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Math Help - Stokes' theorem

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    Stokes' theorem

    I got stuck using the Stokes' theorem, the problem is at the bottom of the pic. I found the Curl of F, and also the normal of the Triangle. As you can see, I ended up with an area integer with 3 variables, how do I solve this? did I do it right? http://img2.tapuz.co.il/forums/1_119833460.jpg
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    Quote Originally Posted by asi123 View Post
    I got stuck using the Stokes' theorem, the problem is at the bottom of the pic. I found the Curl of F, and also the normal of the Triangle. As you can see, I ended up with an area integer with 3 variables, how do I solve this? did I do it right? http://img2.tapuz.co.il/forums/1_119833460.jpg
    A couple of things:

    1. Since you're integrating on the surface (a plane), you can substitute z = 4x + y (which is the equation of the plane). This solves your technical infelicity of three variables in the integrand ......

    2. It looks like you're using a unit vector normal to the surface (plane). That's incorrect. F \cdot dS \neq F \cdot \hat{n} \, dx \, dy in general.

    In fact, F \cdot dS = F \cdot n \, dx \, dy:

    F \cdot dS = F \cdot \left( \frac{\partial z}{\partial x} \, i + \frac{\partial z}{\partial y} \, j - k\right) \, dx \, dy.

    3. Are you OK with the region in the xy-plane you integrate over (and hence the integral terminals) .....? The region is the right angle triangle with vertices at (0, 0), (2, 0) and (2, 2).
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    Quote Originally Posted by mr fantastic View Post
    A couple of things:

    1. Since you're integrating on the surface (a plane), you can substitute z = 4x + y (which is the equation of the plane). This solves your technical infelicity of three variables in the integrand ......

    2. It looks like you're using a unit vector normal to the surface (plane). That's incorrect. F \cdot dS \neq F \cdot \hat{n} \, dx \, dy in general.

    In fact, F \cdot dS = F \cdot n \, dx \, dy:

    F \cdot dS = F \cdot \left( \frac{\partial z}{\partial x} \, i + \frac{\partial z}{\partial y} \, j - k\right) \, dx \, dy.

    3. Are you OK with the region in the xy-plane you integrate over (and hence the integral terminals) .....? The region is the right angle triangle with vertices at (0, 0), (2, 0) and (2, 2).
    Ok, I think I fixed it, is that right?

    Another thing.
    I marked everything in the pic.
    Shouldn't I normalize the normal vector?
    Did I write the boundaries right?

    10x again.
    Attached Thumbnails Attached Thumbnails Stokes' theorem-scan0003.jpg  
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    Quote Originally Posted by asi123 View Post
    Ok, I think I fixed it, is that right?

    Another thing.
    I marked everything in the pic.
    Shouldn't I normalize the normal vector? Mr F says: No. I've already given the correct formula.
    Did I write the boundaries right? Mr F says: I'm not sure why you wanted to switch from xy to uv-coordinates ..... x and y are fine. Using xy-coordinates, the boundary is y = 0 to y = -x + 2 and x = 0 to x = 2 (or alternatively, x = 0 to x = 2 - y and y = 0 to y = 2).

    10x again.
    I get 20/3 (and I stayed with xy-coordinates) but I could be wrong since my careless arithmetic errors are legion.
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    Quote Originally Posted by mr fantastic View Post
    I get 20/3 (and I stayed with xy-coordinates) but I could be wrong since my careless arithmetic errors are legion.
    Why did you choose the boundaries from y=0 to y=-x+2, isn't it from y=0 to y=x?
    Attached Thumbnails Attached Thumbnails Stokes' theorem-scan0001.jpg  
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  6. #6
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    Quote Originally Posted by asi123 View Post
    Why did you choose the boundaries from y=0 to y=-x+2, isn't it from y=0 to y=x?
    Quite right. I accidently read my (2, 2) as (0, 2) when I drew my sketch ......
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