Using the Divergence Theorem, find the outward flux of F across the boundary of the region D.
-$\mathbf { F } = x y \mathbf { i } + \mathrm { y } ^ { 2 } \mathbf { j } - 2 y z \mathbf { k }$; $\mathrm { D }$ : the solid wedge cut from the first quadrant by the plane $\mathrm { y } + \mathrm { z } = 2$ and the parabolic cylinder $x = 1 - 16 y ^ { 2 }$
A) $\frac { 29 } { 960 }$
B) $\frac { 7 } { 240 }$
C) $\frac { 7 } { 120 }$
D) $\frac { 1 } { 30 }$

Question

Quizplus · Accepted Answer

The Divergence Theorem states that the outward flux of a vector field $\mathbf{F}$ across the boundary of a region $D$ is equal to the triple integral over $D$ of the divergence of $\mathbf{F}$. The divergence of $\mathbf{F} = x y \mathbf{i} + y^2 \mathbf{j} - 2 y z \mathbf{k}$ is $\nabla \cdot \mathbf{F} = \frac{\partial}{\partial x}(xy) + \frac{\partial}{\partial y}(y^2) + \frac{\partial}{\partial z}(-2yz) = y + 2y - 2y = y$. The region $D$ is bounded by $x = 1 - 16y^2$, $y + z = 2$, and $y, z \geq 0$. To find the volume integral, we integrate $y$ over the region $D$. The limits for $y$ are from $0$ to $1$, for $x$ from $0$ to $1 - 16y^2$, and for $z$ from $0$ to $2 - y$. Thus, the integral becomes $\int_{0}^{1}\int_{0}^{1-16y^2}\int_{0}^{2-y} y \,dz\,dx\,dy$. Solving this integral gives the outward flux as $\frac{7}{240}$.

Using the Divergence Theorem, Find the Outward Flux of F F=xyi+y2j−2yzk\mathbf { F } = x y \mathbf { i } + \mathrm { y } ^ { 2 } \mathbf { j } - 2 y z \mathbf { k }F=xyi+y2j−2yzk

Using the Divergence Theorem, Find the Outward Flux of F $\mathbf { F } = x y \mathbf { i } + \mathrm { y } ^ { 2 } \mathbf { j } - 2 y z \mathbf { k }$