We want to evaluate the surface integral:
$\iint_S (x^2z + y^2z) \, dS$ where $S$ is the hemisphere $x^2 + y^2 + z^2 = 9$ for $z \geq 0$.

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Step 1: Surface parameterization

The hemisphere can be parameterized in spherical coordinates:
$x = 3\sin\phi\cos\theta, \quad y = 3\sin\phi\sin\theta, \quad z = 3\cos\phi$ where $\phi$ is the polar angle ($0 \leq \phi \leq \pi/2$) and $\theta$ is the azimuthal angle ($0 \leq \theta \leq 2\pi$).

The surface element in spherical coordinates is: $dS = R^2 \sin\phi \, d\phi \, d\theta = 9\sin\phi \, d\phi \, d\theta$

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Step 2: Express the integrand in terms of $\phi$ and $\theta$

Substitute $x$, $y$, and $z$ into $x^2z + y^2z$: $x^2 = (3\sin\phi\cos\theta)^2 = 9\sin^2\phi\cos^2\theta, \quad y^2 = (3\sin\phi\sin\theta)^2 = 9\sin^2\phi\sin^2\theta$ $x^2 + y^2 = 9\sin^2\phi(\cos^2\theta + \sin^2\theta) = 9\sin^2\phi$ $x^2z + y^2z = z(x^2 + y^2) = 3\cos\phi \cdot 9\sin^2\phi = 27\sin^2\phi\cos\phi$

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Step 3: Set up the integral

The surface integral becomes: $\iint_S (x^2z + y^2z) \, dS = \int_0^{2\pi} \int_0^{\pi/2} 27\sin^2\phi\cos\phi \cdot 9\sin\phi \, d\phi \, d\theta$ $= 243 \int_0^{2\pi} \, d\theta \int_0^{\pi/2} \sin^3\phi\cos\phi \, d\phi$

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Step 4: Evaluate the integrals

1. Evaluate $\int_0^{2\pi} d\theta$: $\int_0^{2\pi} d\theta = 2\pi$

2. Simplify $\int_0^{\pi/2} \sin^3\phi\cos\phi \, d\phi$:
   Let $u = \sin\phi$, so $du = \cos\phi \, d\phi$ and when $\phi = 0, u = 0$, and when $\phi = \pi/2, u = 1$.
   $\int_0^{\pi/2} \sin^3\phi\cos\phi \, d\phi = \int_0^1 u^3 \, du = \left[\frac{u^4}{4}\right]_0^1 = \frac{1}{4}$

3. Combine the results: $\iint_S (x^2z + y^2z) \, dS = 243 \cdot 2\pi \cdot \frac{1}{4} = \frac{243\pi}{2}$

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Final Answer:

$\boxed{\frac{243\pi}{2}}$

Would you like details or clarification on any step?

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Related Questions:

1. How do you calculate surface integrals for other surfaces like cylinders or cones?
2. What are the differences between surface integrals and line integrals?
3. How does the divergence theorem relate to surface integrals?
4. Can surface integrals be solved using Cartesian coordinates instead of spherical?
5. How is the Jacobian derived for different parameterizations?

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Tip:

When solving integrals over spheres or hemispheres, spherical coordinates often simplify the computation by leveraging symmetry.