# Conical surface and vector field

• May 14th 2010, 06:47 AM
pdnhan
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.
• May 14th 2010, 05:04 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, 08:23 PM
pdnhan
Thanks, but i also need help with part (b) and (c) as well
• May 15th 2010, 02:48 AM
HallsofIvy
Quote:

Originally Posted by pdnhan
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.

Quote:

(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 $dz= 0$

Quote:

(c) Confirm that $\int \int_{S} \nabla x f.dS$ gives the same value, as asserted by Stokes' theorem.
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.
• May 17th 2010, 06:31 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 $x^2= z^2$ 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.