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Hi,

I'm trying to calculate some line integral with both Gauss' and Stokes' theorems, but for some strange reasons I get different results. Since the solution with Stokes' theorem seems to be somewhat easy I doubt that this question was meant to be solved by Gauss' theorem but I still want to try.

Let C be a curve which is the intersection of a ball [itex]x^2+y^2+z^2=a^2[/itex] and a plain [itex]x-y+z=0[/itex].

The "wanted" line integral is: [itex]\int_C \mathbf{F} \centerdot d \mathbf{x}[/itex] [the orientation doesn't matter since I only interested in absolute numerical value.].

As the intersecting plane has the origin in it and since the center of the ball is the origin as well we can conclude that the plain slices the ball into 2 equal pieces and the curve cuts the ball surface into 2 surfaces with equal area. I will denote S for the whole surface of the ball and S

I calculate and get [itex]\nabla \times \mathbf{F}=(0,-4,0) [/itex]

[itex]\int_C \mathbf{F} \centerdot d \mathbf{x} = \iint_{H} (\nabla \times \mathbf{F}) \centerdot \mathbf{n} ds = \iint_{H} (0,-4,0) \centerdot \frac{(1,-1,1)}{\sqrt{3}}ds= \frac {4} {\sqrt{3}} \iint_{H}ds= \frac {4} {\sqrt{3}} \pi a^2[/itex]

The last surface integral is exactly the surface of H which is a surface of a cycle with a radius of 'a' and this explains the last transition at the above equation.

To conclude:

[itex]\int_C \mathbf{F} \centerdot d \mathbf{x} = \frac{4} {\sqrt{3}} \pi a^2[/itex]

Since Stokes' theorem states that only the boundary of the a surface plays the role I can conclude that:

[itex]\int_C \mathbf{F} \centerdot d \mathbf{x} = \iint_{S_1} (\nabla \times \mathbf{F}) \centerdot \mathbf{n} ds [/itex] and [itex]\int_C \mathbf{F} \centerdot d \mathbf{x} = \iint_{S_S} (\nabla \times \mathbf{F}) \centerdot \mathbf{n} ds [/itex], and since both share the boundary:

[itex]2\int_C \mathbf{F} \centerdot d \mathbf{x} = \iint_{S_1} (\nabla \times \mathbf{F}) \centerdot \mathbf{n} ds + \iint_{S_2} (\nabla \times \mathbf{F}) \centerdot \mathbf{n} ds=\iint_{S} (\nabla \times \mathbf{F}) \centerdot \mathbf{n} ds [/itex]

And I apply the Gauss' theorem:

[itex]\iint_{S} (\nabla \times \mathbf{F}) \centerdot \mathbf{n} ds=\iint_{S} (0,-4,0) \centerdot \mathbf{n} ds=\iiint_{x^2+y^2+z^2\leq a^2} \nabla \centerdot (0,-4,0)dxdydz=0[/itex]

So I get a different answer which means somewhere I did a mistake however I checked myself a million times and I'm sure I did not commit computational mistake and my mistake must be conceptual.

I will appreciate help, thanks in advance.

I'm trying to calculate some line integral with both Gauss' and Stokes' theorems, but for some strange reasons I get different results. Since the solution with Stokes' theorem seems to be somewhat easy I doubt that this question was meant to be solved by Gauss' theorem but I still want to try.

**Setting the stage:**Let C be a curve which is the intersection of a ball [itex]x^2+y^2+z^2=a^2[/itex] and a plain [itex]x-y+z=0[/itex].

**F**is a vector field defined by: [itex] \mathbf{F} = (y-2z,x-z,2x-y)[/itex].The "wanted" line integral is: [itex]\int_C \mathbf{F} \centerdot d \mathbf{x}[/itex] [the orientation doesn't matter since I only interested in absolute numerical value.].

As the intersecting plane has the origin in it and since the center of the ball is the origin as well we can conclude that the plain slices the ball into 2 equal pieces and the curve cuts the ball surface into 2 surfaces with equal area. I will denote S for the whole surface of the ball and S

_{1}, S_{2}for its 2 equal halfes. I will denote the flat surface between the 2 half of the ball with H.I calculate and get [itex]\nabla \times \mathbf{F}=(0,-4,0) [/itex]

**Calculation using Stokes' theorem:**[itex]\int_C \mathbf{F} \centerdot d \mathbf{x} = \iint_{H} (\nabla \times \mathbf{F}) \centerdot \mathbf{n} ds = \iint_{H} (0,-4,0) \centerdot \frac{(1,-1,1)}{\sqrt{3}}ds= \frac {4} {\sqrt{3}} \iint_{H}ds= \frac {4} {\sqrt{3}} \pi a^2[/itex]

The last surface integral is exactly the surface of H which is a surface of a cycle with a radius of 'a' and this explains the last transition at the above equation.

To conclude:

[itex]\int_C \mathbf{F} \centerdot d \mathbf{x} = \frac{4} {\sqrt{3}} \pi a^2[/itex]

**Calculating using the Gauss' theorem:**Since Stokes' theorem states that only the boundary of the a surface plays the role I can conclude that:

[itex]\int_C \mathbf{F} \centerdot d \mathbf{x} = \iint_{S_1} (\nabla \times \mathbf{F}) \centerdot \mathbf{n} ds [/itex] and [itex]\int_C \mathbf{F} \centerdot d \mathbf{x} = \iint_{S_S} (\nabla \times \mathbf{F}) \centerdot \mathbf{n} ds [/itex], and since both share the boundary:

[itex]2\int_C \mathbf{F} \centerdot d \mathbf{x} = \iint_{S_1} (\nabla \times \mathbf{F}) \centerdot \mathbf{n} ds + \iint_{S_2} (\nabla \times \mathbf{F}) \centerdot \mathbf{n} ds=\iint_{S} (\nabla \times \mathbf{F}) \centerdot \mathbf{n} ds [/itex]

And I apply the Gauss' theorem:

[itex]\iint_{S} (\nabla \times \mathbf{F}) \centerdot \mathbf{n} ds=\iint_{S} (0,-4,0) \centerdot \mathbf{n} ds=\iiint_{x^2+y^2+z^2\leq a^2} \nabla \centerdot (0,-4,0)dxdydz=0[/itex]

So I get a different answer which means somewhere I did a mistake however I checked myself a million times and I'm sure I did not commit computational mistake and my mistake must be conceptual.

I will appreciate help, thanks in advance.

**EDIT: I just fixed many errors and typos I found in my question...**[Thanks to Fractal20 for spotting one of them.]
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