Answer to Question #144852 in Calculus for SULEIMAN ABDULAKIM

Question #144852
Determine the \\(curl F\\) at the point (2, 0, 3) given that \\(F=xz i+(2x^2-y)j-yz^2 k\\).
1
Expert's answer
2020-11-17T16:21:59-0500
curlF=×F=ijkxyzxz2x2yyz2=\text{curl}\vec{F}=\nabla\times\vec{F}=\begin{vmatrix} \vec{i} & \vec{j} & \vec{k} \\ \dfrac{\partial}{\partial x} & \dfrac{\partial}{\partial y} & \dfrac{\partial}{\partial z} \\ xz & 2x^2-y & -yz^2 \end{vmatrix}=


=i(z20)j(0x)+k(4x0)==\vec{i}(-z^2-0)-\vec{j}(0-x)+\vec{k}(4x-0)=

=z2i+xj+4xk=-z^2\vec{i}+x\vec{j}+4x\vec{k}

M(2,0,3)M(2,0,3)


curlFM=9i+2j+8k\text{curl}\vec{F}_M=-9\vec{i}+2\vec{j}+8\vec{k}


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