Prove Fr e2y cos x 2xyz3 2e2y sinx x2 z3 3x2yz2 3 has pa

Prove F(r) = [e^2y cos x + 2xyz^3 2e^2y sinx + x^2 z^3 3x^2yz^2 + 3] has path-independent integrals by using Theorem 3 from section10.2. Define path C as r(t) = [2 cos t 4 sin 3t 3 sin 2t] for 0 lesser than equal to y lesser than equal to pi. Evaluate the line integral integral_C F.dr using any method or theorem from section 10.2.

Solution

a) We need to prove that curl of F is 0

F = (e^2yxosx + 2xyz^3)i + (2e^2ysinx +x^2z^3)j + ( 3x^2yz^2 + 3)k = Fx i + Fy j + Fz k
curl F = [d/dx i + d/dy j + d/dz k ]x[Fx i + Fy j + Fz k] = d(Fy)/dx k - d(Fz)/dx j - d(Fx)/dy k + d(Fz)/dy i + d(Fx)/dz j - d(Fy)/dz i
curl F =[ 2e^2y cosx + 2xz^3 - 2e^2ycosx - 2yz^3 ]k + [3x^2z^2 - 3x^2z^2]i + [ 6xyz^2 - 6xyz^2]j = 0

b) r (t) = 2 cost i + 4sin3t j + 3sin2t k
dr (t) = [-2sint i +12cos3t j + 6cos2t k] dt
F.dr = [-2sin(t)*(e^2yxosx + 2xyz^3) + 12cos(3t) * (2e^2ysinx +x^2z^3) + 6cos(2t) * ( 3x^2yz^2 + 3) ] dt
Integral of F.dr = pi[2cos(t)*(e^2yxosx + 2xyz^3) + 4sin(3t) * (2e^2ysinx +x^2z^3) + 3sin(2t) * ( 3x^2yz^2 + 3) ]0
=> [2cos(pi)*(e^2yxosx + 2xyz^3) + 4sin(3pi) * (2e^2ysinx +x^2z^3) + 3sin(2pi) * ( 3x^2yz^2 + 3) ] - [2cos(0)*(e^2yxosx + 2xyz^3) + 4sin(0) * (2e^2ysinx +x^2z^3) + 3sin(0) * ( 3x^2yz^2 + 3) ]
=> [-2(e^2yxosx + 2xyz^3) ] - [2*(e^2yxosx + 2xyz^3) ] = -4(e^2yxosx + 2xyz^3)