The surface integral $\int \int$S $x y z d S ,$ where S is $x ^ { 2 } + y ^ { 2 } + z ^ { 2 } = 9$ and $z \geq 0$ is

A) $- 27 \pi$
B) $27 \pi$
C) $26 \pi$
D) $- 26 \pi$
E) $25 \pi$

Question

The surface integral $\int \int$S $x y z d S ,$ where S is $x ^ { 2 } + y ^ { 2 } + z ^ { 2 } = 9$ and $z \geq 0$ is

A) $- 27 \pi$
B) $27 \pi$
C) $26 \pi$
D) $- 26 \pi$
E) $25 \pi$

Quizplus · Accepted Answer

The given surface S is the upper half of the sphere $x^2 + y^2 + z^2 = 9$. To compute the surface integral $\int \int_S xyz dS$, we can use the spherical coordinates transformation, where $x = \rho\sin\phi\cos\theta$, $y = \rho\sin\phi\sin\theta$, and $z = \rho\cos\phi$, with $\rho = 3$ for the sphere of radius 3. The integral becomes $\int \int_S 3^3\sin^3\phi\cos\theta\sin\theta\cos\phi dS$, since $dS = \rho^2\sin\phi d\phi d\theta$ on a sphere. However, due to the symmetry of the sphere and the nature of the integrand, the integral of $\cos\theta\sin\theta$ over $0$ to $2\pi$ is $0$, because it's an odd function over a symmetric interval. Thus, the correct answer is that the integral evaluates to $0$, which is not among the provided options. This suggests a mistake in the interpretation of the question or the provided options. Given the nature of the question and the options provided, the correct approach would involve recognizing the symmetry and properties of the integrand in spherical coordinates, but it seems there was a misunderstanding in the explanation process.

The Surface Integral ∫∫\int \int∫∫ S xyzdS,x y z d S ,xyzdS,

The Surface Integral $\int \int$ S $x y z d S ,$