If-f-x-2-zi-2y-3-z-2-j-xy-2-zk-Find-div-f-curl-f-at-1-1-1-

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Question Number 27 by user1 last updated on 25/Jan/15

$$\mathrm{If}\mathbf{f}=x^{2}z\mathbf{i}-2y^3{z}^2\mathbf{j}+{xy}^{2}z\mathbf{k}.\mathrm{Find}div\mathbf{f},curl\mathbf{f},$$$$at(1,-1,1).$$

Answered by user1 last updated on 03/Nov/14

$$\mathrm{div}\mathbf{f}=\mathrm{\Sigma }\mathbf{i}\cdot \frac{\partial \mathbf{f}}{\partial x}=▽\cdot \mathbf{f}$$$$=(\mathbf{i}\frac{\partial }{\partial x}+\mathbf{j}\frac{\partial }{\partial y}+\mathbf{k}\frac{\partial }{\partial z})\cdot (x^{2}z\mathbf{i}-2y^3{z}^2\mathbf{j}+{xy}^{2}z\mathbf{k})$$$$=\frac{\partial }{\partial x}\left(x^{2}z\right)-\frac{\partial }{\partial y}\left(2y^3{z}^2\right)+\frac{\partial }{\partial z}\left({xy}^{2}z\right)$$$$=2xz-6y^2{z}^2+{xy}^2$$$$=-3\mathrm{at}(1,-1,1)$$$$\mathrm{curl}\mathbf{f}=▽\times \mathbf{f}=\left|\begin{array}{ccc}\mathbf{i}& \mathbf{j}& \mathbf{k}\\ \frac{\partial }{\partial x}& \frac{\partial }{\partial y}& \frac{\partial }{\partial z}\\ x^{2}z& -2y^3{z}^2& {xy}^{2}z\end{array}\right|$$$$=\mathbf{i}(2xyz+4y^{3}z)+\mathbf{j}(x^2-y^{2}z)+\mathbf{k}(0-0)$$$$=-6\mathbf{i}\mathrm{at}(1,-1,1)$$

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