Find the divergence of $\vec { F } ( x , y , z ) = \left\langle e ^ { y z } , e ^ { x z } , e ^ { x y } \right\rangle$
A) $e ^ { y z } + e ^ { x z } + e ^ { x y }$
B) $y e ^ { y z } + x e ^ { x z } + x e ^ { x y }$
C) $0$
D) $3 e ^ { x y }$

Question

Quizplus · Accepted Answer

The divergence of a vector field $\vec{F} = \langle P, Q, R \rangle$ is given by $\nabla \cdot \vec{F} = \frac{\partial P}{\partial x} + \frac{\partial Q}{\partial y} + \frac{\partial R}{\partial z}$. For $\vec{F}(x, y, z) = \langle e^{yz}, e^{xz}, e^{xy} \rangle$, calculating the partial derivatives gives $\frac{\partial}{\partial x}e^{yz} = 0$, $\frac{\partial}{\partial y}e^{xz} = 0$, and $\frac{\partial}{\partial z}e^{xy} = 0$. Adding these together yields $0$, indicating that the divergence of $\vec{F}$ is $0$.

Find the Divergence Of F⃗(x,y,z)=⟨eyz,exz,exy⟩\vec { F } ( x , y , z ) = \left\langle e ^ { y z } , e ^ { x z } , e ^ { x y } \right\rangleF(x,y,z)=⟨eyz,exz,exy⟩

Find the Divergence Of $\vec { F } ( x , y , z ) = \left\langle e ^ { y z } , e ^ { x z } , e ^ { x y } \right\rangle$