Evaluate the surface integral of G over the surface S.
-$S$ is the hemisphere $x ^ { 2 } + y ^ { 2 } + z ^ { 2 } = 3 , z \geq 0 ; G ( x , y , z ) = z ^ { 2 }$
A) $6 \pi$
B) $4 \pi$
C) $36 \pi$
D) $3 \pi$

Question

Quizplus · Accepted Answer

The surface integral of $G$ over the hemisphere can be calculated using spherical coordinates. In spherical coordinates, $x = \rho \sin \phi \cos \theta$, $y = \rho \sin \phi \sin \theta$, and $z = \rho \cos \phi$, where $\rho$ is the radius, $\phi$ is the polar angle, and $\theta$ is the azimuthal angle. Given $z^2$ as the function to integrate and the hemisphere $x^2 + y^2 + z^2 = 3$, we set $\rho = \sqrt{3}$. The integral becomes:$$\int_{0}^{2\pi} \int_{0}^{\frac{\pi}{2}} (\sqrt{3} \cos \phi)^2 (\sqrt{3}^2 \sin \phi) \, d\phi \, d\theta = 3 \int_{0}^{2\pi} \int_{0}^{\frac{\pi}{2}} \cos^2 \phi \sin \phi \, d\phi \, d\theta$$Solving the integral gives:$$3 \cdot 2\pi \cdot \left[\frac{1}{3} \cos^3 \phi\right]_{0}^{\frac{\pi}{2}} = 2\pi \cdot 1 = 2\pi$$However, the correct calculation involves integrating $z^2$ over the hemisphere surface, which requires correctly applying the surface element in spherical coordinates. The mistake in the previous step was in the integration and interpretation of the function $G(x, y, z) = z^2$ over the hemisphere. Correctly evaluating the integral involves recognizing that $z = \sqrt{3} \cos \phi$ and the surface area element in spherical coordinates is $dS = \rho^2 \sin \phi d\phi d\theta$, leading to the integral:$$\int_{0}^{2\pi} \int_{0}^{\frac{\pi}{2}} 3 \cos^2 \phi \cdot 3 \sin \phi \, d\phi \, d\theta$$This integral evaluates to $6\pi$, which correctly accounts for the surface area of the hemisphere and the function $G(x, y, z) = z^2$.

Evaluate the Surface Integral of G Over the Surface S SSS

Evaluate the Surface Integral of G Over the Surface S $S$