Don't You Love Line Integrals? (Revised, Fixed Answer)

This was a fun problem. The curl of the vector field is equal to $y$ . By Stoke's Theorem, we can calculate the line integral by double-integrating the curl over the enclosed area. Integrate the right side and double the result (that is where the $2$ outside comes from).

$\large{2 \int_{1.25 x - 2.5}^{\sqrt{1 - (x-1)^2}} \int_0^2 y \, dx \, dy = -\frac{17}{6} \approx -2.833}$

Just for fun, I also wrote some code to evaluate the line integral directly, by integrating the dot product of the vector field and the infinitesimal path displacement vector. While tracing the path, I had to avoid the interface points between the curve segments, since the argument of the natural log is zero at those points. The code results (printed at the bottom) agree with the double integral, as expected.

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import math

N = 10**6

I = 0.0

#######################################################

# Right half circle

xc = 1.0
yc = 0.0
R = 1.0

dtheta = math.pi/N

theta = dtheta

while theta <= math.pi - dtheta:

    x = xc + R*math.cos(theta)
    y = yc + R*math.sin(theta)

    Fx = math.sqrt(x**2.0 + y**2.0)

    q1 = x + math.sqrt(x**2.0 + y**2.0)

    Fy = x*y + y*math.log(q1)

    dx = -R*math.sin(theta) * dtheta
    dy = R*math.cos(theta) * dtheta

    dI = Fx*dx + Fy*dy

    I = I + dI

    theta = theta + dtheta

#######################################################

# Left half circle

xc = -1.0
yc = 0.0
R = 1.0

dtheta = math.pi/N

theta = dtheta

while theta <= math.pi - dtheta:

    x = xc + R*math.cos(theta)
    y = yc + R*math.sin(theta)

    Fx = math.sqrt(x**2.0 + y**2.0)

    q1 = x + math.sqrt(x**2.0 + y**2.0)

    Fy = x*y + y*math.log(q1)

    dx = -R*math.sin(theta) * dtheta
    dy = R*math.cos(theta) * dtheta

    dI = Fx*dx + Fy*dy

    I = I + dI

    theta = theta + dtheta

#######################################################

# Left straight line

x1 = -2.0
y1 = 0.0

x2 = 0.0
y2 = -2.5

dx = (x2-x1)/N
dy = (y2-y1)/N

x = x1 + dx
y = y1 + dx

for j in range(0,N-1):

    Fx = math.sqrt(x**2.0 + y**2.0)

    q1 = x + math.sqrt(x**2.0 + y**2.0)

    Fy = x*y + y*math.log(q1)

    dI = Fx*dx + Fy*dy

    I = I + dI

    x = x + dx
    y = y + dy

#######################################################

# Right straight line

x1 = 0.0
y1 = -2.5

x2 = 2.0
y2 = 0.0

dx = (x2-x1)/N
dy = (y2-y1)/N

x = x1+dx
y = y1+dx

for j in range(0,N-1):

    Fx = math.sqrt(x**2.0 + y**2.0)

    q1 = x + math.sqrt(x**2.0 + y**2.0)

    Fy = x*y + y*math.log(q1)

    dI = Fx*dx + Fy*dy

    I = I + dI

    x = x + dx
    y = y + dy

#######################################################

print N
print I

#>>> 
#10000
#-2.83438016141
#>>> ================================ RESTART ================================
#>>> 
#100000
#-2.83343785716
#>>> ================================ RESTART ================================
#>>> 
#1000000
#-2.83334378703
#>>>