For each of the following vector fields, find its curl and determine if it is a gradient field.(a) vec(F)=(4xy+x3)vec(i)+(2x2+z2)vec(j)+(2yz-4z)vec(k)curlvec(F)vec(F) is a gradient field(b) vec(G)=2yzvec(i)+(2xz+z2)vec(j)+(2xy+2yz)vec(k)curlvec(G)=0vec(G)(c) vec(H)=2(xy+z2)vec(i)+4(x2+yz)vec(j)+4(xz+y2)vec(k).curl vec(H)=vec(H)

a) The given vector is vecF= Hence The curl of the vector is It is known by Definition that Curl vecF=vec gradXvecF=[[hati,hatj,hatk],[(del)/(delx),(del)/(dely),(del)/(delz)],[F_1,F_2,F_3]]

Resuelto For each of the following vector fields, find its

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Pregunta: For each of the following vector fields, find its curl and determine if it is a gradient field.(a) vec(F)=(4xy+x3)vec(i)+(2x2+z2)vec(j)+(2yz-4z)vec(k)curlvec(F)vec(F) is a gradient field(b) vec(G)=2yzvec(i)+(2xz+z2)vec(j)+(2xy+2yz)vec(k)curlvec(G)=0vec(G)(c) vec(H)=2(xy+z2)vec(i)+4(x2+yz)vec(j)+4(xz+y2)vec(k).curl vec(H)=vec(H)
For each of the following vector fields, find its curl and determine if it is a gradient field.
$($ a $)$ vec $(F) = (4 x y + x^{3}) v e c (i) + (2 x^{2} + z^{2}) v e c (j) + (2 y z - 4 z) v e c (k)$
curlvec $(F)$
vec $(F)$ is a gradient field
$($ b $)$ vec $(G) = 2$ yzvec $(i) + (2 x z + z^{2}) v e c (j) + (2 x y + 2 y z) v e c (k)$
curlvec $(G) = 0$
vec $(G)$
$($ c $)$ vec $(H) = 2 (x y + z^{2}) v e c (i) + 4 (x^{2} + y z) v e c (j) + 4 (x z + y^{2}) v e c (k) .$
curl vec $(H) =$
vec $(H)$
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Solución
Paso 1
a) The given vector is $\vec{F} =< 4 x y + x^{3}, 2 x^{2} + z^{2}, 2 y z - 4 z >$
Hence The curl of the vector is
$\begin{matrix} \begin{aligned} C u r l \vec{F} & = [\begin{array}{c} \hat{i} & \hat{j} & \hat{k} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ 4 x y + x^{3} & 2 x^{2} + z^{2} & 2 y z - 4 z \end{array}] \\ = \hat{i} (2 z - 2 z) + \hat{j} (0 - 0) + \hat{k} ((4 x - 4 x) \\ = \hat{i} (0) + \hat{j} (0) + \hat{k} (0) \\ = \vec{O} \end{aligned} \end{matrix}$

Explanation:
It is known by Definition that Curl $\begin{matrix} \vec{F} = \vec{\nabla} X \vec{F} = [\begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ F_{1} & F_{2} & F_{3} \end{matrix}] \end{matrix}$
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