Evaluate the line integral, where C is the given curve sin(x)dx + cos(y)dy C consists of the top half of the circle x2 + y2 = 25 from (5, 0) to (-5, 0) and the line segment from (-5, 0) to (-6, 3).

if F is a vector field , if it is defined on a domain D and F=[tex]\nabla f[/tex] for any scalar function f on D then f is known as potential function of F

here, F = sin(x)dx + cos(y)dy

then f = -cos(x) + sin(y)

( because, [tex]\nabla f[/tex] should be F. by applying [tex]\nabla[/tex] operator on f we must obtain F. so to satisfy this condition f must be -cosx + siny.

where, [tex]\nabla g(x,y) =[/tex] partial derive of g(x,y) with respect to x + partial derive of g(x,y) with respect to y )

therefore, line integral of F over the given curve C = f(-6,3)-f(-5,0)

={-cos(-6)}+sin(3) -[{-cos(-5)}+sin(0)]

=-cos(6)+sin(3)-[-cos(5)] [since, sin(0)=0, cos(-a)=cos(a) where, a>0]

= -cos(6)+sin(3)+cos(5)

=-0.53538809312 (using calculator)

jayilych4real jayilych4real · Answer 2 · 2022-04-08T14:40:55+02:00

The evaluation of the line integral is: -0.53538809312 (using a calculator) or -cos(6)+sin(3)+cos(5)

Calculations and Parameters:

If F is a vector field and if it is defined on a domain D and F

= ∇f for any scalar function f on D

Then f is known as the potential function of F

where, F = sin(x)dx + cos(y)dy

then f = -cos(x) + sin(y)

( because ∇f should be F. by applying ∇ operator on f we must obtain F. so to satisfy this condition f must be -cosx + siny.

where, ∇g(x,y) partial derive of g(x,y) with respect to x + partial derive of g(x,y) with respect to y )

Therefore, the line integral of F over the given curve C

= f(-6,3)-f(-5,0)
={-cos(-6)}+sin(3) -[{-cos(-5)}+sin(0)]
=-cos(6)+sin(3)-[-cos(5)] [since, sin(0)=0, cos(-a)=cos(a) where, a>0]
= -cos(6)+sin(3)+cos(5)
=-0.53538809312 (using calculator)