Proof of Zero Value for Vector Field Integral on Closed Surface

[afallingbomb]

What is the value of a surface integral over a closed, continuous surface of a vector field of vectors normal to the surface? The integral of ndS over S.

I believe the answer is zero. Can someone direct me to a proof for an aribitrary closed surface?

 

[LCKurtz]

What is the value of a surface integral over a closed, continuous surface of a vector field of vectors normal to the surface? The integral of ndS over S.

I believe the answer is zero. Can someone direct me to a proof for an aribitrary closed surface?

I presume n is understood to be a unit outward normal and dS the scalar surface element. If so, the resulting integral is a vector, so perhaps you mean the result is the zero vector.

That is true. For example, look at the i component:

i \cdot \iint_S \hat n\, dS = \iint_S i \cdot \hat n\, dS<br /> =\iiint_V \nabla \cdot i \, dV = 0

by the divergence theorem. Similarly for the other two components.

 

[HallsofIvy]

LCKurtz, afallingbomb specifically asked about the integration of "a vector field of vectors normal to the surface" and your choice of \vec{i} does not satisfy that. Certainly the integral of a constant vector, if that was what afallingbomb intended, over a closed smooth surface is 0 because, for each point on the surface, there is a "polar opposite" point where the normal vectors are equal in length and opposite in direction and so cancel.

However, one example of "a vector field of vectors normal to the surface" is the field that, to each point on the surface, assigns the unit normal at that point. In that case,
\int |\vec{n}| dS= \int dS
which is the total area of the surface, not 0.

 

[LCKurtz]

I presume n is understood to be a unit outward normal and dS the scalar surface element. If so, the resulting integral is a vector, so perhaps you mean the result is the zero vector.

That is true. For example, look at the i component:

i \cdot \iint_S \hat n\, dS = \iint_S i \cdot \hat n\, dS<br /> =\iiint_V \nabla \cdot i \, dV = 0

by the divergence theorem. Similarly for the other two components.

LCKurtz, afallingbomb specifically asked about the integration of "a vector field of vectors normal to the surface" and your choice of \vec{i} does not satisfy that.

You have misunderstood my argument. I said above to look at just the i componentof the answer. You are given the vector integral

\vec V = \iint_S \hat n\, dS= V_1i + V_2j + V_3k

where I have assumed that the field on the surface gives unit normals. (The OP hasn't clarified this). Now, if you look at the i component of this you get

V_1 =i \cdot \iint_S \hat n\, dS = \iint_S i \cdot \hat n\, dS<br /> =\iiint_V \nabla \cdot i \, dV = 0

Similarly, dotting j and k into that shows that all three components of V are zero.

Certainly the integral of a constant vector, if that was what afallingbomb intended, over a closed smooth surface is 0 because, for each point on the surface, there is a "polar opposite" point where the normal vectors are equal in length and opposite in direction and so cancel.

However, one example of "a vector field of vectors normal to the surface" is the field that, to each point on the surface, assigns the unit normal at that point. In that case,
\int |\vec{n}| dS= \int dS
which is the total area of the surface, not 0.

But that is a scalar integral, not of the type the OP wrote. Also, if the OP doesn't mean unit vectors on the surface, he needs to specify more details. The vector integral is surely not the zero vector for arbitrary fields.

 

[afallingbomb]

Thanks for the replies and I apologize if I wasn't clear enough.

I agree with LCKurtz. The integral you presented, HallsofIvy, is over a scalar field. I was referring to a vector field which at the boundary only has normal components.

LCKurtz, I did mean unit vectors, but now that I think about it... wouldn't it also be zero for any vector field that is only normal to the surface at the surface?

A surface integral can be viewed as the double integral analog of a line integral and a line integral can be viewed as a summation of all the tangential components across the surface. If the vector field is exclusively normal to the surface, regardless of its magnitude, the integral over the entire closed surface should be zero. Am I correct?

Thanks!

 

[LCKurtz]

If the vector field is exclusively normal to the surface, regardless of its magnitude, the integral over the entire closed surface should be zero. Am I correct?

Thanks!

No. Think about a vector field on a sphere that is directed outward and very large magnitude in the first octant and almost zero magnitude on the rest of the sphere. You need unit vectors, or at least constant magnitude, and uniformly directed either outwards or inwards to guarantee a zero vector answer.

 

[afallingbomb]

So I am wrong to interpret a surface integral as the analog of a line integral to double integrals? I can think of a problem on a plane that gives me a zero line integral over the unit circle with fields of varying magnitudes.

Consider the two vector fields:

<br /> F_1(x,y)=(\frac{x}{\sqrt{x^2+y^2}},\frac{y}{\sqrt{x^2+y^2}}) <br />
<br /> F_2(x,y)=(\frac{2x}{\sqrt{x^2+y^2}},\frac{2y}{\sqrt{x^2+y^2}})<br />

Where F₁ is valid in the domain x>0 (on the unit circle from t = 0 to pi) and F₂ exists in x<0 (on the unit circle from t = \pi to 2\pi).

Integrating over the unit circle, parameterized by x = cos(t) and y = sin(t), using the expression for the line integral

<br /> \oint_C \vec{F}(c(t)) \cdot c&#039;(t) dt<br />

where c(t)= cos(t)\textbf{i}+sin(t)\textbf{j} and c&#039;(t)= -sin(t)\textbf{i}+cos(t)\textbf{j}

<br /> \int_{0}^{\pi} (cos(t)\textbf{i}+sin(t)\textbf{j}) \cdot (-sin(t)\textbf{i}+cos(t)\textbf{j}) dt + \int_{\pi}^{2\pi} (2cos(t)\textbf{i}+2sin(t)\textbf{j}) \cdot (-sin(t)\textbf{i}+cos(t)\textbf{j}) dt <br />

Both those integrals vanish when the dot product operation is performed. The field from 0 to pi (on the right) has half the magnitude of the field from pi to 2*pi (on the left). Yet they are both still exclusively normal to the unit circle.

I must be making some conceptual mistake.

 

[LCKurtz]

A line integral of the form

\oint_C \vec F \cdot d\vec R

can be thought of the work done by a force in moving around a closed curve. The integrand effectively calculates the work using the component of F in the direction tangent to the curve. So, of course, when F is normal to the curve you get zero. No problem there.

Upping the dimension one level, suppose you have a flux integral of the form

\iint_S \vec F \cdot \hat n\, dS

(Note this is not the type of integral you asked about.) This will always give you a positive value if F is in the direction of n, which makes sense since the flux is in the positive orientation direction of the surface. What will make the flux integral equal to zero is if the direction of F is perpendicular to the normal to surface, i.e., parallel to the surface. This means there is no flux flowing through the surface.

I'm afraid it makes me wonder if what you wrote in your original question was what you meant to ask.

Hopefully this helps.

 

[afallingbomb]

Yes it does. Thank you very much for your time!