Question

BOX FINAL ANSWER(S), write neatly, FOLLOW INSTRUCTIONS for THUMBS UP

19stokes theorem2: Problem 5 Previous Problem Problem List Next Problem (1 point) Let F= (2x, 2y, 2x + 22). Use Stokes' theorem to evaluate the integral of F around the curve consisting of the straight lines joining the points (1,0,1),(0,1,0) and (0,0,1). In particular, compute the unit normal vector and the curl of F as well as the value of the integral: n=( (the unit normal vector) VXF=( The value of the integral is

Answer 1

Given the function F = ( 22, 27, 2x+22 ) then by defn of well. cual f is VXF - Toda I fi s close obod 1. %e %y. Az į fo , 2x zy 2x+22 -1 (0-0) -1 (2-0) + f (0-0) - - À és oxF - <0,-2,07 Now to find Plane consistup of lines jong with points. for (1,0,1) (910) recter line in 20-1,1-0, 0-1) -<-,1,-1) similarly for (1,01) (0,0,1) is <-1,0,0) & their cross product will f ester the required normal weeter test a inil = ĝt ê 1- 0.01 conormalizing it to unit recter gives requed Namal vector let see How , CScanned vath CamScanner

Hence the equation of plane - which contains straight lines joing points (10),(0,1,0), (09) is y+z=d antain o loi passes in here 8 ofl ad »d=1 OD & Eqn of plane is y+z =1 let let f 2 - g= y + z-1 Pet them. vg = hons Couets 4 n z v llogll z ñ (required Normal vedter to contents : By store's thesem SS, vxFi ñds = 5S;2dA of a ) e -Masky ==2(4-3). CS$canned with CamScanner