Question

If vector A=xz³î-2x²yj+2yz⁴k find CURL A at the point [1, -1, 1]​

Answer 1

Given that

`\vec A = xz^3 \, \vec\imath - 2x^2 y \, \vec\jmath + 2yz^4 \, \vec k`

its curl is

`\displaystyle \\abla*\vec A = \left((\partial\left(2yz^4\right))/(\partial y) - (\partial\left(-2x^2y\right))/(\partial z)\right) \, \vec\imath - \left((\partial\left(2yz^4\right))/(\partial x) - (\partial\left(xz^3\right))/(\partial z)\right) \, \vec\jmath \\ ~~~~~~~~~~~~ + \left((\partial\left(-2x^2y\right))/(\partial x) - (\partial\left(xz^3\right))/(\partial y)\right) \, \vec k`

`\\abla*\vec A = 2z^4 \, \vec\imath + 3xz^2 \, \vec\jmath - 4xy \, \vec k`

so that at the point (1, -1, 1), the curl is

`\\abla*\vec A \bigg|_((x,y,z)=(1,-1,1)) = \boxed{2 \, \vec\imath + 3 \, \vec\jmath + 4 \, \vec k}`