Thread: Conical surface and vector field

1. 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.

2. Do you just need help on the surface S? It's an upside down cone with the vertex at the origin.

3. Thanks, but i also need help with part (b) and (c) as well

4. 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.

(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 $\displaystyle dz= 0$

(c) Confirm that $\displaystyle \int \int_{S} \nabla x f.dS$ gives the same value, as asserted by Stokes' theorem.
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.

5. 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.