Question

Consider the given vector field. f(x, y, z) = 8xyezi + yzexk (a) find the curl of the vector field. curl f = $$zexi+(8xyez−yzex)j−8xezk correct: your answer is correct. (b) find the divergence of the vector field. div f = $$8yez+yex correct: your answer is correct.

Answer 1

Answer:

(a)

$curl\ f=ze^xi+(8xye^z-yze^x)j-8xe^zk$

(b)

$div\ f=8ye^z+ye^x$

Step-by-step explanation:

We are given a vector function f(x,y,z) as:

$f(x,y,z)=8xye^zi+yze^xk$

where,

$f=f_1i+f_2j+f_3k$

(a)

Hence, the curl f is calculated as:

$curl\ f=\begin{vmatrix}i &j &k \\ d/dx &d/dy &d/dz \\ f_1 &f_2 &f_3 \end{vmatrix}$

Hence, we have:

$f_1=8xye^z,\ f_2=0,\ f_3=yze^x$

Hence,

$curl\ f=\begin{vmatrix}i &j &k \\ d/dx &d/dy &d/dz \\ 8xye^z &0 &yze^x \end{vmatrix}$

Hence,

$curl\ f=\dfrac{d}{dy}(yze^x)i-j(\dfrac{d}{dx}(yze^x)-\dfrac{d}{dz}(8xye^z))+k(\dfrac{d}{dx}(0)-\dfrac{d}{dy}(8xye^z))\\\\\\curl\ f=ze^xi-j(yze^x-8xye^z))+k(0-8xe^z)\\\\\\curl\ f=ze^xi+(8xye^z-yze^x)j-8xe^zk$

(b)

Divergence f is calculated as:

$div f=\dfrac{d}{dx}(f_1)+\dfrac{d}{dy}(f_2)+\dfrac{d}{dz}(f_3)\\\\\\div\ f=\dfrac{d}{dx}(8xye^z)+\dfrac{d}{dy}(0)+\dfrac{d}{dz}(yze^x)\\\\\\div\ f=8ye^z+ye^x$

Answer 2

Let check your computations:
Recall the formula: 
$Curl(f)=(R_y-Q_z)i+(P_z-R_x)j+(Q_x-P_y)k\\\text{Wherein:}\\P=8xye^z\\Q=0\\R=yze^x\\\text{Compute the partial derivative:}\\R_y-Q_z=ze^x-0=ze^x\\P_z-R_x=8xye^z-yze^x\\Q_x-P_y=-8xe^z\\\textbf{Conclusion:}\\Curl(f)=ze^xi+(8xye^z-yze^x)j-8xe^zk\\\textbf{Divergence:}\\\nabla f=P_x+Q_y+R_z=8ye^z+ye^x$