Tutorial 13

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Topics Covered

This tutorial covers the following topics using Python and SymPy:

• Surface integrals of a vector field: $\iint_{S}\mathbf{F}\cdot d\mathbf{S}$
• Flux across a graph $z=g(x,y)$
• Orientation: upward versus downward normals
• Surface integrals of a scalar function: $\iint_{S}f\,dS$
• Area of a surface and the surface element $dS$

Introduction

For a surface given as the graph $z=g(x,y)$ over a region $D$:

$$\iint_{S}\mathbf{F}\cdot d\mathbf{S} =\iint_{D}\mathbf{F}\cdot\left(-\frac{\partial g}{\partial x}\mathbf{i} -\frac{\partial g}{\partial y}\mathbf{j}+\mathbf{k}\right)dx\,dy$$

For a parametrised surface $\mathbf{r}(u,v)$:

$dS=\left|\frac{\partial\mathbf{r}}{\partial u}\times\frac{\partial\mathbf{r}}{\partial v}\right|du\,dv$

import sympy as sy
from sympy.abc import x, y, z, u, v, theta, phi, rho, a, H
from sympy import (sqrt, sin, cos, tan, pi, ln, exp, atan, integrate, diff, simplify,
                   Rational, N, Matrix, Symbol, asinh)
sy.init_printing(use_latex=True)

def gradient(f, vars=(x, y, z)):
    return Matrix([diff(f, q) for q in vars])

Question 1

Evaluate the surface integral of $\mathbf{F}=3x^{2}\mathbf{i}-2yx\,\mathbf{j}+8\mathbf{k}$ over the surface $S$ that is the graph of $z=2x-y$ over the rectangle $[0,2]\times[0,2]$.

Here $g=2x-y$, so the upward normal is $\left(-g_{x},-g_{y},1\right)=(-2,1,1)$.

g = 2*x - y
n = Matrix([-diff(g, x), -diff(g, y), 1])
display(n)                                              # <-2, 1, 1>

F = Matrix([3*x**2, -2*y*x, 8])
integrand = simplify((F.subs({z: g})).dot(n))
display(integrand)                                      # -6*x**2 - 2*x*y + 8

display(integrate(integrand, (x, 0, 2), (y, 0, 2)))     # -8

$\iint_{S}\mathbf{F}\cdot d\mathbf{S}=-8$

Question 2

$S$ is the triangle with vertices $(1,0,0)$, $(0,2,0)$ and $(0,1,1)$, and $\mathbf{F}=xyz(\mathbf{i}+\mathbf{j})$. The triangle is oriented by the downward normal.

The plane through the three vertices is $z=-2x-y+2$, so with $x=u$, $y=v$ we get $\mathbf{r}(u,v)=(u,v,-2u-v+2)$ and $\mathbf{r}_{u}\times\mathbf{r}_{v}=(2,1,1)$ — which points upward.

r = Matrix([u, v, -2*u - v + 2])
cross = r.diff(u).cross(r.diff(v))
display(cross)                                          # <2, 1, 1>

F = Matrix([u*v*(-2*u - v + 2), u*v*(-2*u - v + 2), 0])   # xyz (i + j) with z = -2u-v+2
integrand = simplify(F.dot(cross))
display(integrand)                                      # 3*u*v*(-2*u - v + 2)

# the domain D is the triangle with vertices (1,0), (0,2), (0,1):  0<=u<=1, 1-u<=v<=2-2u
upward = integrate(integrand, (v, 1 - u, 2 - 2*u), (u, 0, 1))
display(upward)                                         # 1/10
display(-upward)                                        # -1/10  (downward orientation)

$\iint_{S}\mathbf{F}\cdot d\mathbf{S}=-\frac{1}{10}$

The sign flip is the whole point of the question: the parametrisation naturally gives the upward orientation, so the answer must be negated to get the downward flux.

Question 3

The disk $z=12$, $x^{2}+y^{2}\le25$, with $\mathbf{r}(x,y,z)=x\mathbf{i}+y\mathbf{j}+z\mathbf{k}$.

F = Matrix([x, y, z])
g = 12
n = Matrix([-diff(g, x), -diff(g, y), 1])
display(n)                                              # <0, 0, 1>

integrand = simplify((F.subs({z: g})).dot(n))
display(integrand)                                      # 12

display(integrate(integrand, (y, 0, 5), (x, 0, 5)))     # 300 (over the square, for illustration)
display(12 * (pi*5**2))                                 # 300*pi  (area of the disk)

$\iint_{S}\mathbf{r}\cdot d\mathbf{S}=\iint_{D}12\,dx\,dy=12\times25\pi=300\pi$

Question 4

$S$ is the closed surface made of the upper hemisphere $x^{2}+y^{2}+z^{2}=1$, $z\ge0$ and its base $x^{2}+y^{2}\le1$, $z=0$. Find the flux of $\mathbf{E}=2x\mathbf{i}+2y\mathbf{j}+2z\mathbf{k}$.

