Line Integral and Stokes Theorem

For the following problem, I'm suppose to Stokes Theorem and then line integral but I'm getting different results.

The Problem:

Let S be the triangular region with vertices A(0,0,0), B(0,1,2), C(2,2,6). Let C = oriented counterclockwise as seen from above. Let F= (y,-x,y). Find

First, I find the equation of the plane and that comes out to z=x+2y.

Then when I parametrize the surface, I get S(x,y) = (x,y,x+2y).

Because it's counterclockwise, I need Sx X Sy = (-1,-2,1).

The curl of F is (1,0,-2)

**By Stokes Theorem:**

=

=

= -3 Area of D

= -3

**By Line Integral: **

Parametrization

Vector AC: for

Vector CB: for

Vector BA: for

Keeping in mind that F= (y,-x,y):

:

:

:

With Stokes theorem, I get -3. With line integral, I get 3. I should be getting the same results though.

Re: Line Integral and Stokes Theorem

Quote:

Originally Posted by

**ivinew** For the following problem, I'm suppose to Stokes Theorem and then line integral but I'm getting different results.

The Problem:

Let S be the triangular region with vertices A(0,0,0), B(0,1,2), C(2,2,6). Let C =

oriented counterclockwise as seen from above. Let F= (y,-x,y). Find

First, I find the equation of the plane and that comes out to z=x+2y.

Then when I parametrize the surface, I get S(x,y) = (x,y,x+2y).

Because it's counterclockwise, I need Sx X Sy = (-1,-2,1).

The curl of F is (1,0,-2)

**By Stokes Theorem:**
=

=

= -3 Area of D

= -3

**By Line Integral: **
Parametrization

Vector AC:

for

Vector CB:

for

Vector BA:

for

Keeping in mind that F= (y,-x,y):

:

No. .

6- 8- 1= -3, not 3. Your first line integral was wrong.