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Hey, I've rechecked this : Plotting y=x , y=-x would show that

D[sub]f[/sub] = { (x,y) belong to R[sup]2[/sup] : y > 0 , (1-y[sup]2[/sup] ) [sup]1/2[/sup] < x < - (1-y[sup]2[/sup] ) [sup]1/2[/sup] } U { (x,y) belong to R[sup]2[/sup] :

y < 0 , - (1-y[sup]2[/sup] ) [sup]1/2[/sup] < x < (1-y[sup]2[/sup] ) [sup]1/2[/sup] }

The domain is on the right bounded by y=x , y=-x. y changes from x to -x. The left part of the domain is bounded by y=x, y=-x, y changes from -x to x.

Yes, (1-y²)[sup]1/2[/sup]< x < -(1-y²)[sup]1/2[/sup] doesn't make sense, in the solutions guide, it's written :

- (1-y²)[sup]1/2[/sup]< x < (1-y²)[sup]1/2[/sup] , which is definitely wrong!

Perhaps three sets , as you suggested, is more logical.

**Chemist**- Replies: 3

Can someone verify if this is the domain of def of the following function?

f(x,y) = Ln [ y / ( x[sup]2[/sup] + y[sup]2[/sup] -1 ) ]

D[sub]f[/sub] = { (x,y) belong to R[sup]2[/sup] : y > 0 , (1-y[sup]2[/sup] ) [sup]1/2[/sup] < x < - (1-y[sup]2[/sup] ) [sup]1/2[/sup] } U { (x,y) belong to R[sup]2[/sup] :

y < 0 , - (1-y[sup]2[/sup] ) [sup]1/2[/sup] < x < (1-y[sup]2[/sup] ) [sup]1/2[/sup] }

Thanks,

We can obtain Green's theorem area formula by going backward :

A = ∫ ∫ dydx = ∫ ∫ ( 1/2 + 1/2 ) dydx = 1/2 ∫ ( xdy - ydx )

, ofcourse the domain is closed.

But it seems sometimes we can compute the area directly from A = ∫ xdy

Example:

Determine the area b/w y=x[sup]2[/sup] and y = x across C[sup]+[/sup]

A = 1/2 ( ∫ xdy - ydx )

I divided the line integral into arc OA and a straight line OA where A (1,1), the point of intersection of the parabola and the straight line, and O is the origin ofcourse.

I chose the parametirc equations,

For [OA] : y = x , x=x 0 ≤ x ≤ 1

For arc OA : x=y[sup]2[/sup] , y=y 1 ≤ y ≤ 0

I[sub]1[/sub] = 1/2 ∫ (x-x)dx = 0 ----> for [OA]

I[sub]2[/sub] = 1/2 ∫ (y[sup]2[/sup] - 2y[sup]2[/sup])dy = 1/6 ---> for arc OA

A = I[sub]1[/sub] + I[sub]2[/sub] = 1/6

Now notice that the area could have been calculated directly from:

A = ∫ xdy ( line int across C[sup]+[/sup] ) = ∫ xdy ( across [OA] ) + ∫ xdy ( across arc OA )

Using the same variables, A = 1/6

The second way is quicker, but I can't seem to figure out when to use it.

George,Y wrote:

The second integral you mentioned is mainly for computing work done by conservative force. As you can see, the integrated is a dot product of work on a small piece F.dr, where F and dr are both vectors.

The first integral can be used to integrate f(t,t,t) |v(t)|, f(t,t,t) is displacement from origin, and if v(t) stands for velocity, then|v(t)| stands for speed, or the magnitude of velocity. i cannot figure out what the product means.

So you'd better use the second defaultly.

So far, i haven't encountered the situation to use your first integral.

Thanks for the info , I think I got the difference between the two. And yes |V(t)| is the velocity magnitude or speed.

Well, take the example: f(x, y, z) = x + y + z

That's from R³⇒R. You could define it from R³⇒R³ by:

f(x, y, z) = (x + y + z, 0, 0)

I'm not sure what this is, what's the 0,0 for?

But these are two completely different things with completely different properties. They look the same, but they aren't.

Is this what you are talking about?

No , I was simply saying that if we have a scalar function f(x,y,z) = x +y +z , we can change it to a vector field or to its total differential form. This is what I did in calculating the path integral of a function f, I changed it to its vector form and calculated the scalar product.

They are pretty much the same exact thing. That is, the both "sum up" the value of a function along a certain path. But one sums up vector values, such as a force, and one sums up constant values, such as heat or density.

If you have f(x, y, z) as the density of a wire at a point in 3d space, and c(t) be the path of that wire. Then the path integral will mass of that wire.

Thanks for the explanation.

If we assume that the total differential form of a vector field is exact, the line integral is always dependent on the path, can we relate this to a conservative field? That is, a line integral exists only for conservative vector fields, which means if a vector field has a line integral over a certain path ( does it make a difference if it's differentiable or not here? ) , it must have a scalar potential function. Does this make any sense? I'm just wondering ...

I think I may be right here. I was just solving some problems, it turns out that whenever the total diff form of a vector field is exact ( rotor = 0 ), the line integral is independent of the path, and I can simply compute it F(B) - F(A) if we assume the line integral is from A to B across an arc. This really makes the computation easy for a line integral.

