Math Help - Stokes' Theorem

1. Stokes' Theorem

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) = $, 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.

2. 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) = $, 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.

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

3. 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 $d\vec S$.

4. 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.

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.