Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 17 - Line and Surface Integrals - 17.1 Vector Fields - Exercises - Page 919: 27

Answer

\begin{align*} \mathbf{div(F)}&=0 \\ \text{curl} \mathbf{(F)}&=\left( 1-3z^2 \right)i - \left( 2x-1 \right)j + \left(1+2y \right) k \end{align*}

Work Step by Step

Given $$\mathbf{F}=\left\langle z-y^{2}, x+z^{3}, y+x^{2}\right\rangle$$ Since \begin{align*} \mathbf{div(F)}&=\frac{\partial \mathbf{F}_x}{\partial x}+\frac{\partial \mathbf{F}_y }{\partial y}+\frac{\partial \mathbf{F}_z}{\partial z}\\ &=\frac{\partial ( z-y^{2})}{\partial x}+\frac{\partial (x+z^{3}) }{\partial y}+\frac{\partial ( y+x^{2})}{\partial z}\\ &= 0 \end{align*} and \begin{align*} \text{curl} \mathbf{(F)}&=\begin{array}{|||} i&j &k\\ \frac{\partial}{\partial x}& \frac{\partial}{\partial y}& \frac{\partial}{\partial z}\\ z-y^{2}& x+z^{3}& y+x^{2} \end{array} \\ &= \left( \frac{\partial( y+x^{2})}{\partial y}- \frac{\partial( x+z^{3})}{\partial z}\right)i - \left( \frac{\partial( y+x^{2} )}{\partial x}- \frac{\partial( z-y^{2})}{\partial z}\right)j + \left( \frac{\partial( x+z^{3} )}{\partial x}- \frac{\partial( z-y^{2} )}{\partial y}\right) k\\ &= \left( 1-3z^2 \right)i - \left( 2x-1 \right)j + \left(1+2y \right) k\\ \end{align*}
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