Answer to Question #210674 in Electricity and Magnetism for Hal

Question

1) If a vector field A(x,y,z)=xz³e_x - 2x²yze_y+2yz⁴e_z , calculate the curl ∇ x A at point M(1,-1,-1).

Answer 1

Gives

$$A(x,y,z)=xz^3\hat{e_x}+2x^2yz\hat{e_y}+2yz^4\hat{e_z}$$

PoinM(1,-1,-1)

$$\nabla\times A=\begin{bmatrix} \hat{e_x}&\hat{e_y} & \hat{e_z}\\ \frac{d}{dx}&\frac{d}{dx}&\frac{d}{dx}\\ xz^3&-2x^2yz&2yz^4 \end{bmatrix}$$

$$\hat{e_x}(\frac{d}{dy}(2yz^4)+\frac{d}{dz}(2x^2yz))-\hat{e_y}(\frac{d}{dx}(2yz^4)-\frac{d}{dx}(xz^3))+\hat{e_z}(\frac{d}{dx}(-2x^2yz)-\frac{d}{dy}(xz^3))$$

$$\nabla\times A=\hat{e_x}(2z^4+2x^2y)+\hat{e_y}(3xz^2)-\hat{e_z}(4xyz)$$