A particle starts at the point ,moves along the x-axis to (3,0)and then along the semicircle to the starting point.Use Green's Theorem to find the work done on this particle by the force field
D) 0

Question

A particle starts at the point  ,moves along the x-axis to (3,0)and then along the semicircle  to the starting point.Use Green's Theorem to find the work done on this particle by the force field 
D) 0

Quizplus · Accepted Answer

Using Green's Theorem, we have:
$$\int_C \vec{F} \cdot d\vec{r} = \iint_R \left(\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}ight) dA$$
where $\vec{F} = (x+y, x-y)$ and $C$ is the closed curve consisting of the line segment from $(11eaae29_f432_987d_87fe_955b383a432f_TB2067_11,0)$ to $(3,0)$ followed by the semicircle centered at the origin with radius 3.

We can parameterize the line segment as $\vec{r}(t) = (11eaae29_f432_987d_87fe_955b383a432f_TB2067_11 - 8t, 0)$ where $0 \leq t \leq 1$. Then, $d\vec{r} = (-8,0) dt$ and
$$\int_{(11eaae29_f432_987d_87fe_955b383a432f_TB2067_11,0)}^{(3,0)} \vec{F} \cdot d\vec{r} = \int_0^1 \vec{F}(\vec{r}(t)) \cdot d\vec{r}(t) = \int_0^1 (-88t) dt = -44$$

Next, we can parameterize the semicircle as $\vec{r}(t) = (3\cos t, 3\sin t)$ where $0 \leq t \leq \pi$. Then, $d\vec{r} = (-3\sin t, 3\cos t) dt$ and
$$\int_{	ext{semicircle}} \vec{F} \cdot d\vec{r} = \int_0^\pi \vec{F}(\vec{r}(t)) \cdot d\vec{r}(t) = \int_0^\pi (3\cos t + 3\sin t)(-3\sin t, 3\cos t) dt = 0$$
because the integrand is an odd function of $t$.

Therefore, the total work done is
\begin{align*}
\int_C \vec{F} \cdot d\vec{r} &= \int_{(11eaae29_f432_987d_87fe_955b383a432f_TB2067_11,0)}^{(3,0)} F_1 dx + F_2 dy + \int_{	ext{semicircle}} \vec{F} \cdot d\vec{r} \
&= -44 + 0 \
&= -44
\end{align*}
Since work is a scalar quantity, we take the absolute value to get the total work done as $|-44| = 44$. Converting $44$ to hexadecimal gives the answer A: 11eaae29_f432_bf90_87fe_730f14e8ee72_TB2067_11.

A Particle Starts at the Point ,Moves Along the X-Axis