Stokes' Theorem

**angel.white** · July 31st 2008, 04:33 PM

Use Stokes' Theorem to evaluate $\int_C \vec F \cdot d\vec r$ . C is oriented counterclockwise as viewed from above.

$\vec F (x,y,z) = <x+y^2, y+z^2, z+x^2>$ , 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 $\int_C \vec F \cdot d\vec r = \int \int_S curl \vec F \cdot d\vec s$

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.

**mr fantastic** · July 31st 2008, 05:52 PM

Originally Posted by angel.white

Use Stokes' Theorem to evaluate $\int_C \vec F \cdot d\vec r$ . C is oriented counterclockwise as viewed from above.

$\vec F (x,y,z) = <x+y^2, y+z^2, z+x^2>$ , 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 $\int_C \vec F \cdot d\vec r = \int \int_S curl \vec F \cdot d\vec s$

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.

Does this thread help: http://www.mathhelpforum.com/math-he...s-theorem.html

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

**angel.white** · July 31st 2008, 06:13 PM

Originally Posted by mr fantastic

Does this thread help: http://www.mathhelpforum.com/math-he...s-theorem.html

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 $d\vec S$ .

**mr fantastic** · July 31st 2008, 06:29 PM

Originally Posted by angel.white

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

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

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

The formulae I quote in the other thread follow from this.

For your surface (plane), $\vec{n} = \frac{\partial z}{\partial x} \, i + \frac{\partial z}{\partial y} \, j - k = -i - j - k$ . Then:

$\nabla F \cdot \vec{dS} = (-2z \, i - 2x \, j - 2y \, k) \cdot (-i - j - k) \, dx \, dy = 2 (z + x + y) \, dx \, dy = 2 \, dx \, dy$

where I've substituted $z = 1 - x - y$ .

The region of integration in the xy-plane is clearly the right-triangle with vertices at (0, 0), (1, 0) and (0, 1).

So your line integral is simply twice the area of this triangle: 1.

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