MATH 2110Q Lecture Notes - Lecture 18: Multiple Integral, Unit Vector

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26 Oct 2018
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Math 2110q lecture 18 line integrals of scalar functions: warm up on arc length: the vector function (cid:1870) (t) = (cid:1855)(cid:1867)(cid:1871)(cid:4666)(cid:884)(cid:1872)(cid:4667),(cid:2869)(cid:3047)(cid:3119) defines a curve or path in, = |(cid:1870) (cid:4666)(cid:1872)(cid:4667)|(cid:1856)(cid:1872) space. Integrating f over c gives area of a curtain": approximate area by dividing into pieces where c is the portion of x + y = 25 where y 0 traced counterclockwise (cid:3030) Example: step 1: parametrize c, the line integral of a scalar function f (x, y) over c: (cid:1870) (t), (cid:1853) (cid:1872) (cid:1854) is (cid:4666)(cid:1876),(cid:1877)(cid:4667)(cid:1856) Compute (cid:884)(cid:1876)(cid:1856) (cid:3030: (cid:1870) (cid:4666)(cid:1872)(cid:4667)=(cid:1766)(cid:1871)(cid:1855)(cid:1867)(cid:1871)(cid:1872),(cid:1871)(cid:1871)(cid:1866)(cid:1872)(cid:1767, 0 t (to give where y 0, step 2: find |(cid:1870) (cid:1872)| = |(cid:1766) (cid:1871)(cid:1855)(cid:1867)(cid:1871)(cid:1872),(cid:1871)(cid:1871)(cid:1866)(cid:1872)(cid:1767), = (cid:884)5(cid:1871)(cid:1866)(cid:2870)(cid:1872)+(cid:884)5(cid:1855)(cid:1867)(cid:1871)(cid:2870)(cid:1872, = (cid:884)5(cid:4666)(cid:1871)(cid:1866)(cid:2870)(cid:1872)+(cid:1855)(cid:1867)(cid:1871)(cid:2870)(cid:1872)(cid:4667) = 5. = (cid:884)(cid:1876) 5(cid:1856)(cid:1872: (cid:884)(cid:1876)(cid:1856) (cid:2868, we need to write 2x in terms of t, x = 5cost comes from our parametrization. = ((cid:1870) (cid:4666)(cid:1872)(cid:4667))|(cid:1870) (cid:4666)(cid:1872)(cid:4667)|(cid:1856)(cid:1872: the line integral of a scalar function f (x,y) is (cid:4666)(cid:1876),(cid:1877)(cid:4667)(cid:1856) (cid:3029)(cid:3028)

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