Question

Consider the given vector field. F(x, y, z) = 8x^2 + y^2 + z^2 (x i + y j + z k)(a) Find the curl of the vector field. curl F = _______ (b) Find the divergence of the vector field. div F = _________

Answer 1

Answer:

a) Curl F=0

b) div F=16x+2y+2z

Step-by-step explanation:

Assuming the given vector field is $\vec F=8x^2i +y^2j+z^2k$, then the curl of F is

$Curl F=\nabla \times F$

$Curl F=\left|\begin{array}{ccc}i&j&k\\\frac{\partial}{\partial x} &\frac{\partial}{\partial y} &\frac{\partial}{\partial z} \\8x^2&y^2&z^2\end{array}\right|$

$\implies Curl F=(\frac{\partial (z^2)}{\partial y} -\frac{\partial (y^2)}{\partial z} )i-(\frac{\partial (z^2)}{\partial x} -\frac{\partial (8x^2)}{\partial z} )j+(\frac{\partial (y^2)}{\partial y} -\frac{\partial (8x^2)}{\partial y} )k$

$\implies Curl F=0i+0j+0k$=0

b) The divergence of F is:

$div F=\nabla \cdot \vec F$

$div F=(\frac{\partial}{\partial x}i+\frac{\partial}{\partial y}j+\frac{\partial}{\partial z}k)\cdot(8x^2i+y^2j+z^2k)$

$\implies div F=\frac{\partial}{\partial x} (8x^2)+\frac{\partial}{\partial y} (y^2)+\frac{\partial}{\partial z} (z^2)=16x+2y+2z$