[amp_mcq option1=”2z cos z2″ option2=”sin xy + 2z cos z2″ option3=”x sin xy – cos z” option4=”None of these” correct=”option4″]<\/p>\n
The divergence of a vector field $\\overrightarrow{V}(x, y, z)$ is defined as:<\/p>\n
$$\\nabla \\cdot \\overrightarrow{V} = \\frac{\\partial P}{\\partial x} + \\frac{\\partial Q}{\\partial y} + \\frac{\\partial R}{\\partial z}$$<\/p>\n
where $P$, $Q$, and $R$ are the $x$, $y$, and $z$ components of $\\overrightarrow{V}$, respectively.<\/p>\n
In this case, we have:<\/p>\n
$$\\overrightarrow{V}(x, y, z) = -(x \\cos xy + y) \\hat{\\imath} + (y \\cos xy) \\hat{\\jmath} + \\left[ (\\sin z^2) + x^2 + y^2 \\right] \\hat{k}$$<\/p>\n
Therefore, the divergence of $\\overrightarrow{V}$ is:<\/p>\n
$$\\begin{align} Here is a brief explanation of each option:<\/p>\n [amp_mcq option1=”2z cos z2″ option2=”sin xy + 2z cos z2″ option3=”x sin xy – cos z” option4=”None of these” correct=”option4″]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[690],"tags":[],"class_list":["post-20249","post","type-post","status-publish","format-standard","hentry","category-calculus","no-featured-image-padding"],"yoast_head":"\n
\n\\nabla \\cdot \\overrightarrow{V} &= \\frac{\\partial}{\\partial x} \\left[ -(x \\cos xy + y) \\right] + \\frac{\\partial}{\\partial y} \\left[ (y \\cos xy) \\right] + \\frac{\\partial}{\\partial z} \\left[ (\\sin z^2) + x^2 + y^2 \\right] \\
\n&= -\\cos xy – \\frac{\\partial}{\\partial y} \\left[ y \\cos xy \\right] + 2z \\cos z^2 \\
\n&= -\\cos xy – y \\sin xy + 2z \\cos z^2 \\
\n&= \\boxed{-\\cos xy + y \\sin xy + 2z \\cos z^2}.
\n\\end{align<\/em>}$$<\/p>\n
\n\n