But if the total diff form is exact, continuous and its 1st derivative is continuous too, the line integral in a closed domain would be zero. Why's that?

And can anyone prove Green-Rieman's formula? I have an idea but I think it's incorrect.

Thanks a lot Ricky, I know that a line integral is used to compute the work or circulation depending on what the vector field is, but what's the path integral used for?

Tried to integrate a vector

or

Tried to take the dot product of a scalar.

Of course, neither makes sense.

Nah , I actually changed the scalar function to a vector field ... why can't we do that? Can't every scalar function have its vector components too?

**Chemist**- Replies: 11

I just can't figure out what's the difference b/w the following two methods for solving line integrals.

In the book, they say for evaluating a line integral for a function f(x,y,z) = x - 3y[sup]2[/sup] + z over the line segment C joining the origin and the point (1,1,1) , we must 1st determine the parametric equations in terms of a parameter t.

r = t i + t j + t k 0 ≤ t ≤ 1 ( smooth parameter , 1st derivative is cont and never zero )

they give the line integral form : ∫ f(x,y,z)ds = ∫ f(t,t,t) |v(t)| dt 0 ≤ t ≤ 1

In class, however, this method is never used.

We start from ∫ F.dr ( both vectors ) where F = vector field = M i + N j + O k

∫ F.Tds ( T is the tangent unit vector, F.T----> dot product ) = ∫ F.dr = ∫ F.drdt /dt = ∫ ( M dx/dt + N dy/dt + O dz/dt ) dt

We've used this form to determine the line integral in all the problems we have encountered so far. If I try the 1st way, the answer is not always the same for the 2nd, actually, rarely is it the same. I tried to see the common point between the two:

∫ F.Tds = ∫ F.dr = ∫ F.dr dt / dr = ∫ F.V(t) dt , here obviously V is a velocity vector.

so, ∫ F.V dt = ∫ F |V(t)| cosθ dt ( θ b/w F and V )

In the book , they're using ∫ f(t,t,t) |V(t)| dt , so perhaps they assume that V is parallel to F ? Why would they do that, and what's the significance of either method?

Thanks in advance

Ricky, but notice that if I use y=0 for f(x,y) = yx^2 / ( x^4 - 2yx^2 + 3y^2 ) , the limit will be 0.

If y=x, the limit will be lim x--->0 (x / ( x^2 - 2x + 3 ) ) = 0

What can I conclude from this?

Sure thing krassi_holmz,

f(x,y) = Siny / x

It's required to find the limit of this function as (x,y) ---> (0,y)

So if I take y=mx, lim f(x,y) = Sin(mx) / x = m since Sin(mx) / mx = 1 ( maclaurin series )

Here I think this function has no limit since m is a parameter which belongs to R, it can take any value.

Most often, the question is to find whether a function has a limit. In this case, we're using y=x^2 as well. What we tend to pove is that the function has no limit at all.

f(x,y) = yx^2 / ( x^4 - 2yx^2 + 3y^2 ) if (x,y) ≠ (0,0)

f(x,y) = 0 if (x,y) = (0,0)

Study the limit of this function as ( x,y) ---> (0,0) using y=mx and y=x^2 where m is a real parameter. Show that f doesn't have a limit at the origin O.

for y =mx, the limit of f(x,y) = 0

for y=x^2, the limit of f(x,y) = 1/2

We deduce that the limit is not unique, hence, there's no limit.

**Chemist**- Replies: 9

In studying the limits of multi-variable functions, is it reliable to use y=mx and y=x^2 to evaluate the limit of f(x,y) ? Why do we use only these two anyway? And what does a "level curve" mean?

P.S. I know we can use polar coordinates, which is easier.

Thanks a lot gnitsuk

Maybe You mean 10p^2 - 2p - 1 = 0 or some constant ..?

well just how exactly did you get your answer?

**Chemist**- Replies: 4

Can anyone solve this for me ∫(2^x)dx ?

Thanks

hahaha . I thought that was i,j,k,l,m,&n !

Ah just what I need, thanks man.

Then integrate just the top half, which is y = √( r²-x² )

I think it gets a little complicated after that.

Indeed I tried that, and yes the poblem is with the integration.

mathsyperson, your method is good, but since I'm doing this from scratch, just why is the area (0,2π)∫1/2r² dθ ?

Thank your both for your time.

**Chemist**- Replies: 12

Can anyone please show me , using simple integration, how the area of a circle turns out to be (Pi)r^2 ? I've tried many ways and sometimes I end up with 2(Pi)r or completely irrelevant answers ... :S

Thanks

Call f anything you wish k, n, l, ... it is not , a function.

My post wasn't very clear, so let me first clarify that it is not given that f(Ax) = f(By). I should add that f denotes an activity coefficient, so it's a constant.

And yes MathsIsFun , A is in base x, and B is in base y. Hope you figure it out cos I've been told it's a difficult problem.

Thank you for your time.

**Chemist**- Replies: 4

Can anyone prove the validity of this equation by deriving it :

f(AxBy) = (fA^xfB^y)^[1/(x+y)] ?