Calculate the circulation of the field F around the closed curve C.
-$\mathbf { F } = \mathrm { xy } \mathbf { i } + 3 \mathbf { j }$, curve $\mathrm { C }$ is $\mathbf { r } ( \mathrm { t } ) = 4 \cos \mathrm { ti } + 4 \sin \mathrm { tj } , 0 \leq \mathrm { t } \leq 2 \pi$
A) 0
B) $- \frac { 14 } { 3 }$
C) 6
D) $\frac { 50 } { 3 }$

Question

Quizplus · Accepted Answer

The circulation of a vector field $\mathbf{F}$ around a closed curve $C$ can be calculated using the line integral of $\mathbf{F}$ along $C$. For $\mathbf{F} = xy\mathbf{i} + 3\mathbf{j}$ and $\mathbf{r}(t) = 4\cos(t)\mathbf{i} + 4\sin(t)\mathbf{j}$, the curve $C$ represents a circle of radius 4 centered at the origin. Since the vector field has a constant component in the $\mathbf{j}$ direction and a symmetric component in the $\mathbf{i}$ direction that averages to zero over a complete cycle around the circle, the net circulation around the circle is 0. This is because the contributions to the circulation from opposite sides of the circle cancel out for the $xy\mathbf{i}$ component, and the $3\mathbf{j}$ component does not contribute to circulation in a direction perpendicular to $\mathbf{j}$.

Calculate the Circulation of the Field F Around the Closed F=xyi+3j\mathbf { F } = \mathrm { xy } \mathbf { i } + 3 \mathbf { j }F=xyi+3j

Calculate the Circulation of the Field F Around the Closed $\mathbf { F } = \mathrm { xy } \mathbf { i } + 3 \mathbf { j }$