Curls and divergences * Calculate the curl and the divergence of each of the following vec- tor fields. If the curl turns out to be zero, try to discover a scalar function φ of which the vector field is the gradient. (a) F=(x+y,−x+y,−2z); (b) G=(2y,2x+3z,3y); (c) H=(x2 −z2,2,2xz).

Answer:a) [tex]\nabla F=0, \nabla\times F=(0,0,-2)[/tex]b)  [tex]\nabla F=0, \nabla\times F=(0,0,0)[/tex], [tex]f=2xy+3yz+C, C\in\mathbb{R}[/tex].c)  [tex]\nabla F=4x, \nabla\times F=(0,-4z,0)[/tex]Step-by-step explanation:Remember, if F= <f,g,h> is a vector field and [tex]\nabla[/tex] is the operator [tex]\nabla=<\frac{\partial}{\partial x}, \frac{\partial}{\partial y}, \frac{\partial}{\partial z}>[/tex]the divergence of F is [tex]\nabla F=<\frac{\partial}{\partial x}, \frac{\partial}{\partial y}, \frac{\partial}{\partial z}>\cdot F= \frac{\partial f}{\partial x}+\frac{\partial g}{\partial y} +\frac{\partial h}{\partial z}[/tex]the curl of F is [tex]\nabla\times F=det(\left[\begin{array}{ccc}i&j&k\\\frac{\partial}{\partial x}&\frac{\partial}{\partial y}&\frac{\partial}{\partial z}\\f&g&h\end{array}\right] )[/tex]a) [tex]F=(x+y,-x+y,-2z)[/tex]The divergence of F is [tex]\nabla F=\frac{\partial}{\partial x}(x+y)+\frac{\partial}{\partial y}(-x+y)+\frac{\partial}{\partial z} (-2z)=1+1-2=0[/tex]The curl of F is[tex]\nabla\times F=det(\left[\begin{array}{ccc}i&j&k\\\frac{\partial}{\partial x}&\frac{\partial}{\partial y}&\frac{\partial}{\partial z}\\x+y&-x+y&-2z\end{array}\right])\\=i(\frac{\partial}{\partial y}(-2z)-\frac{\partial}{\partial z}(-x+y))-j(\frac{\partial}{\partial x}(-2z)-\frac{\partial}{\partial z}(x+y))+k(\frac{\partial}{\partial x}(-x+y)-\frac{\partial}{\partial y}(x+y))=0i-0j-2k=(0,0,-2)[/tex]b) [tex]F=(2y,2x+3z,3y)[/tex]The divergence of F is [tex]\nabla F=\frac{\partial}{\partial x}(2y)+\frac{\partial}{\partial y}(2x+3z)+\frac{\partial}{\partial z} (3y)=0+0+0=0[/tex]The curl of F is[tex]\nabla\times F=det(\left[\begin{array}{ccc}i&j&k\\\frac{\partial}{\partial x}&\frac{\partial}{\partial y}&\frac{\partial}{\partial z}\\2y&2x+3z&3y\end{array}\right])\\=i(\frac{\partial}{\partial y}(3y)-\frac{\partial}{\partial z}(2x+3z))-j(\frac{\partial}{\partial x}(3y)-\frac{\partial}{\partial z}(2y))+k(\frac{\partial}{\partial x}(2x+-3z)-\frac{\partial}{\partial y}(2y))=0i-0j+0k=(0,0,0)[/tex]Since the curl of F is 0 the we will try find f such that the gradient of f be F.Since [tex]f_x=2y[/tex], [tex]f=2xy+g(y,z)[/tex]. Since [tex]f_y=2x+3z[/tex], [tex]2x+3z=f_y=2x+g_y(y,z)\\3z=g_y[/tex]. Since [tex]f_z=3y[/tex] and [tex]f_z=3y+h'(z)[/tex], then [tex]h'(z)=0[/tex]. Thins means that [tex]h(z)=C, C\in\mathbb{R}[/tex]Therefore,[tex]f=2xy+3yz+C, C\in\mathbb{R}[/tex].c) [tex]H=(x^2-z^2,2,2xz)[/tex]The divergence of F is [tex]\nabla F=\frac{\partial}{\partial x}(x^2-z^2)+\frac{\partial}{\partial y}(2)+\frac{\partial}{\partial z} (2xz)=2x+0+2x=4x[/tex]The curl of F is[tex]\nabla\times F=det(\left[\begin{array}{ccc}i&j&k\\\frac{\partial}{\partial x}&\frac{\partial}{\partial y}&\frac{\partial}{\partial z}\\x^2-z^2&2&2xz\end{array}\right])\\=i(\frac{\partial}{\partial y}(2xz)-\frac{\partial}{\partial z}(2))-j(\frac{\partial}{\partial x}(2xz)-\frac{\partial}{\partial z}(x^2-z^2))+k(\frac{\partial}{\partial x}(2)-\frac{\partial}{\partial y}(x^2-z^2))=0i-(2z-(-2z))j-0k=(0,-4z,0)[/tex]