Find the work done by F over the curve in the direction of increasing t.
-$\mathbf { F } = x \mathrm { e } ^ { 3 \mathrm { x } ^ { 2 } } \mathbf { i } + \mathrm { e } ^ { 7 \mathrm { y } _ { j } } + \mathrm { e } ^ { 3 \mathrm { z } } \mathbf { k }$; the path is $\mathrm { C } _ { 1 } \cup \mathrm { C } _ { 2 }$ where $\mathrm { C } _ { 1 }$ is the straight line from $( 0,0,0 )$ to $( 1,1,0 )$ and $\mathrm { C } _ { 2 }$ is the straight line from $( 1,1,0 )$ to $( 1,1,1 )$
A) $W = \frac { e ^ { 3 } } { 6 } + \frac { e ^ { 7 } } { 7 } + \frac { e ^ { 3 } } { 3 }$
B) $W = \frac { e ^ { 3 } - 1 } { 6 } + \frac { e ^ { 7 } - 1 } { 7 }$
C) $W = \frac { e ^ { 3 } - 1 } { 6 } + \frac { e ^ { 7 } - 1 } { 7 } + \frac { e ^ { 3 } - 1 } { 3 }$
D) $W = \frac { e ^ { 3 } - 1 } { 3 } + \frac { 2 e ^ { 7 } - 2 } { 7 } + \frac { e ^ { 3 } - 1 } { 3 }$

Question

Quizplus · Accepted Answer

The work done by a force field $\mathbf{F}$ along a path $C$ is given by the line integral of $\mathbf{F}$ along $C$. For the given force field $\mathbf{F} = x e^{3x^2} \mathbf{i} + e^{7y} \mathbf{j} + e^{3z} \mathbf{k}$ and the piecewise path $C_1 \cup C_2$, the work done can be calculated by evaluating the line integral over each segment and summing the results. For $C_1$, parameterize the line from $(0,0,0)$ to $(1,1,0)$ with $x=t$, $y=t$, $z=0$, and for $C_2$, from $(1,1,0)$ to $(1,1,1)$ with $x=1$, $y=1$, $z=t$. The correct calculation involves integrating the dot product of $\mathbf{F}$ with the differential displacement vector along each path, resulting in $\frac{e^3 - 1}{6}$ for the $x$ component along $C_1$, $\frac{e^7 - 1}{7}$ for the $y$ component along $C_1$, and $\frac{e^3 - 1}{3}$ for the $z$ component along $C_2$, summing up to the total work done.

Find the Work Done by F Over the Curve in the Direction