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📅 April 30, 2026 👤 scraface

Apr 30, 2026

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W = \int_C \vec{F} \cdot d\vec{r}
Let \( \vec{r}(t) = \langle 3\cos t, 3\sin t \rangle \), so:
\vec{r}'(t) = \langle -3\sin t, 3\cos t \rangle, \quad t \in [0, 2\pi]
Now, evaluate \( \vec{F}(\vec{r}(t)) = \langle -3\sin t, 3\cos t \rang...
\vec{F} \cdot \vec{r}' = (-3\sin t)(-3\sin t) + (3\cos t)(3\cos t) = 9...
Thus, the integral becomes:
W = \int_0^{2\pi} 9 \, dt = 9 \cdot 2\pi = 18\pi
The work done by the field is:
\boxed{18\pi}
Question:** An electrical engineer studies a alternating current signa...

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