Question

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(1) Let A - (0,0), B- (1,1) and consider the veetor field f(r, y,z)vi+aj. Evaluate the line integral J f.dr )along the parabola y from A to B and (i)along the straight line from A to B. Is the vector field f conservative? (2) For the vector feld f # 22(r1+ gd) + (x2 + y2)k use the definition of line integral to (3) You are given that the vector field f in Q2 is conservative. Find the corresponding (4) Consider the vector field F(x,y) = xy-Hj (-Fi + Fj) and let C be the closed evaluate the line integral Jf.dr along the helical path r - costi + sintj +tk, 0 sts potential function and use this to check the line integral evaluated in Q2 curve consisting of three segments: the straight line from (0,0) to (1,0) followed by the circular arc from (1,0) to (0, 1) followed by the straight line from (0,1) to the origin (oriented anticlockwise). Let n denote the unit normal vector to C directed out of the region D bounded by C. (a) Calculate the line integral φ F-dr directly without using Green's theorem. FFdA without using Green's theo- (b) Calculate the double integral rem and compare your answer with part (a). (c) Calculate the flux out of D by evaluating the line integral ф F-nds directly without using the flux form of Green's theorem. (d) Calculate the flux out of D by applying Green's Theorem and evaluating the double integral | | ▽ . F dA. a across a curve C rt) (5) Calculate the 2-D flux of a constant vector field F 2(t)i + y(tj, a t b, indicating the positive direction. what is the flux when the curve C is closed? Explain.

Answer 1

This problem is from Vector Calculus.

Prerequisite : Green's Theorem , Flux Integrals, unit normal vector to a curve in a plane.

Given rightarrow F (x, y) = xy i^^- x^2 j^^= F_1 i^^+ F_2 j^^Rightarrow F_1 = xy & F_2 = -x^2 C rightarrow closed curve shown in figure rightarrow n^^rightarrow unit normal to C directed out of the region D enclosed by C a) To find rightarrow contourintegral_c F^rightarrow dr^rightarrow without using Green's Theorem We start by parametrizing the curve C: - 1^st part rightarrow OA: r(t) = (1 - t) r_0 + tr_1, 0 lessthanorequalto t lessthanorequalto 1 where r_0 = < 0, 0 > , r_1 = < 1, 0 > Therefore r(t) = (1 - t) < 0, 0 > + t < 1, 0 > = < 0, 0 > + < t, 0 > = < t, 0 > Therefore OA: - r(t) = ti^^+ 0j^^= ti^^, 0 lessthanorequalto t lessthanorequalto 1 i.e. x = t, y = 0 dx = dt, dy = 0 dt & F^rightarrow (t) = t (0) i^^- t^2 j^^= -t^2 j^^& dr^rightarrow = dx i^^+ dy j^^= dt i^^+ 0 dt j^^= dt i^^Therefore integral_OA F^rightarrow dr^rightarrow = integral^1_0 (- t^2 j^^) (d + i^^) = 0 (Because j^^i^^= 0)