# Find curl F for the vector field at the given point. F(x, y, z) = x2zi − 2xzj + yzk; (7, −9, 3)?

## I'm missing something. I got 17i-14j+6k for my answer which was wrong.

Jul 29, 2018

#### Answer:

The curl is $= 21 \vec{i} + 49 \vec{j} - 14 \vec{k}$

#### Explanation:

$F \left(x , y , z\right) = {x}^{2} z \vec{i} - 2 x z \vec{j} + y z \vec{k}$

The curl of a vector field is defined as

$c u r l F = \nabla \times F$

$= | \left(\vec{i} , \vec{j} , \vec{k}\right) , \left(\frac{\partial}{\partial x} , \frac{\partial}{\partial y} , \frac{\partial}{\partial z}\right) , \left({x}^{2} z , - 2 x z , y z\right) |$

$= \vec{i} | \left(\frac{\partial}{\partial y} , \frac{\partial}{\partial z}\right) , \left(- 2 x z , y z\right) | - \vec{j} | \left(\frac{\partial}{\partial x} , \frac{\partial}{\partial z}\right) , \left({x}^{2} z , y z\right) | + \vec{k} | \left(\frac{\partial}{\partial x} , \frac{\partial}{\partial y}\right) , \left({x}^{2} z , - 2 x z\right) |$

$= \vec{i} \left(z + 2 x\right) - \vec{j} \left(0 - {x}^{2}\right) + \vec{k} \left(- 2 z - 0\right)$

$= \vec{i} \left(z + 2 x\right) + \vec{j} {x}^{2} - \vec{k} 2 z$

At the point $\left(7 , - 9 , 7\right)$, the curl is

$= \vec{i} 21 + \vec{j} 49 - \vec{k} 14$