MATH215 Lecture Notes - Lecture 2: Polar Regions Of Earth

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violethare11

13 Mar 2017

School

University of Alberta

Department

Mathematics

Course

MATH215

Professor

Alexander Litvak

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Document Summary

Double integrals: let f be a continuous function over the rectangle r = {(x, y) : a x b, c y d}. Zzq f (x, y) da = z b a z g2(x) g1(x) f (x, y) dy dx. (b) if q = {(x, y) : c y d, h1(y) x h2(y)} is x-simple, then. Z b g1(x) f (x, y) dy dx = z d c z h2(y) h1(y) f (x, y) dx dy: let f be a continuous function on a polar region of the form. Q = {(r, ) : , h1( ) r h2( )}, where 2 and h1( ) 0 for all between and . Zzq f (x, y) da = z . Z h2( ) h1( ) f (r cos , r sin ) r dr d : suppose that t is a one-to-one c 1 transform which maps a region s in the uv-plane onto a region r in the xy-plane.

Related Questions

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skylion693

Now you'll evaluate the integral â«C6ysin(2x)dx+5xydy â« C 6 y sin â¡ ( 2 x ) d x + 5 x y d y on the closed curve C consisting of the line segments from (0,0) to (5,1) to (0,1) to (0,0) using Green's Theorem. Green's Theorem says this integral can be rewritten in the form â«â«Df(x,y)dA â« â« D f ( x , y ) d A In this integral, f(x,y) =

Setting up the double integral over the region D, you get

â«BAâ«DCf(x,y)dxdyâ«ABâ«CDf(x,y)dxdy

(Note that the order of integration is specified--for this integral it will turn out that this is the easier order of integration). In this,
A =

Evaluting this integral,
â«C6ysin(2x)dx+5xydy=â«â«Df(x,y)dA=â«C6ysinâ¡(2x)dx+5xydy=â«â«Df(x,y)dA=