If $\vec{F}$ Is a Path-Independent Field, Then $\int _ { C } \vec { F } \cdot d \vec { r } = 0$

question 69

True/False

If $\vec{F}$ is a path-independent field, then $\int _ { C } \vec { F } \cdot d \vec { r } = 0$ where C has the parameterization $\vec { r } ( t ) = \cos t \vec { i } + \sin t \vec { j } , 0 < t < 3 \pi$

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