[amp_mcq option1=”\\[ – 2\\pi \\]” option2=”\\[ – \\pi \\]” option3=”\\[\\pi \\]” option4=”\\[2\\pi \\]” correct=”option3″]<\/p>\n
The correct answer is $\\boxed{\\pi}$.<\/p>\n
The line integral of a vector field $F$ over a curve $C$ is given by the formula<\/p>\n
$$\\oint_C F \\cdot dr = \\int_a^b F(x, y, z) \\cdot (dx, dy, dz)$$<\/p>\n
where $a$ and $b$ are the endpoints of $C$.<\/p>\n
In this case, the vector field $F = yzi$ and the curve $C$ is the circle $x^2 + y^2 = 1$ at $z = 1$. To evaluate the line integral, we can use the parameterization<\/p>\n
$$x = \\cos t, \\quad y = \\sin t, \\quad z = 1$$<\/p>\n
for $0 \\leq t \\leq 2\\pi$. Then,<\/p>\n
$$F(x, y, z) = yzi = \\sin t \\cdot 1 \\cdot i = \\sin t i$$<\/p>\n
and<\/p>\n
$$dr = dx \\times dy = -\\sin t \\, dt \\times \\cos t \\, dt = -\\sin^2 t \\, dt$$<\/p>\n
Therefore, the line integral is<\/p>\n
$$\\oint_C F \\cdot dr = \\int_0^{2\\pi} -\\sin^2 t \\, dt = \\int_0^{2\\pi} (1 – \\cos 2t) \\, dt = \\pi$$<\/p>\n
The other options are incorrect because they do not take into account the direction of the curve $C$. The curve $C$ is a circle in the counter-clockwise direction, so the line integral must be positive.<\/p>\n","protected":false},"excerpt":{"rendered":"
[amp_mcq option1=”\\[ – 2\\pi \\]” option2=”\\[ – \\pi \\]” option3=”\\[\\pi \\]” option4=”\\[2\\pi \\]” 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-20238","post","type-post","status-publish","format-standard","hentry","category-calculus","no-featured-image-padding"],"yoast_head":"\n