# Stokes' Theorem

• Jul 31st 2008, 04:33 PM
angel.white
Stokes' Theorem
Use Stokes' Theorem to evaluate $\displaystyle \int_C \vec F \cdot d\vec r$. C is oriented counterclockwise as viewed from above.

$\displaystyle \vec F (x,y,z) = <x+y^2, y+z^2, z+x^2>$, C is the triangle with vertices (1,0,0), (0,1,0), and (0,0,1)

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I've computed the curl as <-2z, -2x, -2y> and the equation of the plane as z=1-x-y

Stokes' theorem says $\displaystyle \int_C \vec F \cdot d\vec r = \int \int_S curl \vec F \cdot d\vec s$

But I can't figure out how to evaluate the second part, since the curl is a vector, and the surface has three variables, I don't know how to get it into a form I can integrate.
• Jul 31st 2008, 05:52 PM
mr fantastic
Quote:

Originally Posted by angel.white
Use Stokes' Theorem to evaluate $\displaystyle \int_C \vec F \cdot d\vec r$. C is oriented counterclockwise as viewed from above.

$\displaystyle \vec F (x,y,z) = <x+y^2, y+z^2, z+x^2>$, C is the triangle with vertices (1,0,0), (0,1,0), and (0,0,1)

-----

I've computed the curl as <-2z, -2x, -2y> and the equation of the plane as z=1-x-y

Stokes' theorem says $\displaystyle \int_C \vec F \cdot d\vec r = \int \int_S curl \vec F \cdot d\vec s$

But I can't figure out how to evaluate the second part, since the curl is a vector, and the surface has three variables, I don't know how to get it into a form I can integrate.

By the way, your curl and equation for the plane look OK.
• Jul 31st 2008, 06:13 PM
angel.white
Quote:

Originally Posted by mr fantastic

By the way, your curl and equation for the plane look OK.

It is still unclear to me, perhaps I do not understand what is meant by $\displaystyle d\vec S$.
• Jul 31st 2008, 06:29 PM
mr fantastic
Quote:

Originally Posted by angel.white
It is still unclear to me, perhaps I do not understand what is meant by $\displaystyle d\vec S$.

$\displaystyle \vec{dS}$ is an infinitesimal element of directed surface.

$\displaystyle \vec{dS} = \hat{n} dS$ where $\displaystyle \hat{n}$ is a unit normal vector to S.

For your surface (plane), $\displaystyle \vec{n} = \frac{\partial z}{\partial x} \, i + \frac{\partial z}{\partial y} \, j - k = -i - j - k$. Then:
$\displaystyle \nabla F \cdot \vec{dS} = (-2z \, i - 2x \, j - 2y \, k) \cdot (-i - j - k) \, dx \, dy = 2 (z + x + y) \, dx \, dy = 2 \, dx \, dy$
where I've substituted $\displaystyle z = 1 - x - y$.