[amp_mcq option1=”0″ option2=”\\[\\frac{2}{{\\sqrt 3 }}\\]” option3=”1″ option4=”\\[2\\sqrt 3 \\]” correct=”option3″]<\/p>\n
The correct answer is $\\boxed{0}$.<\/p>\n
The circulation of a vector field $\\overrightarrow{A}$ around a closed path $\\Gamma$ is given by the following integral:<\/p>\n
$$\\oint_\\Gamma \\overrightarrow{A} \\cdot d\\overrightarrow{l}$$<\/p>\n
In this case, the vector field is $\\overrightarrow{A} = xy\\hat{\\imath} + x^2\\hat{\\jmath}$ and the path is a triangle with vertices $(0,0)$, $(1,0)$, and $(1,1)$. The line integral over this path can be evaluated as follows:<\/p>\n
$$\\oint_\\Gamma \\overrightarrow{A} \\cdot d\\overrightarrow{l} = \\int_0^1 xy\\,dx + \\int_0^1 x^2\\,dy$$<\/p>\n
The first integral can be evaluated using the substitution $u = x$, which gives:<\/p>\n
$$\\int_0^1 xy\\,dx = \\int_0^1 u^2\\,du = \\frac{1}{3}$$<\/p>\n
The second integral can be evaluated using the substitution $v = y$, which gives:<\/p>\n
$$\\int_0^1 x^2\\,dy = \\int_0^1 v^2\\,dv = \\frac{1}{3}$$<\/p>\n
Therefore, the total line integral is:<\/p>\n
$$\\oint_\\Gamma \\overrightarrow{A} \\cdot d\\overrightarrow{l} = \\frac{1}{3} + \\frac{1}{3} = 0$$<\/p>\n
The other options are incorrect because they do not represent the correct value of the line integral.<\/p>\n","protected":false},"excerpt":{"rendered":"
[amp_mcq option1=”0″ option2=”\\[\\frac{2}{{\\sqrt 3 }}\\]” option3=”1″ option4=”\\[2\\sqrt 3 \\]” correct=”option3″]<\/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-20208","post","type-post","status-publish","format-standard","hentry","category-calculus","no-featured-image-padding"],"yoast_head":"\n