To evaluate $\iint_S (\nabla \times \mathbf{F}) \cdot \mathbf{dS}$ using Stokes' Theorem, we will follow these steps:

Stokes' Theorem

Stokes' Theorem relates the surface integral of the curl of a vector field to the line integral of the vector field over the boundary curve of the surface: $\iint_S (\nabla \times \mathbf{F}) \cdot \mathbf{dS} = \oint_{\partial S} \mathbf{F} \cdot \mathbf{dr}.$

Here:

• $\mathbf{F}(x, y, z) = z e^y \, \mathbf{i} + x \cos(y) \, \mathbf{j} + xz \sin(y) \, \mathbf{k}$,
• $S$ is the hemisphere $x^2 + y^2 + z^2 = 4$, $y \geq 0$,
• $\partial S$ is the boundary circle of the hemisphere in the plane $y = 0$, which is the circle $x^2 + z^2 = 4$, $y = 0$.

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Step 1: Parameterize the boundary curve $\partial S$

The boundary curve lies in the plane $y = 0$, and the circle $x^2 + z^2 = 4$. We parameterize this circle as: $\mathbf{r}(t) = 2 \cos(t) \, \mathbf{i} + 0 \, \mathbf{j} + 2 \sin(t) \, \mathbf{k}, \quad t \in [0, 2\pi].$

The derivative is: $\mathbf{r}'(t) = -2 \sin(t) \, \mathbf{i} + 0 \, \mathbf{j} + 2 \cos(t) \, \mathbf{k}.$

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Step 2: Evaluate $\mathbf{F} \cdot \mathbf{dr}$ along $\partial S$

Substitute $\mathbf{r}(t)$ into $\mathbf{F}$: $\mathbf{F}(\mathbf{r}(t)) = (2 \sin(t) e^0) \, \mathbf{i} + (2 \cos(t) \cos(0)) \, \mathbf{j} + (2 \cos(t) \cdot 2 \sin(t) \sin(0)) \, \mathbf{k}.$ Simplify: $\mathbf{F}(\mathbf{r}(t)) = 2 \sin(t) \, \mathbf{i} + 2 \cos(t) \, \mathbf{j} + 0 \, \mathbf{k}.$

The differential $\mathbf{dr}$ is: $\mathbf{dr} = \mathbf{r}'(t) \, dt = (-2 \sin(t) \, \mathbf{i} + 0 \, \mathbf{j} + 2 \cos(t) \, \mathbf{k}) \, dt.$

Dot product: $\mathbf{F} \cdot \mathbf{dr} = (2 \sin(t))(-2 \sin(t)) + (2 \cos(t))(0) + (0)(2 \cos(t)) \, dt.$ $\mathbf{F} \cdot \mathbf{dr} = -4 \sin^2(t) \, dt.$

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Step 3: Evaluate the line integral

$\oint_{\partial S} \mathbf{F} \cdot \mathbf{dr} = \int_0^{2\pi} -4 \sin^2(t) \, dt.$

Use the identity $\sin^2(t) = \frac{1}{2}(1 - \cos(2t))$: $\int_0^{2\pi} -4 \sin^2(t) \, dt = \int_0^{2\pi} -4 \cdot \frac{1}{2}(1 - \cos(2t)) \, dt.$ $= \int_0^{2\pi} -2 (1 - \cos(2t)) \, dt.$ $= \int_0^{2\pi} -2 \, dt + \int_0^{2\pi} 2 \cos(2t) \, dt.$

The first integral: $\int_0^{2\pi} -2 \, dt = -2 \cdot 2\pi = -4\pi.$

The second integral: \int_0^{2\pi} 2 \cos(2t) \, dt = 0 \quad \text{(since \cos(2t)$is periodic over$0$to$2\pi)}.

Thus: $\oint_{\partial S} \mathbf{F} \cdot \mathbf{dr} = -4\pi.$

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Final Answer

$\iint_S (\nabla \times \mathbf{F}) \cdot \mathbf{dS} = \boxed{-4\pi}.$

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Do you have any questions, or would you like additional details?

Here are 5 related questions to expand on this topic:

1. How do we compute the curl of a vector field?
2. Can we apply Stokes’ Theorem for surfaces with non-circular boundaries?
3. How would the solution change if the surface orientation were reversed?
4. What are the conditions required for Stokes’ Theorem to hold?
5. How can Stokes’ Theorem be used in physics, particularly electromagnetism?

Tip: Always check the orientation of the surface and boundary when applying Stokes' Theorem to ensure consistent results.