Let $\mathbf { F } ( x , y , z ) = x \mathbf { j }$ . Evaluate the line integral $\int _ { C } \mathbf { F } \cdot d \mathbf { r }$ along the rectangular path from $( 0,0,0 )$ to $( 1,0,1 )$ to $( 1,1,1 )$ to $( 0,1,0 )$ to $( 0,0,0 )$ .
A) $1$

B) $\sqrt { 2 }$
C) $\sqrt { 3 }$
D) $2$
E)$\frac { \sqrt { 2 } } { 2 }$
F) $\frac { \sqrt { 3 } } { 2 }$

G) $\frac { 1 } { 2 }$

H) $\frac { 1 } { 4 }$

Question

Let $\mathbf { F } ( x , y , z ) = x \mathbf { j }$ . Evaluate the line integral $\int _ { C } \mathbf { F } \cdot d \mathbf { r }$ along the rectangular path from $( 0,0,0 )$ to $( 1,0,1 )$ to $( 1,1,1 )$ to $( 0,1,0 )$ to $( 0,0,0 )$ .
A) $1$

B) $\sqrt { 2 }$
C) $\sqrt { 3 }$
D) $2$ 
E)$\frac { \sqrt { 2 } } { 2 }$ 
F) $\frac { \sqrt { 3 } } { 2 }$ 
 
G) $\frac { 1 } { 2 }$

H) $\frac { 1 } { 4 }$

Quizplus · Accepted Answer

The vector field $\mathbf{F}(x, y, z) = x\mathbf{j}$ only has a component in the $y$-direction. The line integral $\int_C \mathbf{F} \cdot d\mathbf{r}$ along a path $C$ essentially measures the work done by the field along that path. Since the field is only in the $y$-direction, any movement in the $x$ or $z$ direction does not contribute to the integral. The path from $(0,0,0)$ to $(1,0,1)$ to $(1,1,1)$ to $(0,1,0)$ and back to $(0,0,0)$ only has a segment from $(1,0,1)$ to $(1,1,1)$ that aligns with the $y$-direction, over which the field is constant ($x=1$). The length of this segment is 1 (from $y=0$ to $y=1$), so the integral over this path is $1 \times 1 = 1$, since the field strength is $1$ ($x=1$) and the displacement in the $y$-direction is $1$. All other segments contribute $0$ to the integral because they are perpendicular to the field direction.

Let F(x,y,z)=xj\mathbf { F } ( x , y , z ) = x \mathbf { j }F(x,y,z)=xj

Let $\mathbf { F } ( x , y , z ) = x \mathbf { j }$