Question

Find \(\operatorname{curl}(\mathbf{F} \times \mathbf{G})\) $$ \begin{array}{l} \mathbf{F}(x, y, z)=\mathbf{i}+2 x \mathbf{j}+3 y \mathbf{k} \\ \mathbf{G}(x, y, z)=x \mathbf{i}-y \mathbf{j}+z \mathbf{k} \end{array} $$

Short answer

The curl of the cross product of the given vectors \(\mathbf{F}\) and \(\mathbf{G}\) is -\(\mathbf{i}\) + 3\(\mathbf{j}\) - 2\(\mathbf{k}\)

Step-by-step solution

Compute the cross product \(\mathbf{F} \times \mathbf{G}\)

To compute the cross product between the vectors, use the determinant formula: \[\mathbf{H} = \mathbf{F} \times \mathbf{G} = \left| \begin{array}{ccc} \mathbf{i} & \mathbf{j} & \mathbf{k} \ 1 & 2x & 3y \ x & -y & z \end{array} \right| = \mathbf{i}(2x*z-(-y*3y))-\mathbf{j}(1*z-x*3y)+\mathbf{k}(1*(-y)-2x*x)\]This simplifies to:\[\mathbf{H} = \mathbf{i}(2xz+3y^2)+\mathbf{j}(z-3xy)+\mathbf{k}(-y-2x^2)\]

Compute the curl of \(\mathbf{H}\)

Find the curl of the resulting vector \(\mathbf{H}\) from the cross product by using the determinant method again:\[curl(\mathbf{H}) = \left| \begin{array}{ccc} \mathbf{i} & \mathbf{j} & \mathbf{k} \ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \ 2xz+3y^2 & z-3xy & -y-2x^2 \end{array} \right|\]This becomes:\[curl(\mathbf{H}) = \mathbf{i}\left( \frac{\partial (-y-2x^2)}{\partial y} - \frac{\partial (z-3xy)}{\partial z} \right) - \mathbf{j}\left( \frac{\partial (-y-2x^2)}{\partial x} - \frac{\partial (2xz+3y^2)}{\partial z} \right) + \mathbf{k}\left( \frac{\partial (z-3xy)}{\partial x} - \frac{\partial (2xz+3y^2)}{\partial y} \right)\]After differentiation, we get:\[curl(\mathbf{H}) = -\mathbf{i} + 3\mathbf{j} - 2\mathbf{k}\]

Conclusion

Therefore, the curl of the cross product of the given vectors is -\mathbf{i} + 3\mathbf{j} - 2\mathbf{k}