Why is the vector field tangent to the surface?

[flyingpig]

Homework Statement

I got this question right, but when I looked at it again, I think both the book and I are wrong.

Evaluate the surface integral $$\iint_S \mathbf{F} \cdot d\mathbf{S}$$ for the given vector field F and the oriented surface S.

For a closed surface, use the positive orientation

$$\mathbf{F}(x,,y,z) = <xy,yz,zx>$$ S is the part of the paraboloid $$z = 4 - x^2 - y^2$$ that lies above the square $$0 \leq x \leq 1$$, $$0\leq y \leq 1$$ and has a upward orientation


I will just post what the book did because I did exactly the same thing

[PLAIN]http://img37.imageshack.us/img37/2959/33371017.jpg

Now here is the problem

Notice how they substitute the parabolid into z for the vector field? I did that too, but what I don't understand is, how could that z be the same z in the vector field?

Isn't that a different z?

 

[magicarpet512]

F(x,,y,z)=<xy,yz,zx>F(x,,y,z)=<xy,yz,zx>
S is the part of the paraboloid
z=4−x2−y2

z is defined as a function of x and y here.
Why or how are you thinking that it is something different?

 

[flyingpig]

No, the vector field is a function of x, y, z. How is it that the z in the vector field the same as the paraboloid?

 

[Dick]

To compute F.dS you want to evaluate F on the surface of the paraboloid and dot with dS on the surface of the paraboloid, right? x, y and z are all the same in F and dS. That's what the 'S' means under the first double integral. You are evaluating these quantities on 'S', the paraboloid.

 

[flyingpig]

What if the vector field is F = <x,y,z>? It certainly doesn't make sense to me if we substitute z = 4 - x² - y² into the vector field

F = <x,y, 4 - x² - y²>, doesn't this change the vector field completely?

 

[Dick]

What if the vector field is F = <x,y,z>? It certainly doesn't make sense to me if we substitute z = 4 - x² - y² into the vector field

F = <x,y, 4 - x² - y²>, doesn't this change the vector field completely?

To integrate F.dS, you have to evaluate F(x,y,z) and dS(x,y,z) somewhere. It would make no sense whatsoever to evaluate them at different points, would it? If so given dS(x,y,z) what point would you want to evaluate F at? I've got to admit your confusion isn't making much sense to me.

 

[Dick]

What if the vector field is F = <x,y,z>? It certainly doesn't make sense to me if we substitute z = 4 - x² - y² into the vector field

F = <x,y, 4 - x² - y²>, doesn't this change the vector field completely?

Yes, it changes the vector field. But it DOESN'T CHANGE IT ON S. Does that help? Because z=4-x^2-y^2 on S.

 

[Ray Vickson]

Homework Statement

I got this question right, but when I looked at it again, I think both the book and I are wrong.

Evaluate the surface integral $$\iint_S \mathbf{F} \cdot d\mathbf{S}$$ for the given vector field F and the oriented surface S.

For a closed surface, use the positive orientation

$$\mathbf{F}(x,,y,z) = <xy,yz,zx>$$ S is the part of the paraboloid $$z = 4 - x^2 - y^2$$ that lies above the square $$0 \leq x \leq 1$$, $$0\leq y \leq 1$$ and has a upward orientation


I will just post what the book did because I did exactly the same thing

[PLAIN]http://img37.imageshack.us/img37/2959/33371017.jpg

Now here is the problem

Notice how they substitute the parabolid into z for the vector field? I did that too, but what I don't understand is, how could that z be the same z in the vector field?

Isn't that a different z?

$$d\mathbf{S}$$ is a vector whose direction is perpendicular to the surface and points upward, and whose magnitude is $$dA$$ = area of the surface patch. So, for a rectangle $$dx \times dy$$ in the (x,y) plane we have $$dx \times dy = cos(\theta) dA,$$ where $$\theta$$ is the angle between the normal to the surface and the z-axis. If you write the equation of the surface as $$h(x,y,z) = 0,$$ the normal is the direction of $$\nabla h .$$

RGV

 

[flyingpig]

Yes, it changes the vector field. But it DOESN'T CHANGE IT ON S. Does that help? Because z=4-x^2-y^2 on S.

No because when we take the dot product, we are dotting a completely different vector field.

 

[Dick]

No because when we take the dot product, we are dotting a completely different vector field.

Yes, because the vector field <x,y,z> and <x,y,4-x^2-y^2> are equal on S where z=4-x^2-y^2. I'm probably not going to have the energy to repeat this again.

 

[flyingpig]

x,y,4-x^2-y^2> are equal on S where z=4-x^2-y^2

What does that even mean?

 

[cepheid]

What does that even mean?

