Answer:
a) Curl F=0
b) div F=16x+2y+2z
Step-by-step explanation:
Assuming the given vector field is [tex]\vec F=8x^2i +y^2j+z^2k[/tex], then the curl of F is
[tex]Curl F=\nabla \times F[/tex]
[tex]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|[/tex]
[tex]\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[/tex]
[tex]\implies Curl F=0i+0j+0k[/tex]=0
b) The divergence of F is:
[tex]div F=\nabla \cdot \vec F[/tex]
[tex]div F=(\frac{\partial}{\partial x}i+\frac{\partial}{\partial y}j+\frac{\partial}{\partial z}k)\cdot(8x^2i+y^2j+z^2k)[/tex]
[tex]\implies div F=\frac{\partial}{\partial x} (8x^2)+\frac{\partial}{\partial y} (y^2)+\frac{\partial}{\partial z} (z^2)=16x+2y+2z[/tex]
