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Math Help - Surface Integral

  1. #1
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    Surface Integral

    Calculate  \iint (x+y+z) dS
    If S is part of the plane 2x + y + z =1

    Ok I think i have done this but want to double check, here's my working:
    Sub in z = 1 - 2x - y to eliminate Z
    ds = \frac{dA}{cos(\theta)} = \frac{dx dy}{\vec{n}\vec{k}}
    \vec{N} = \nabla g{x,y,z}
    \vec{n} = \vec{N}/\abs{\vec{N}}

    Do all the subbing, i get
    \iint \sqrt{6}(-x+1) dxdy

    Limits are between [1-2x..0] for y and [1..0] for x.

    Thus i obtain \sqrt{1/6}

    Is this correct?
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  2. #2
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    Quote Originally Posted by imagemania View Post
    If S is part of the plane 2x + y + z =1
    I assume you mean in the first ocatant for the plane

    If that is the cases the x limit should go from zero to one half.
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  3. #3
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    Ohh of course it's 2x, thanks
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    Quote Originally Posted by imagemania View Post
    Ok I've come across a similar question which I am having troubles with, this time its the method
    A vector field exists so that:
    \vec{F(r)}=f(r)\vec{r}
    where
    r = \sqrt{{x}^{2}+{y}^{2}+{z}^{2}}

    C is the path on the sphere which is {x}^{2}+{y}^{2}+{z}^{2}=1

    Prove that \int \vec{F}.d\vec{r} = 0

    Now im not entirely sure how to best approach this, i can easily see:

    \int \vec{F}.d\vec{r} = \int f(r)\vec{r}.d\vec{r}

    Though i am unsure how to encorporate the path to deduce zero.

    I guess the main thing throwing me here is there f(r) i have r, and i could use sphereical coordinates, but I do not know f(r), just part of it (the r). Do i assume f(r) = r?? Then convert into sphereical coordinates, work out normal etc ?
    Use Stokes Theorem and note that

    \text{Curl} f(r)\mathbf{r}=\nabla  f(r) \times  \mathbf{r}+f(r)\nabla \times \mathbf{r}= \nabla  f(r) \times  \mathbf{r}
    Last edited by TheEmptySet; May 19th 2011 at 06:54 AM. Reason: Used wrong vector identity
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    Ugh Oh yeah, forgot about that theory! Thanks
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    Ok had another look, and it would seem straight foward ie do a matrix between the derivatives and \vec{r}. The issue is do we know \vec{r}. I know r as in f(r). Stokes theorem:
    \oint \vec{F}.d\vec{r} = \int \nabla \times(f(r)\vec{r}.d\vec{s})

    As you quite rightly said we can simply look at the cross product, but if i imply use r despite it not being the vector r, will the path C not be used anywhere in this formula (Which if its not would be logical, considering it would show its independent of path) perhaps on my sheet it should say:
    \vec{r} = \sqrt{{x}^{2}+{y}^{2}+{z}^{2}}
    Instead of
    r = \sqrt{{x}^{2}+{y}^{2}+{z}^{2}}
    ??
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  7. #7
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    Quote Originally Posted by imagemania View Post
    Ok had another look, and it would seem straight foward ie do a matrix between the derivatives and \vec{r}. The issue is do we know \vec{r}. I know r as in f(r). Stokes theorem:
    \oint \vec{F}.d\vec{r} = \int \nabla \times(f(r)\vec{r}.d\vec{s})

    As you quite rightly said we can simply look at the cross product, but if i imply use r despite it not being the vector r, will the path C not be used anywhere in this formula (Which if its not would be logical, considering it would show its independent of path) perhaps on my sheet it should say:
    \vec{r} = \sqrt{{x}^{2}+{y}^{2}+{z}^{2}}
    Instead of
    r = \sqrt{{x}^{2}+{y}^{2}+{z}^{2}}
    ??
    The above does not make sense the supposed "vector" does not have any components!

    What you have is the position vector and by defintion it is

    \mathbf{r}=x\mathbf{i}+y\mathbf{j}+z\mathbf{k}
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  8. #8
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    Ok so i do the cross product between that and nabla, but then is the surface
    {x}^{2}+{y}^{2}+{z}^{2}=1
    and r,
     r = \sqrt{{x}^{2}+{y}^{2}+{z}^{2}}

    Used at all in this question :s
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  9. #9
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    Quote Originally Posted by imagemania View Post
    Ok so i do the cross product between that and nabla, but then is the surface
    {x}^{2}+{y}^{2}+{z}^{2}=1
    and r,
     r = \sqrt{{x}^{2}+{y}^{2}+{z}^{2}}

    Used at all in this question :s
    By Stokes theorem you have that

    \int_{C}f(r)\mathbf{r}\cdot d \mathbf{s}=\iint_{S}(\nabla \times (f(r)\mathbf{r}))\cdot d\mathbf{S}=\iint_{S}(\nabla f(r) \times (\mathbf{r}))\cdot  \mathbf{n}dS

    The normal vector to the sphere is some multiple of the position vector

    This gives as the integrand

    (\nabla f(r) \times (\mathbf{r}))\cdot  \mathbf{r}

    but these two vectors are perpendicular so their dot product is zero.
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