))) - (\sin(\cos(0)) - \cos(\sin(0))) = -2\sin(1)$$