91Ó°ÊÓ

We can use spherical coordinates to parametrize the hemisphere, given by the equation: \(x^{2}+y^{2}+z^{2}=36\). We will use the variables \(r\), \(\theta\), and \(\phi\), where \(r=6\), \(0 \leq \theta \leq 2\pi\), and \(0 \leq \phi \leq \frac{\pi}{2}\). Then, we can find the parametric equations for \(x\), \(y\), and \(z\) as follows: $$ x = r\sin{\phi}\cos{\theta} \\ y = r\sin{\phi}\sin{\theta} \\ z = r\cos{\phi} $$ Since \(r = 6\), we can substitute it in the equations to get: $$ x = 6\sin{\phi}\cos{\theta} \\ y = 6\sin{\phi}\sin{\theta} \\ z = 6\cos{\phi} $$

Next, we need the differential area element \(dS\). We can find this by computing the cross product of the partial derivatives of the parametric equations with respect to \(\theta\) and \(\phi\). First, we find these partial derivatives: $$ \frac{\partial(x, y, z)}{\partial\theta} = \left(-6\sin{\phi}\sin{\theta}, 6\sin{\phi}\cos{\theta}, 0\right) \\ \frac{\partial(x, y, z)}{\partial\phi} = \left(6\cos{\phi}\cos{\theta}, 6\cos{\phi}\sin{\theta}, -6\sin{\phi}\right) $$ Next, we compute the cross product of these partial derivatives: $$ \frac{\partial(x, y, z)}{\partial\theta} \times \frac{\partial(x, y, z)}{\partial\phi} = \left(-36\sin^2{\phi}\cos{\theta}, -36\sin^2{\phi}\sin{\theta}, 36\sin{\phi}\cos{\phi}\right) $$ Now, we need to find the magnitude of this vector, which will give us the differential area element \(dS\): $$ dS = \sqrt{(-36\sin^2{\phi}\cos{\theta})^2 + (-36\sin^2{\phi}\sin{\theta})^2 + (36\sin{\phi}\cos{\phi})^2}d\theta d\phi $$ Simplifying, we get: $$ dS = 36\sin\phi\, d\theta d\phi $$

Now, we substitute the values for \(x\), \(y\), \(z\), and \(dS\) into our integral, along with the function \(f(x, y, z) = x^2 + y^2\). We get: $$ \iint_{S} f(x, y, z) d S = \int_0^{2\pi} \int_0^{\frac{\pi}{2}} (6\sin{\phi}\cos{\theta})^2 + (6\sin{\phi}\sin{\theta})^2\cdot 36\sin{\phi} \, d\theta d\phi $$

We can now proceed to evaluate the integral. First, we notice that the integrand is separable, so we can rewrite the integral as a product of two separate integrals: $$ \int_0^{2\pi} \int_0^{\frac{\pi}{2}} (36\sin^2{\phi}\cos^2{\theta} + 36\sin^2{\phi}\sin^2{\theta})\cdot 36\sin{\phi} \, d\theta d\phi \\ = 36^2\int_0^{2\pi}(\cos^2\theta+\sin^2\theta)\, d\theta \int_0^{\frac{\pi}{2}}\sin^3\phi\, d\phi $$ We will now evaluate the two separate integrals: 1) \(\int_0^{2\pi} (\cos^2\theta+\sin^2\theta) \, d\theta\): $$ =\int_0^{2\pi} 1 \, d\theta \\ = \theta\Bigg|_0^{2\pi} \\ = 2\pi $$ 2) \(\int_0^{\frac{\pi}{2}} \sin^3\phi \, d\phi\). To compute this integral, use the power reduction formula: \(\sin^2\phi = \frac{1}{2}(1 - \cos{2\phi})\) $$ \int_0^{\frac{\pi}{2}} \sin\phi\sin^2\phi \, d\phi = \int_0^{\frac{\pi}{2}} \sin\phi\cdot \frac{1}{2}(1 - \cos{2\phi}) \, d\phi \\ = \frac{1}{2}\int_0^{\frac{\pi}{2}} \sin\phi - \sin\phi\cos{2\phi} \, d\phi \\ = \frac{1}{2}\left[(-\cos\phi)\Bigg|^{\frac{\pi}{2}}_0 + \frac{1}{2}\sin{2\phi}\Bigg|^{\frac{\pi}{2}}_0\right]\\ = \frac{1}{2}\left[ \left(-\cos{\frac{\pi}{2}} + \cos{0} \right) + \frac{1}{2}\left(\sin{2\cdot\frac{\pi}{2}} - \sin{0}\right)\right] \\ = \frac{1}{2}(1)\\\ By multiplying the two integrals, we find the value of the surface integral: $$ \iint_{S} f(x, y, z) d S = 36^2(2\pi)\left(\frac{1}{2}\right) = 36^2\pi $$ Therefore, the surface integral \(\iint_{S} f(x, y, z) d S = 36^2\pi\).