Evaluate $\int _ { C } \mathbf { F } d \mathbf { r }$ where $\mathbf { F } ( x , y , z ) = \left\langle \frac { z } { x - y } , \frac { z } { y - x } , \ln ( x - y ) \right\rangle$ and C is the circle $( x + 2 ) ^ { 2 } + ( y + 3 ) ^ { 2 } = 1$ in the clockwise direction.
A) $6$
B)$- 2$
C) $- 4$
D) $1$
E) $- 6$
F) $2$
G)$4$
H)$0$

Question

Evaluate $\int _ { C } \mathbf { F } d \mathbf { r }$ where $\mathbf { F } ( x , y , z ) = \left\langle \frac { z } { x - y } , \frac { z } { y - x } , \ln ( x - y ) \right\rangle$ and C is the circle $( x + 2 ) ^ { 2 } + ( y + 3 ) ^ { 2 } = 1$ in the clockwise direction.
A) $6$ 
B)$- 2$ 
C) $- 4$
D) $1$ 
E) $- 6$
F) $2$ 
G)$4$ 
H)$0$

Quizplus · Accepted Answer

The vector field $\mathbf { F } ( x , y , z ) = \left\langle \frac { z } { x - y } , \frac { z } { y - x } , \ln ( x - y ) \right\rangle$ is not conservative, but the integral over a closed path (like the circle $C$) can still be evaluated using various methods. However, a key observation here is that the circle $C$ lies entirely in the $xy$-plane, where $z = 0$. This means that the first two components of $\mathbf{F}$, which are multiplied by $z$, will be zero along the path $C$. The third component, $\ln(x - y)$, does not contribute to the line integral directly in the context of the given path, which is a closed loop in the $xy$-plane, and does not involve changes in $z$. Therefore, the integral simplifies significantly and effectively evaluates to zero because the components of the vector field that would contribute to the integral are nullified by the condition $z = 0$ on the path $C$.

Evaluate ∫CFdr\int _ { C } \mathbf { F } d \mathbf { r }∫C​Fdr

Evaluate $\int _ { C } \mathbf { F } d \mathbf { r }$