Answer
See proof
Work Step by Step
Given: \[ \vec{F} = \oint_{\vec{C}} (y \, dx + x \, dy) \] where \(\vec{C}\) is the unit circle oriented counterclockwise. Solution Using Green's Theorem: We can use Green's Theorem to evaluate the line integral: \[ \vec{F} = \oint_{\vec{C}} (y \, dx + x \, dy) = \iint_R \left(\frac{\partial}{\partial x} - \frac{\partial}{\partial y}\right)(y \, dx + x \, dy) \, dA = \iint_R (0) \, dA = 0 \] Here, \(R\) is the region enclosed by \(\vec{C}\). Direct Evaluation: To verify our result, let's evaluate the given integral directly using parametrization. We have the parametric equations for \(\vec{C}\): \[ x = \cos(t), \quad y = \sin(t), \quad 0 \leq t \leq 2\pi \] Then, we calculate \(\vec{r}(t)\) and \(\frac{d\vec{r}}{dt}\): \[ \vec{r}(t) = \cos(t) \, \hat{i} + \sin(t) \, \hat{j}, \quad \frac{d\vec{r}}{dt} = -\sin(t) \, \hat{i} + \cos(t) \, \hat{j} \] Now, we compute the line integral: \[ \begin{aligned} \vec{F} &= \oint_{\vec{C}} (y \, dx + x \, dy) \\ &= \int_0^{2\pi} \left(\sin(t) \, (-\sin(t)) + \cos(t) \, \cos(t)\right) \, dt \\ &= \int_0^{2\pi} (-\sin^2(t) + \cos^2(t)) \, dt \\ &= \int_0^{2\pi} \cos(2t) \, dt \\ &= \frac{1}{2} \left[\sin(2t)\right]_0^{2\pi} \\ &= \frac{1}{2} \left(\sin(4\pi) - \sin(0)\right) \\ &= \frac{1}{2} (0 - 0) \\ &= 0 \end{aligned} \] Conclusion: By both using Green's Theorem and directly evaluating the line integral, we have shown that \[ \oint_{\vec{C}} (y \, dx + x \, dy) = 0 \] This confirms that the line integral of \(\vec{F}\) over the given unit circle \(\vec{C}\) is indeed 0.
