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Question

Find div F\mathbf{F} and curl F\mathbf{F} for the vector field F\mathbf{F} at the given point.

F(x,y,z)=(exsiny)i+(excosy)j+k at (1,3,2)\begin{array}{l}{\mathbf{F}(x, y, z)=\left(e^{-x} \sin y\right) \mathbf{i}+\left(e^{-x} \cos y\right) \mathbf{j}+\mathbf{k} \text { at }} \\ {(1,3,-2)}\end{array}

Solution

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Answered 2 years ago
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Here F(x,y)=(exsiny)i^+(excosy)j^+k^F(x,y)=( e^{-x}\sin{y})\hat{i}+(e^{-x}\cos{y})\hat{j}+\hat{k} So

divF=(exsiny)x+(excosy)y+(1)z=exsinyexsiny=2exsiny\begin{align*} div F &= \frac{\partial (e^{-x}\sin{y})}{\partial x}+\frac{\partial(e^{-x}\cos{y})}{\partial y}+\frac{\partial(1)}{\partial z}\\ &=-e^{-x}\sin{y}-e^{-x}\sin{y}\\ &=-2e^{-x}\sin{y}\\ \end{align*}

Hence at the point (1,3,2)(1,3,-2) divergence is 2e1sin3-2e^{-1}\sin{3}.

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