Use Green's Theorem to find the work done by the force $\mathbf { F } ( x , y ) = \left( 6 x - 7 y ^ { 2 } \right) \mathbf { i } + 3 y \mathbf { j }$ in moving a particle in the positive direction once around the triangle with vertices $( 0,0 )$ , $( 1,0 )$ , and $( 0,1 )$ .
A) 7
B) $\frac { 7 } { 6 }$
C) $\frac { 7 } { 3 }$
D) 42

Question

Quizplus · Accepted Answer

Green's Theorem states that given a vector field $\mathbf{F} = P\,\mathbf{i} + Q\,\mathbf{j}$ and a closed curve C in the plane, oriented counterclockwise, the line integral of $\mathbf{F}$ around C is equal to the double integral over the region D enclosed by C of the curl of $\mathbf{F}$.

Mathematically, this can be written as:
$$\oint_{C}\mathbf{F}\cdot d\mathbf{r} = \iint_{D} \left( \frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}\right) dA$$

In our case,
$\mathbf{F}(x,y) = (6x - 7y^2)\,\mathbf{i} + 3y\,\mathbf{j}$

We can rewrite this as $P(x,y) = 6x - 7y^2$ and $Q(x,y) = 3y$.

To apply Green's Theorem, we need to find the curl of $\mathbf{F}$:
\begin{align*}
\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}&=\frac{\partial}{\partial x}(3y)-\frac{\partial}{\partial y}(6x-7y^2)\
&=0-(-14y)\
&=14y
\end{align*}

Now, we need to find the region D enclosed by the triangle. This is the triangle itself, with vertices $(0,0)$, $(1,0)$ and $(0,1)$.

We can split the triangle into two parts, using the line $x=y$. Let's focus on the left part of the triangle, which is the region enclosed by the x-axis, the line $x=y$ and the line $x=0$.

Using Green's Theorem, we have:
\begin{align*}
\oint_{C}\mathbf{F}\cdot d\mathbf{r} &= \iint_{D} \left( \frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}\right) dA\
&= \iint_{D} 14y\,dA\
&=\int_{0}^{1}\int_{0}^{x}14y\,dydx\
&= \int_{0}^{1}\left[7y^2\right]_{0}^{x}dx\
&=\int_{0}^{1}7x^2\,dx\
&=\left[\frac{7}{3}x^3\right]_{0}^{1}\
&=\frac{7}{3}
\end{align*}

Similarly, we can calculate the line integral for the right part of the triangle, which is the region enclosed by the x-axis, the line $x=y$ and the line $y=1$, and get the same result of $\frac{7}{3}$.

Therefore, the total work done by the force in moving a particle once around the triangle in the positive direction is $\frac{7}{3}+\frac{7}{3}=\frac{14}{3}\approx4.67$, which is closest to $\frac{7}{3}$. So the correct answer is (C).

Use Green's Theorem to Find the Work Done by the Force