Answer to Question #225911 in Chemical Engineering for Lokika

Question

If V = x²z²i-2y²z²j+ xy²zk then Curl V at (3,2,1) is

A) 20i + 14k

B)2i + 14k

C)14j+20k

D) 20i + 14j

Answer 1

By definition, $curl(x^2z^2,2y^2z^2,xy^2z)=∇×(x^2z^2,2y^2z^2,xy^2z)$ or, equivalently,

$∇×(x^2z^2,2y^2z^2,xy^2z)= \begin{vmatrix} i & j & k\\ \frac{\partial }{\partial x} & \frac{\partial }{\partial y} & \frac{\partial }{\partial z}\\ x^2z^2 & 2y^2z^2 & xy^2z \end{vmatrix}$

$curl(x^2z^2,2y^2z^2,xy^2z) =(\frac{∂}{∂y}(xy^2z)−\frac{∂}{∂z}(2y^2z^2),\frac{∂}{∂z}(x^2z^2)−\frac{∂}{∂x}(xy^2z),\frac{∂}{∂x}(2y^2z^2)−\frac{∂}{∂y}(x^2z^2))\\$

Now, just plug in the found partial derivatives to get the curl:

$curl(x^2z^2,2y^2z^2,xy^2z)=(2yz(x−2y),z(2x^2−y^2),0).$

Finally, find the curl at the specific point.

$(curl(x^2z^2,2y^2z^2,xy^2z))|((x_0,y_0,z_0)=(3,2,1))=(−4,14,0)$

The answer should be -4i + 14j