[amp_mcq option1=”0″ option2=”2″ option3=”4″ option4=”6″ correct=”option1″]<\/p>\n
The correct answer is $\\boxed{\\text{B) }2}$.<\/p>\n
The line integral is a way of calculating the work done by a force along a curve. In this case, the force is given by the vector field $\\mathbf{F}(x, y, z) = (2xy^2, 2x^2y, 0)$, and the curve is the line segment from the origin to the point $(1, 1, 1)$.<\/p>\n
To calculate the line integral, we first need to find a parametrization of the curve. A convenient parametrization is $x = t$, $y = t$, and $z = t$, for $0 \\leq t \\leq 1$. This parametrization tells us that the point $(x, y, z)$ is on the curve when $t = 0$, it is at the origin, and it is at the point $(1, 1, 1)$ when $t = 1$.<\/p>\n
Now we can calculate the line integral:<\/p>\n
$$\\int\\limits_{\\text{c}} {\\left( {2{\\text{x}}{{\\text{y}}^2}{\\text{dx}} + 2{{\\text{x}}^2}{\\text{ydy}} + {\\text{dz}}} \\right)} = \\int\\limits_0^1 {\\left( {2t^2} \\right)} dt = 2.$$<\/p>\n
Therefore, the value of the line integral is $\\boxed{\\text{B) }2}$.<\/p>\n
The other options are incorrect because they do not give the correct value of the line integral.<\/p>\n","protected":false},"excerpt":{"rendered":"
[amp_mcq option1=”0″ option2=”2″ option3=”4″ option4=”6″ 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-20253","post","type-post","status-publish","format-standard","hentry","category-calculus","no-featured-image-padding"],"yoast_head":"\n