# Line Integral and Stokes Theorem

• Dec 1st 2011, 06:57 PM
ivinew
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 = $\partial S$ oriented counterclockwise as seen from above. Let F= (y,-x,y). Find $\oint_{c} F dS$

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:

$\oint_{c} F dS = \int \int_{S} curl F \bullet dS$

= $\int \int_{D}(1,0,-2) \bullet (-1,-2,1) dA$

= $\int \int_{D} -3 dA$

= -3 Area of D
= -3

By Line Integral:

Parametrization

Vector AC: $c_{1}(t) = (2t,2t,6t)$ for $0\leq t\leq 1$
Vector CB: $c_{2}(t) = (2-2t,2-t,6-4t)$ for $0\leq t\leq 1$
Vector BA: $c_{3}(t) = (0,1-t,2-2t)$ for $0\leq t\leq 1$

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

$c_{1}$:

$\int^1_{0} (2t,-2t,2t)\bullet(2,2,6)dS$

$= \int^1_{0} 4t-4t+12t dS$

$= 12$

$c_{2}$:

$\int^1_{0} (2-t,-2+2t,2-t)\bullet(-2,-1,-4)dS$

$= \int^1_{0} -4+2t+2-2t-8+4t dS$

$= -8$

$c_{3}$:

$\int^1_{0} (1-t,0,1-t)\bullet(0,-1,-2)dS$

$= \int^1_{0} -2+2t dS$

$= -1$

$c_{1}+c_{2}+c_{3} = 12 -8-1 = 3$

With Stokes theorem, I get -3. With line integral, I get 3. I should be getting the same results though.
• Dec 2nd 2011, 04:55 AM
HallsofIvy
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 = $\partial S$ oriented counterclockwise as seen from above. Let F= (y,-x,y). Find $\oint_{c} F dS$

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:

$\oint_{c} F dS = \int \int_{S} curl F \bullet dS$

= $\int \int_{D}(1,0,-2) \bullet (-1,-2,1) dA$

= $\int \int_{D} -3 dA$

= -3 Area of D
= -3

By Line Integral:

Parametrization

Vector AC: $c_{1}(t) = (2t,2t,6t)$ for $0\leq t\leq 1$
Vector CB: $c_{2}(t) = (2-2t,2-t,6-4t)$ for $0\leq t\leq 1$
Vector BA: $c_{3}(t) = (0,1-t,2-2t)$ for $0\leq t\leq 1$

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

$c_{1}$:

$\int^1_{0} (2t,-2t,2t)\bullet(2,2,6)dS$

$= \int^1_{0} 4t-4t+12t dS$

$= 12$

No. $\int_0^1 12t dt= 6t^2|_0^1= 6$.

Quote:

$c_{2}$:

$\int^1_{0} (2-t,-2+2t,2-t)\bullet(-2,-1,-4)dS$

$= \int^1_{0} -4+2t+2-2t-8+4t dS$

$= -8$

$c_{3}$:

$\int^1_{0} (1-t,0,1-t)\bullet(0,-1,-2)dS$

$= \int^1_{0} -2+2t dS$

$= -1$

$c_{1}+c_{2}+c_{3} = 12 -8-1 = 3$

With Stokes theorem, I get -3. With line integral, I get 3. I should be getting the same results though.
6- 8- 1= -3, not 3. Your first line integral was wrong.