# Conical surface and vector field

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• May 14th 2010, 05:47 AM
pdnhan
Conical surface and vector field
Consider the conical surface
S={$\displaystyle {(x,y,z) \in R^3 |x^2+y^2=z^2,0 \leq z \leq 1}$}
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 $\displaystyle \oint f.dr$ by direct integration.
(c) Confirm that $\displaystyle \int \int_{S} \nabla x f.dS$ gives the same value, as asserted by Stokes' theorem.
Thanks in advance
• May 14th 2010, 04:04 PM
ANDS!
Do you just need help on the surface S? It's an upside down cone with the vertex at the origin.
• May 14th 2010, 07:23 PM
pdnhan
Thanks, but i also need help with part (b) and (c) as well
• May 15th 2010, 01:48 AM
HallsofIvy
Quote:

Originally Posted by pdnhan
Consider the conical surface
S={$\displaystyle {(x,y,z) \in R^3 |x^2+y^2=z^2,0 \leq z \leq 1}$}
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 $\displaystyle x^2= z^2$ 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 $\displaystyle \theta= 0$ plane and rotate it around the z-axis.

Quote:

(b) Evaluate the path integral $\displaystyle \oint f.dr$ 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 $\displaystyle x= cos(\theta)$, $\displaystyle y= sin(\theta)$, $\displaystyle z= 1$ so that $\displaystyle dx= -sin(\theta)d\theta$, $\displaystyle dy= cos(\theta)d\theta$, and $\displaystyle dz= 0$

Quote:

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

The derivatives with respect to r and $\displaystyle \theta$ are $\displaystyle \vec{X}_r= cos(\theta)\vec{i}+ sin(\theta)\vec{j}+ \vec{k}$ and $\displaystyle \vec{X}_\theta= -rsin()\theta)\vec{i}+ rcos(\theta)\vec{j}$ 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 $\displaystyle -r cos(\theta)\vec{i}- r sin(\theta)\vec{j}+ r\vec{k}$ and so the "vector differential of surface area is $\displaystyle (-r cos(\theta)\vec{i}- r sin(\theta)\vec{j}+ r\vec{k})drd\theta$.

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

The actual integrals are almost trivial.
• May 17th 2010, 05:31 AM
pdnhan
Quote:

Originally Posted by HallsofIvy
You are told the figure is a cone. In the x,z plane (y= 0), the equation reduces to $\displaystyle x^2= z^2$ 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 $\displaystyle \theta= 0$ 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 $\displaystyle x= cos(\theta)$, $\displaystyle y= sin(\theta)$, $\displaystyle z= 1$ so that $\displaystyle dx= -sin(\theta)d\theta$, $\displaystyle dy= cos(\theta)d\theta$, and $\displaystyle dz= 0$

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

The derivatives with respect to r and $\displaystyle \theta$ are $\displaystyle \vec{X}_r= cos(\theta)\vec{i}+ sin(\theta)\vec{j}+ \vec{k}$ and $\displaystyle \vec{X}_\theta= -rsin()\theta)\vec{i}+ rcos(\theta)\vec{j}$ 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 $\displaystyle -r cos(\theta)\vec{i}- r sin(\theta)\vec{j}+ r\vec{k}$ and so the "vector differential of surface area is $\displaystyle (-r cos(\theta)\vec{i}- r sin(\theta)\vec{j}+ r\vec{k})drd\theta$.

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

The actual integrals are almost trivial.

Thank you so much. You're a legend.