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Find the divergence of vector function $\bar{A}$ = x$^{2}$ $\bar i$+x$^{2}$y$^{2}$ $\bar j$+24x$^{2}$y$^{2}$z$^{3}$ $\bar k$
divergence of vector field • 291  views
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Solution:

$\begin{aligned} \bar{A} &= x^{2} \bar i + x^{2}y^{2} \bar j + 24 x^{2} y^{2} z^{3} \bar k \\ \bar{\nabla} . \bar{A} &= \bigg( \frac{\partial}{\partial x}. \bar{i} + \frac{\partial}{\partial y}. \bar{j} + \frac{\partial}{\partial z}. \bar{k} \bigg) . \bigg(x^2 \bar{i} + x^2y^2\bar j + 24 x^2y^2z^3 \bar k \bigg) \\ &= \frac{\partial}{\partial x}(x^2) + \frac{\partial}{\partial y} (x^2y^2) + \frac{\partial}{\partial z} (24x^2y^2z^3) \\ &= 2x + 2yx^2 + 24 x^2y^2(3z^2) \\ &= 2x + 2yx^2 + 72 x^2y^2 z^2 \\ &= 2x \big[ 1+yx+36xy^2z^2 \big] \end{aligned}$

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