[amp_mcq option1=”7″ option2=”4″ option3=”3″ option4=”0″ correct=”option1″]<\/p>\n
The divergence of a vector field is a measure of how much the vector field spreads out or converges at a point. It is calculated by taking the sum of the partial derivatives of the three components of the vector field with respect to their respective coordinates.<\/p>\n
In this case, the vector field is given by<\/p>\n
$$\\mathbf{F}(x, y, z) = 3xz\\hat{\\imath} + 2xy\\hat{\\jmath} – yz^2\\hat{k}$$<\/p>\n
The divergence of $\\mathbf{F}$ is then given by<\/p>\n
$$\\nabla \\cdot \\mathbf{F} = \\frac{\\partial}{\\partial x} (3xz) + \\frac{\\partial}{\\partial y} (2xy) + \\frac{\\partial}{\\partial z} \\left(-yz^2\\right) = 3x + 2y – 2yz$$<\/p>\n
Evaluating this at the point $(1, 1, 1)$ gives<\/p>\n
$$\\nabla \\cdot \\mathbf{F}(1, 1, 1) = 3 + 2 – 2 = 3$$<\/p>\n
Therefore, the divergence of the vector field $\\mathbf{F}$ at the point $(1, 1, 1)$ is $\\boxed{3}$.<\/p>\n
The other options are incorrect because they do not represent the correct value of the divergence of the vector field $\\mathbf{F}$ at the point $(1, 1, 1)$.<\/p>\n","protected":false},"excerpt":{"rendered":"
[amp_mcq option1=”7″ option2=”4″ option3=”3″ option4=”0″ correct=”option1″]<\/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-20204","post","type-post","status-publish","format-standard","hentry","category-calculus","no-featured-image-padding"],"yoast_head":"\n