The surface integral $\int \int$S $x y z d S \text {, }$ where S is $z ^ { 2 } = x ^ { 2 } + y ^ { 2 }$ between $z = 1$ and $z = 2$ is

A) $\frac { 5 \sqrt { 5 } - 1 } { 6 }$
B) $\frac { 15 \pi } { \sqrt { 2 } }$
C) $\frac { 17 \sqrt { 17 } - 1 } { 4 }$
D) $\frac { \sqrt { 3 } } { 6 }$
E)0

Question

The surface integral $\int \int$S $x y z d S \text {, }$ where S is $z ^ { 2 } = x ^ { 2 } + y ^ { 2 }$ between $z = 1$ and $z = 2$ is

A) $\frac { 5 \sqrt { 5 } - 1 } { 6 }$
B) $\frac { 15 \pi } { \sqrt { 2 } }$
C) $\frac { 17 \sqrt { 17 } - 1 } { 4 }$
D) $\frac { \sqrt { 3 } } { 6 }$
E)0

Quizplus · Accepted Answer

The surface integral involves the product $xyz$, and since the surface is symmetric with respect to the x and y axes, for every point $(x, y, z)$ with a positive contribution to the integral, there is a corresponding point $(-x, y, z)$ or $(x, -y, z)$ with an equal but negative contribution. This symmetry causes all contributions to cancel out, resulting in an integral value of 0.

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

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