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

User Rjss
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1 Answer

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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}

User Oussama
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