Answer to Question #158034 in Calculus for Padma vasa

Question

If a=(xz^(3))i-(2x^(2)yz)j+(2yz^(4))k find curl a at (1 -1 1)

Answer 1

$$\vec a=xz^3\vec i-2x^2yz\vec j+2yz^4\vec k$$

$$\text{curl } \vec a=\nabla\times\vec a=\begin{vmatrix} \vec i & \vec j & \vec k \\ \\ \dfrac{\partial}{\partial x} & \dfrac{\partial}{\partial y} & \dfrac{\partial}{\partial z} \\ \\ xz^3& -2x^2yz & 2yz^4 \end{vmatrix}$$

$$=\vec i(2z^4-(-2x^2y)-\vec j(0-3xz^2)+\vec k(-4xyz-0)$$

$$=(2z^4+2x^2y)\vec i+3xz^2\vec j-4xyz\vec k$$

$$\text{curl } \vec a\big|_{(1,-1,1)}=3\vec j+4\vec k$$