. Condition for an Irrotational FieldA vector field is irrotational if its curl is zero: (nabla times vec{F} = vec{0}).2. Calculating the CurlThe curl is calculated using the following determinant:(nabla times vec{F}=left|begin{matrix}^{i}&^{j}&^{k}\ frac{partial }{partial x}&frac{partial }{partial y}&frac{partial }{partial z}\ axy+z^{3}&3x^{2}-z&bxz^{2}-yend{matrix}right|)Expanding this, we solve for each component:(^{i}) component:(frac{partial}{partial y}(bxz^2 – y) – frac{partial}{partial z}(3x^2 – z) = (-1) – (-1) = 0)(^{j}) component:(-left[frac{partial }{partial x}(bxz^{2}-y)-frac{partial }{partial z}(axy+z^{3})right]=-left[bz^{2}-3z^{2}right]=-z^{2}(b-3))(^{k}) component:(frac{partial}{partial x}(3x^2 – z) – frac{partial}{partial y}(axy + z^3) = 6x – ax = x(6 – a))3. Solving for constantsFor (nabla times vec{F} = vec{0}) to hold true, each component must be zero:From the (^{j}) component:(-z^2(b – 3) = 0 implies b – 3 = 0 implies mathbf{b = 3})From the (^{k}) component:(x(6 – a) = 0 implies 6 – a = 0 implies mathbf{a = 6})Final AnswerThe constants are:(a = 6) and (b = 3) Dive deeper in AI Mode Footer linksResults are personalised-Try without personalisation
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