Question

Let
A=x^2yi−2xzj+2yzk
, find Curl curl A.

3j+4k

2x+2)k

(2x+2)j

3j−4k

Answer 1

Answer on Question #56126 – Math – Vector Calculus

Let

A=x^2yi-2xzj+2yzk

, find Curl curl A.

3j+4k

2x+2)k

(2x+2)j

3j-4k

Solution

$$\begin{array}{l} \curl A = \left| \begin{array}{ccc} i & j & k \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ x^2 y & -2xz & 2yz \end{array} \right| = i \left( \frac{\partial}{\partial y}(2yz) - \frac{\partial}{\partial z}(-2xz) \right) - j \left( \frac{\partial}{\partial x}(2yz) - \frac{\partial}{\partial z}(-2xz) \right) \\ \frac{\partial}{\partial z}(x^2 y) + k \left( \frac{\partial}{\partial x}(-2xz) - \frac{\partial}{\partial y}(x^2 y) \right) = (2z + 2x)i - (0 - 0)j + \\ + (-2z - x^2)k = < 2z + 2x, 0, -2z - x^2 > \end{array}$$ $$\begin{array}{l} \text{curl curl } A = \left| \begin{array}{ccc} i & j & k \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ 2z + 2x & 0 & -2z - x^2 \end{array} \right| \\ = i \left( \frac{\partial}{\partial y} (-2z - x^2) - \frac{\partial}{\partial z} (0) \right) \\ - j \left( \frac{\partial}{\partial x} (-2z - x^2) - \frac{\partial}{\partial z} (2z + 2x) \right) \\ + k \left( \frac{\partial}{\partial x} (0) - \frac{\partial}{\partial y} (2z + 2x) \right) = 0 \cdot i - (-2x - 2)j + 0 \cdot k = \\ = (2x + 2)j = < 0, 2x + 2, 0 > \end{array}$$

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