It means that the vector field is always equal to <x,y,z> throughout space. On the surface of the paraboloid, it just so happens that z = 4 - x² - y² -- this is what defines the paraboloid's surface. Therefore, if you evaluate the vector field at points that lie ON THE PARABOLOID, you will get the result F = <x, y, (4 - x² - y²)>, since ON the surface of the paraboloid, z is not a free variable, but is constrained by the values of x and y. You haven't changed the vector field, you're just restricting yourself to determining its value at very specific points in space (the ones that lie on the surface of the paraboloid).

 

[HallsofIvy]

Dick is correct. The entire integration is taking place on the paraboloid z= 4- x^2- y^2. On that paraboloid, it is perfectly correct to replace z with 4- x^2- y^2.

 

[flyingpig]

But even on the surface, wouldn't substituting 4 - x^ - y^2 into z "bend" the vector?

 

[HallsofIvy]

No, the vectors themselves aren't in the surface. The vector equations defines a vector at each point on the surface but the vectors don't lie in the surface. For example, at every point on the surface of a sphere there are tangent vectors that lie in the tangent plane to the sphere.

 

[Dick]

If your vector field is <x,y,z> and your surface S is z=4-x^2-y^2, then for example at the point <1,1,2> the vector field <x,y,z> is <1,1,2> and the vector field <x,y,4-x^2-y^2> is also <1,1,2>. It doesn't look 'bent' to me. Can you suggest a point where it might be 'bent'?

 

[SammyS]

I see Dick found some more energy! (I admit, it's not exactly a repeat of his earlier posts.)

 

[flyingpig]

At point <5,2,1>

On the surface it isn't <5,2,1> anymore it is, <5,2, 4 - 25 - 4> = <5,2,-25>

it has been "bent"

 

[vela]

So you have the vector field F(x,y,z) = <x,y,z>. This holds for all x, y, and z.

On the paraboloid, you have z=4-x²-y², so the vector field at a point on the paraboloid is <x,y,4-x²-y²>.

The point <5,2,1> isn't on the paraboloid.

 

[flyingpig]

How do I find a point on it then?

 

[HallsofIvy]

What kind of point are you looking for? Vela did not say anything about a point, he said the vector <5, 2, 1> does not lie on the paraboloid. vectors at a point lie in the tangent plane to the surface, not the surface itself.

I understand your confusion. The difficulty is that we learn about "position vectors", vectors from the origin to a point, in "Euclidean space", R², and R³. I used to worry about whether "position vectors" on a sphere went through the sphere or curved around the sphere. All mention of "position vectors" should be banned! There are no position vectors in surfaces and spaces that are not flat. All vectors are "tangent vectors" at some point on the surface. In a very real sense "all vectors are derivatives".

 

[flyingpig]

I don't understand if all the vectors on the surface are tangent to it, then doesn't that mean no matter what vector field I use, it will still be tangent?

 

[vela]

How do I find a point on it then?

Do you see why the point <5,2,1> isn't in S?

 

[flyingpig]

Because 5 - 25 - 4 does not equal 1...

 

[vela]

Because 5 - 25 - 4 does not equal 1...

You mean $4-25-4 \ne 1$, right?

How do I find a point on it then?

If you understand why a point does not lie in S, you must surely understand which points do lie in S. So why are you asking this?

 

[flyingpig]

You mean $4-25-4 \ne 1$, right?

Yeah

If you understand why a point does not lie in S, you must surely understand which points do lie in S. So why are you asking this?

They lie on the surface and I am guessing what HallsofIvy is trying to tell me is that that point has a vector that is tangent to the surface. But this is my point, if my vector field is defined such that it is NOT even tangent to the surface, then where does the tangent vector on the surface comes from?

 

[vela]

They lie on the surface and I am guessing what HallsofIvy is trying to tell me is that that point has a vector that is tangent to the surface. But this is my point, if my vector field is defined such that it is NOT even tangent to the surface, then where does the tangent vector on the surface comes from?

You can think of it this way: For every point x=(x,y,z) in R³, which I'll call M, there's a copy of R³, which I'll call T_x, attached to it. This copy is called the tangent space of M at x because it is, well, tangent to M at x. The vector F(x)=(F_x,F_y,F_z) assigned to the point x lives in T_x, not M. This is what HallsofIvy meant when he said the vectors lie in the tangent plane T_x to the surface M, not the surface M itself. He just referred to the paraboloid when he should have been talking about R³, for this problem. If this doesn't make sense to you, don't worry about it.

All you need to know for now is that for each point x=(x,y,z) in R³, you have a vector F=(F_x,F_y,F_z) assigned to it, and the direction F points has nothing to do with being tangent to the paraboloid.