is there a vector field g on ℝ³ such that curl(g) = ⟨x sin(y), cos(y), z - 3xy⟩?\nyes\nno\nexplain.\nthere…

is there a vector field g on ℝ³ such that curl(g) = ⟨x sin(y), cos(y), z - 3xy⟩?\nyes\nno\nexplain.\nthere ---select--- such g because div(curl g) ? 0.\nneed help? read it watch it master it\nsubmit answer

is there a vector field g on ℝ³ such that curl(g) = ⟨x sin(y), cos(y), z - 3xy⟩?\nyes\nno\nexplain.\nthere ---select--- such g because div(curl g) ? 0.\nneed help? read it watch it master it\nsubmit answer

Answer

Explanation:

Step1: Record la definición de divergencia

La divergencia de un campo vectorial $\mathbf{F}=\langle P,Q,R\rangle$ en $\mathbb{R}^3$ está dada por $\text{div}\mathbf{F}=\frac{\partial P}{\partial x}+\frac{\partial Q}{\partial y}+\frac{\partial R}{\partial z}$. Aquí, $\text{curl}\mathbf{G}=\langle x\sin(y),\cos(y),z - 3xy\rangle$, entonces $P = x\sin(y)$, $Q=\cos(y)$ y $R=z - 3xy$.

Step2: Calcular las derivadas parciales

Calculamos $\frac{\partial P}{\partial x}=\sin(y)$, $\frac{\partial Q}{\partial y}=-\sin(y)$ y $\frac{\partial R}{\partial z}=1$.

Step3: Calcular la divergencia del curl

Sumamos las derivadas parciales: $\text{div}(\text{curl}\mathbf{G})=\frac{\partial P}{\partial x}+\frac{\partial Q}{\partial y}+\frac{\partial R}{\partial z}=\sin(y)-\sin(y)+1 = 1\neq0$.

Step4: Aplicar el teorema

Sabemos que para cualquier campo vectorial $\mathbf{G}$ suave en $\mathbb{R}^3$, $\text{div}(\text{curl}\mathbf{G}) = 0$. Pero en este caso $\text{div}(\text{curl}\mathbf{G})\neq0$.

Answer:

No There is no such $\mathbf{G}$ because $\text{div}(\text{curl}\mathbf{G})\neq0$.