[amp_mcq option1=”\\[\\frac{1}{2}\\]” option2=”1″ option3=”\\[\\frac{3}{2}\\]” option4=”\\[\\frac{5}{2}\\]” correct=”option3″]<\/p>\n
The correct answer is $\\boxed{\\frac{1}{2}}$.<\/p>\n
Green’s theorem states that the line integral of a vector field over a simple closed curve $C$ is equal to the double integral of the curl of the vector field over the region enclosed by $C$. In this case, the vector field is $F(x, y) = xy \\hat{\\imath} – y^2 \\hat{\\jmath}$ and the curve $C$ is the square cut from the first quadrant by the lines $x = 1$ and $y = 1$.<\/p>\n
The curl of $F$ is $F_y = x$ and $F_x = -y$. Therefore, the double integral of the curl of $F$ over the region enclosed by $C$ is<\/p>\n
$$\\iint_R (-y) \\, dx \\, dy = \\int_0^1 \\int_0^1 -y \\, dx \\, dy = \\int_0^1 \\left[ -\\frac{y^2}{2} \\right]_0^1 dy = \\frac{1}{2}.$$<\/p>\n
By Green’s theorem, the line integral of $F$ over $C$ is also equal to $\\frac{1}{2}$.<\/p>\n
The other options are incorrect because they do not take into account the orientation of the curve $C$. The curve $C$ is positively oriented, which means that we are integrating in the counterclockwise direction. If the curve $C$ were negatively oriented, the line integral would be equal to $-\\frac{1}{2}$.<\/p>\n","protected":false},"excerpt":{"rendered":"
[amp_mcq option1=”\\[\\frac{1}{2}\\]” option2=”1″ option3=”\\[\\frac{3}{2}\\]” option4=”\\[\\frac{5}{2}\\]” 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-20213","post","type-post","status-publish","format-standard","hentry","category-calculus","no-featured-image-padding"],"yoast_head":"\n