To compute the flux integral $I = \iint_{\omega} F \cdot d\mathbf{S}$, where $\omega$ is the upper hemisphere of the unit sphere in $\mathbb{R}^3$ with outward orientation and $F(x, y, z) = (0, x, 0)$, we can proceed as follows:

1. Understand the Geometry of $\omega$: The region $\omega$ is defined as the upper hemisphere of the unit sphere, which is the set $\omega = \{ (x, y, z) \in \mathbb{R}^3 \mid x^2 + y^2 + z^2 = 1, \, z \geq 0 \}.$ Since the outward orientation is specified, the surface normal $d\mathbf{S}$ will point away from the surface of the hemisphere.

2. Express the Surface Integral: We want to compute the surface integral: $I = \iint_{\omega} F \cdot d\mathbf{S},$ where $d\mathbf{S} = \mathbf{n} \, dS$, with $\mathbf{n}$ being the outward-pointing unit normal vector on $\omega$, and $dS$ representing the surface area element.

3. Parameterize $d\mathbf{S}$: Since $\omega$ is the upper unit hemisphere, we can use spherical coordinates: $x = \sin \theta \cos \phi, \quad y = \sin \theta \sin \phi, \quad z = \cos \theta,$ where $\theta \in [0, \pi/2]$ (as it is the upper hemisphere), and $\phi \in [0, 2\pi]$.

   The surface area element $dS$ on the unit sphere is $dS = \sin \theta \, d\theta \, d\phi$, and the outward unit normal vector $\mathbf{n}$ is simply $(x, y, z) = (\sin \theta \cos \phi, \sin \theta \sin \phi, \cos \theta)$ because $\omega$ lies on the unit sphere.

4. Compute $F \cdot d\mathbf{S}$: Given $F(x, y, z) = (0, x, 0)$, we have $F(x, y, z) \cdot \mathbf{n} = (0, x, 0) \cdot (x, y, z) = 0 \cdot x + x \cdot y + 0 \cdot z = xy.$ So, $F \cdot d\mathbf{S} = xy \, dS = \sin^2 \theta \cos \phi \sin \phi \, d\theta \, d\phi.$

5. Set Up the Integral: Substitute $F \cdot d\mathbf{S}$ into the integral $I$: $I = \iint_{\omega} F \cdot d\mathbf{S} = \int_0^{2\pi} \int_0^{\pi/2} \sin^2 \theta \cos \phi \sin \phi \, d\theta \, d\phi.$

6. Evaluate the Integral: Separate the integrals over $\theta$ and $\phi$: $I = \left( \int_0^{\pi/2} \sin^2 \theta \, d\theta \right) \left( \int_0^{2\pi} \cos \phi \sin \phi \, d\phi \right).$ Notice that $\cos \phi \sin \phi = \frac{1}{2} \sin 2\phi$. Then, $\int_0^{2\pi} \cos \phi \sin \phi \, d\phi = \int_0^{2\pi} \frac{1}{2} \sin 2\phi \, d\phi = 0,$ because the integral of $\sin 2\phi$ over a full period $[0, 2\pi]$ is zero.

7. Conclusion: Since the $\phi$-integral evaluates to zero, the entire surface integral $I$ is zero: $I = 0.$

Thus, the answer is: $\boxed{I = 0}.$

Would you like a breakdown of each step or have any questions?

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Related Questions:

1. How would the result change if $F(x, y, z) = (0, y, 0)$ instead?
2. What is the physical interpretation of the flux integral in this context?
3. How does the divergence theorem apply to surface integrals like this?
4. How would you set up this problem using a different coordinate system?
5. What is the significance of choosing an outward versus inward orientation?

Tip: For integrals over symmetric regions, look for terms that may integrate to zero due to symmetry—it often simplifies the problem greatly.