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Math Help - Conical surface and vector field

  1. #1
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    Post Conical surface and vector field

    Consider the conical surface
    S={ {(x,y,z) \in R^3 |x^2+y^2=z^2,0 \leq z \leq 1}}
    and the vector field f(x,y,z) = (-y,x,z)
    (a) Carefully sketch S (can anyone please send me a picture?)
    (b) Evaluate the path integral  \oint f.dr by direct integration.
    (c) Confirm that  \int \int_{S} \nabla x f.dS gives the same value, as asserted by Stokes' theorem.
    Thanks in advance
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  2. #2
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    Do you just need help on the surface S? It's an upside down cone with the vertex at the origin.
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  3. #3
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    Thanks, but i also need help with part (b) and (c) as well
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  4. #4
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    Quote Originally Posted by pdnhan View Post
    Consider the conical surface
    S={ {(x,y,z) \in R^3 |x^2+y^2=z^2,0 \leq z \leq 1}}
    and the vector field f(x,y,z) = (-y,x,z)
    (a) Carefully sketch S (can anyone please send me a picture?)
    You are told the figure is a cone. In the x,z plane (y= 0), the equation reduces to x^2= z^2 or the two lines z= x and z=-x. In the y,z plane (x= 0), it is z= y and z= -y. Use those to sketch it. Alternatively, use the cylindrical coordinates, z= r. Draw the line z= r in the \theta= 0 plane and rotate it around the z-axis.

    (b) Evaluate the path integral  \oint f.dr by direct integration.
    ??? There is no path mentioned in this problem! Do you mean the boundary at z= 1? That is a circle with center at (0, 0, 1), in the z= 1 plane, with radius 1. It is given by x= cos(\theta), y= sin(\theta), z= 1 so that dx= -sin(\theta)d\theta, dy= cos(\theta)d\theta, and dz= 0

    (c) Confirm that  \int \int_{S} \nabla x f.dS gives the same value, as asserted by Stokes' theorem.
    Thanks in advance
    On the surface, z= \sqrt{x^2+ y^2}, so the position vector of a point on the cone is \vec{X}= x\vec{i}+ y\vec{j}+ z\vec{k}= x\vec{i}+ y\vec{j}+ \sqrt{x^2+ y^2}\vec{k} which, in cylindrical coordinates, is r cos(\theta)\vec{i}+ r sin(\theta)\vec{j}+ r\vec{k}.

    The derivatives with respect to r and \theta are \vec{X}_r= cos(\theta)\vec{i}+ sin(\theta)\vec{j}+ \vec{k} and \vec{X}_\theta= -rsin()\theta)\vec{i}+ rcos(\theta)\vec{j} and are vectors in the tangent plane at each point.

    Their cross product, the "fundamental vector product" for the surface, oriented in the positive z direction, is -r cos(\theta)\vec{i}- r sin(\theta)\vec{j}+ r\vec{k} and so the "vector differential of surface area is (-r cos(\theta)\vec{i}- r sin(\theta)\vec{j}+ r\vec{k})drd\theta.

    Calculate \nabla\times \vec{f}, put it in polar coordinates with x= r cos(\theta), y= r sin(\theta), and z= r, take the dot product with d\vec{S} and integrate. r will run from 0 to 1, \theta from 0 to 2\pi.

    The actual integrals are almost trivial.
    Last edited by HallsofIvy; May 15th 2010 at 02:59 AM.
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  5. #5
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    Quote Originally Posted by HallsofIvy View Post
    You are told the figure is a cone. In the x,z plane (y= 0), the equation reduces to x^2= z^2 or the two lines z= x and z=-x. In the y,z plane (x= 0), it is z= y and z= -y. Use those to sketch it. Alternatively, use the cylindrical coordinates, z= r. Draw the line z= r in the \theta= 0 plane and rotate it around the z-axis.


    ??? There is no path mentioned in this problem! Do you mean the boundary at z= 1? That is a circle with center at (0, 0, 1), in the z= 1 plane, with radius 1. It is given by x= cos(\theta), y= sin(\theta), z= 1 so that dx= -sin(\theta)d\theta, dy= cos(\theta)d\theta, and dz= 0


    On the surface, z= \sqrt{x^2+ y^2}, so the position vector of a point on the cone is \vec{X}= x\vec{i}+ y\vec{j}+ z\vec{k}= x\vec{i}+ y\vec{j}+ \sqrt{x^2+ y^2}\vec{k} which, in cylindrical coordinates, is r cos(\theta)\vec{i}+ r sin(\theta)\vec{j}+ r\vec{k}.

    The derivatives with respect to r and \theta are \vec{X}_r= cos(\theta)\vec{i}+ sin(\theta)\vec{j}+ \vec{k} and \vec{X}_\theta= -rsin()\theta)\vec{i}+ rcos(\theta)\vec{j} and are vectors in the tangent plane at each point.

    Their cross product, the "fundamental vector product" for the surface, oriented in the positive z direction, is -r cos(\theta)\vec{i}- r sin(\theta)\vec{j}+ r\vec{k} and so the "vector differential of surface area is (-r cos(\theta)\vec{i}- r sin(\theta)\vec{j}+ r\vec{k})drd\theta.

    Calculate \nabla\times \vec{f}, put it in polar coordinates with x= r cos(\theta), y= r sin(\theta), and z= r, take the dot product with d\vec{S} and integrate. r will run from 0 to 1, \theta from 0 to 2\pi.

    The actual integrals are almost trivial.
    Thank you so much. You're a legend.
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