Calculate the flow in the field F along the path C.
-$\mathbf { F } = ( \mathrm { x } - \mathrm { y } ) \mathbf { i } - \left( \mathrm { x } ^ { 2 } + \mathrm { y } ^ { 2 } \right) \mathbf { j } ; \mathrm { C }$ is curve from $( 4,0 )$ to $( - 4,0 )$ on the upper half of the circle $\mathrm { x } ^ { 2 } + \mathrm { y } ^ { 2 } = 16$
A) $8 \pi$
B) $- 8 \pi$
C) $\frac { 16 \pi - 1 } { 4 }$
D) $16 \pi$

Question

Quizplus · Accepted Answer

Parametrize the curve $C$ by $\boldsymbol{r}(t) = (4\cos t, 4\sin t)$ for $0\le t\le \pi$ so that $\boldsymbol{r}(0)=(4,0)$ and $\boldsymbol{r}(\pi)=(-4,0)$. Then the line integral of $\boldsymbol{F}$ over $C$ is given by
\begin{align*}
\int_C \boldsymbol{F}\cdot d\boldsymbol{r} &= \int_0^\pi \boldsymbol{F}(\boldsymbol{r}(t))\cdot \boldsymbol{r}'(t)\,dt \
&= \int_0^\pi (4\cos t - 4\sin t)(-\sin t)\,dt \
&\quad + \int_0^\pi (16\cos^2 t + 16\sin^2 t)(4\cos t)\,dt \
&= \int_0^\pi (-4\cos t\sin t - 64\cos^2 t)\,dt \
&= \int_0^\pi (-2\sin 2t - 32\cos^2 t + 32)\,dt \
&= -32\pi/2 + 32\pi/2 \
&= 8\pi.
\end{align*}

Calculate the Flow in the Field F Along the Path