Let R be the rectangle with vertices (0,0), (3,0), (0,9), (3,9), and let C be the boundary of R traversed?

Let R be the rectangle with vertices (0,0), (3,0), (0,9), (3,9), and let C be the boundary of R traversed counterclockwise. For the vector field F(x,y) = 3yI+xJ,





find the integral:





⌠


⌡C (F·dX)





Where C is the lower bound.

Let R be the rectangle with vertices (0,0), (3,0), (0,9), (3,9), and let C be the boundary of R traversed?
since the integral is taken over x the whole boundary can be dividad into two domains


1) (0,3) whr y = 9


2) (3,0) whr y = 0


now its just simple integration
Reply:Let C1 be the part of C from (0,0) to (3,0). Along C1, y has the constant value of 0, and dX = (dx)I. So along C1,





F·dX = (3y evaluated at y=0) = 0 So





∫ F·dX = 0


C1





Let C2 be the part of C from (3,0) to (3,9). Along C2, x has the constant value of 3, and dX = (dy)J So along C2,





F·dX = (x evaluated at x=3) = 3 So





∫ F·dX =


C2





9


∫ 3 dy = 27


0





Let C3 be the part of C from (3,9) to (0,9). Along C3, y has the constant value of 9, and dX = (dx)I So along C3,





F·dX = (3y evaluated at y=9) = 27 So





∫ F·dX =


C3





0


∫ 27 dx= -81


3





I'll let you do C4. You will find that the integral over C4 is zero.





So





∫ F·dX = 0 + 27 - 81 + 0 = -54


C





You could also do this problem using Green's Theorem, which says this line integral is equal to





∫∫ (∂(x)/∂x - ∂(3y)/∂y) dx dy


A





where A is the region enclosed by C. Since





(∂(x)/∂x - ∂(3y)/∂y = -2, the integral is equal to -2 * area enclosed by C





= -2*9*3 = -54