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Use Stokes' Theorem to evaluate $\iint_{S} \text {curl} \mathbf{F} \cdot d \mathbf{S} \cdot \mathbf{F}(x, y, z)=z e^{y} \mathbf{i}+x \cos y \mathbf{j}+x z \sin y \mathbf{k},$
$S$ is the hemisphere $x^{2}+y^{2}+z^{2}=16, y \geq 0,$ oriented in the direction of the positive $y$ -axis

$x^{2}+y^{2}+z^{2}=16 \quad y \geq 0$

$y=0 \quad x^{2}+z^{2}=16$

$r=4$

$\left\{\begin{array}{l}{x=4 \cos (-t)} \\ {z=4 \sin (-t)}\end{array}\right.$

$\mathbf{F}(x, y, z)=z e^{y} \mathbf{i}+x \cos y \mathbf{j}+x z \sin y \mathbf{k}$

$f(x, y, z)=z \hat{i}+x \hat{j}$

$\iint_{s} \text {curl} f d s=\int_{c} f(r(t)) r^{\prime}(t) d t$

$r(t)=4 \cos (-t) \hat{i}+4 \sin (-t) \hat{k}$

$r^{\prime} (t)=-4 \sin t \hat{i}+(-4 \cos t) \hat{k}$

$f(r(t))=4 \sin (-t) \hat{i}+4 \cos (-t) \hat{j}$

$\int_{0}^{2 \pi}\left[4 \sin (-t)\hat{i}-4 \cos(-t) \hat{j}\right]\left [-4 \sin t \hat{i}-4 \cos t \hat{k}\right]$

$=\int_{0}^{2 \pi} 16 \sin ^{2}(t)d t=16 \pi$

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