Flux (7/3/2020)

There are three ways to do this one:

1) Integrate $\vec{B}$ over the given surface
2) Integrate $\vec{B}$ over some other surface with the same boundary curve as the given surface
3) Integrate $\vec{A}$ over the boundary curve

The given surface is too complicated to deal with conveniently, so we can integrate over the interior of the flat ellipse at height $z = 1$ to accomplish procedure $(2)$ . It is also convenient to implement procedure $(3)$ . The first segment of the attached code implements procedure $(2)$ , and the second segment of code implements procedure $(3)$ .

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import math

Num1 = 2000

a = 2.0
b = 1.0
c = 3.0

z = 1.0

right = 1.0 - (z/c)**2.0

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

phi1 = 0.0

xmax = a*math.sqrt(right)
dx = 2.0*xmax/Num1

x = -xmax + dx

while x <= xmax - dx:

    right2 = right - (x/a)**2.0 
    ymax = b*math.sqrt(right2)

    dy = 2.0*ymax/Num1

    y = -ymax + dy

    while y <= ymax - dy:

        dSx = 0.0
        dSy = 0.0
        dSz = dx*dy

        Bx = x - x**2.0
        By = 1.0 - y
        Bz = 2.0*x*z - 2.0

        dphi1 = Bx*dSx + By*dSy + Bz*dSz

        phi1 = phi1 + dphi1

        y = y + dy

    x = x + dx

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

print Num1
print phi1
print ""

#2000
#-11.1615146814

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

Num2 = 10**6

dtheta = 2.0*math.pi/Num2

phi2 = 0.0

theta = 0.0

while theta <= 2.0*math.pi:

    theta1 = theta
    theta2 = theta + dtheta

    cos1 = math.cos(theta1)
    sin1 = math.sin(theta1)

    cos2 = math.cos(theta2)
    sin2 = math.sin(theta2)

    denom1 = (cos1/a)**2.0 + (sin1/b)**2.0
    denom2 = (cos2/a)**2.0 + (sin2/b)**2.0

    r1 = math.sqrt(right/denom1)
    r2 = math.sqrt(right/denom2)

    x1 = r1*cos1
    y1 = r1*sin1

    x2 = r2*cos2
    y2 = r2*sin2

    dx = x2 - x1
    dy = y2 - y1
    dz = 0.0

    Ax = 2.0*y1 + z
    Ay = (x1**2.0)*z
    Az = x1*y1 + z

    dphi2 = Ax*dx + Ay*dy + Az*dz

    phi2 = phi2 + dphi2

    theta = theta + dtheta

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

print Num2
print phi2
print ""

#1000000
#-11.1700650878

The boundary curve of the given surface is:

$\frac{x^2}{4} + y^2 + \frac{1}{9} = 1$

$\implies \frac{x^2}{4} + y^2 = \frac{8}{9}$

The curve can be parameterised as such:

$\vec{r} = \sqrt{\frac{32}{9}} \cos{t} \ \hat{i} + \sqrt{\frac{8}{9}} \sin{t} \ \hat{j} + \hat{k}$

$\implies d\vec{r} = \left(-\sqrt{\frac{32}{9}} \sin{t} \ \hat{i} + \sqrt{\frac{8}{9}} \cos{t} \ \hat{j} + 0 \ \hat{k}\right) \ dt$

The magnetic flux through the surface is (Using Stoke's theorem):

$\int_{S} \vec{B} \cdot d\vec{S} = \int_{S} (\nabla \times \vec{A}) \cdot d\vec{S} = \oint_{C} \vec{A} \cdot d\vec{r}$

Here,

$\vec{A} = (2y+z) \ \hat{i} + x^2z \ \hat{j} + (xy+z) \ \hat{k}$ $\implies \vec{A} = (2y(t)+1) \ \hat{i} + x(t)^2 \ \hat{j} + (x(t)y(t)+1) \ \hat{k}$

The required integrand is, therefore, after simplification:

$d\Phi = \left(\frac{64\,\sqrt{2}\,{\cos\left(t\right)}^3}{27}-\frac{4\,\sqrt{2}\,\sin\left(t\right)\,\left(\frac{4\,\sqrt{2}\,\sin\left(t\right)}{3}+1\right)}{3}\right) \ dt$

$\Phi = \int_{0}^{2 \pi} \left(\frac{64\,\sqrt{2}\,{\cos\left(t\right)}^3}{27}-\frac{4\,\sqrt{2}\,\sin\left(t\right)\,\left(\frac{4\,\sqrt{2}\,\sin\left(t\right)}{3}+1\right)}{3}\right) \ dt$ $\Phi = -\frac{32 \pi}{9}$