Find the flux $\vec { F }$ of through $S$, $\iint _ { S } \vec { F } \cdot \vec { N } d S$, where $\vec { N }$ is the upward unit normal vector to S. $\overrightarrow { \mathbf { F } } ( x , y , z ) = 2 z \hat { \mathbf { i } } - 8 \hat { \mathbf { j } } + y \hat { \mathbf { k } }$
$S : x + y + z = 9$, first octant
A) $- \frac { 81 } { 2 }$
B) $\frac { 81 } { 2 }$
C) $\frac { 99 } { 2 }$
D) $\frac { 93 } { 2 }$
E) $- \frac { 87 } { 2 }$

Question

Quizplus · Accepted Answer

To find the flux of $\vec{F}$ through $S$, we use the formula $\iint_S \vec{F} \cdot \vec{N} dS$. The surface $S$ is defined by $x + y + z = 9$ in the first octant, which means $x, y, z \geq 0$.First, we find the normal vector $\vec{N}$. Since the surface is defined by $x + y + z = 9$, a normal vector to the surface can be derived from the gradient of $x + y + z - 9 = 0$, which gives $\vec{N} = \hat{i} + \hat{j} + \hat{k}$. For a unit normal vector pointing upwards, we normalize this, but since the calculation of flux involves a dot product, the magnitude of $\vec{N}$ will be adjusted by the area element $dS$, so we can directly use $\vec{N} = \hat{i} + \hat{j} + \hat{k}$ for simplicity in calculation.The vector field is $\vec{F}(x, y, z) = 2z\hat{i} - 8\hat{j} + y\hat{k}$.The dot product $\vec{F} \cdot \vec{N} = (2z)\cdot(1) + (-8)\cdot(1) + (y)\cdot(1) = 2z - 8 + y$.To find the flux, we integrate this over the surface. However, since the surface is in the first octant bounded by $x + y + z = 9$, we can use the projection of the surface onto the $xy$-plane for easier integration. In the $xy$-plane, the boundary is given by $x + y = 9$ (setting $z = 0$).The flux integral becomes a double integral over the region in the $xy$-plane. We need to express $z$ in terms of $x$ and $y$ using $z = 9 - x - y$ for the integration.Substituting $z = 9 - x - y$ into the dot product gives $2(9 - x - y) - 8 + y = 18 - 2x - 2y - 8 + y = 10 - 2x - y$.The double integral over the region $D$ in the $xy$-plane is $\iint_D (10 - 2x - y) dA$.To solve this integral, we can use the limits for $x$ from $0$ to $9$ and for $y$ from $0$ to $9 - x$, resulting in the integral $\int_0^9 \int_0^{9-x} (10 - 2x - y) dy dx$.Solving this integral gives the total flux, which is $\frac{81}{2}$, indicating that the correct answer is B.

Find the Flux F⃗\vec { F }F Of Through SSS ∬SF⃗⋅N⃗dS\iint _ { S } \vec { F } \cdot \vec { N } d S∬S​F⋅NdS

Find the Flux $\vec { F }$ Of Through $S$ $\iint _ { S } \vec { F } \cdot \vec { N } d S$