Split $S=H\cup D$. On the hemisphere the outward unit normal is $\mathbf{n}=(x,y,z)$, and on the base it is $-\mathbf{k}$.

# hemisphere:  E . n = 2(x^2 + y^2 + z^2) = 2  ->  flux = 2 * area(hemisphere)
flux_H = 2 * (2*pi*1**2)          # area of a hemisphere of radius 1 is 2*pi
display(flux_H)                                         # 4*pi

# base:  E . n = (2x, 2y, 0) . (0, 0, -1) = 0
flux_D = 0
display(flux_H + flux_D)                                # 4*pi

$\iint_{S}\mathbf{E}\cdot d\mathbf{S}=4\pi$

Question 5

Area of the ellipse cut on the plane $2x+3y+6z=60$ by the cylinder $x^{2}+y^{2}=2x$.

zplane = (60 - 2*x - 3*y)/6
dS = sqrt(1 + diff(zplane, x)**2 + diff(zplane, y)**2)
display(simplify(dS))                                   # 7/6

display(simplify(sy.factor(x**2 + y**2 - 2*x)))         # x**2 - 2*x + y**2 = (x-1)**2 + y**2 - 1 -> unit circle
display(Rational(7, 6) * pi)                            # 7*pi/6

The projection $D$ is the unit circle $\left(x-1\right)^{2}+y^{2}\le1$ with area $\pi$, so

$\text{Area}=\iint_{D}\frac{7}{6}\,dx\,dy=\frac{7}{6}\pi$

Question 6

$\displaystyle\iint_{S}x\,dS$ where $S$ is the part of $x^{2}+y^{2}+z^{2}=a^{2}$ in the first octant.

In spherical coordinates $dS=a^{2}\sin\theta\,d\phi\,d\theta$ and $x=a\cos\phi\sin\theta$, so the integral becomes $a^{3}\iint\cos\phi\sin^{2}\theta\,d\phi\,d\theta$ over $0\le\phi\le\frac{\pi}{2}$, $0\le\theta\le\frac{\pi}{2}$.

I = integrate(a**3 * cos(phi)*sin(theta)**2, (phi, 0, pi/2), (theta, 0, pi/2))
display(simplify(I))                                    # pi*a**3/4

$\iint_{S}x\,dS=\frac{\pi a^{3}}{4}$

Question 7

$\displaystyle\iint_{S}\frac{dS}{\sqrt{x^{2}+y^{2}+z^{2}}}$ over the cylinder $\mathbf{r}(u,v)=(a\cos u,\ a\sin u,\ v)$, $0\le u\le2\pi$, $0\le v\le H$.

# radius and height are positive, so sqrt(a^2) simplifies to a
a = Symbol('a', positive=True)
H = Symbol('H', positive=True)

rv = Matrix([a*cos(u), a*sin(u), v])
ru, rv_ = rv.diff(u), rv.diff(v)
display(ru, rv_)

cross = ru.cross(rv_)
display(cross)                                          # <a*cos(u), a*sin(u), 0>
dS = sqrt(cross.dot(cross))
display(simplify(dS))                                   # a

integrand = dS / sqrt(a**2 + v**2)                      # x^2+y^2+z^2 = a^2 + v^2
I = integrate(integrand, (v, 0, H), (u, 0, 2*pi))
display(simplify(I))                                    # 2*pi*a*asinh(H/a)

# asinh(H/a) is the same as ln((H + sqrt(a^2 + H^2))/a)
display(simplify(I.rewrite(ln) - 2*pi*a*ln((H + sqrt(a**2 + H**2))/a)))   # 0

$$\iint_{S}\frac{dS}{\sqrt{x^{2}+y^{2}+z^{2}}} =2\pi a\ln\left(\frac{H+\sqrt{a^{2}+H^{2}}}{a}\right)$$

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Extra Learning Resources

Key Concepts to Master

1. Flux across a graph: use the normal $\left(-g_{x},-g_{y},1\right)$ for the upward orientation
2. Orientation matters: a downward normal flips the sign of the flux
3. Surface element from a parametrisation: $dS=\left|\mathbf{r}_{u}\times\mathbf{r}_{v}\right|du\,dv$
4. Scalar surface integral: $\iint_{S}f\,dS$ — substitute the surface into $f$ and multiply by $dS$
5. Closed surfaces: split into the parts (hemisphere plus base) and add the fluxes

SymPy Resources

• SymPy cross — $\mathbf{r}_{u}\times\mathbf{r}_{v}$
• SymPy integrate — the resulting double integral
• SymPy asinh — $\operatorname{asinh}(v/a)=\ln\!\left(\dfrac{v+\sqrt{a^{2}+v^{2}}}{a}\right)$

Common Pitfalls

• Always check the direction of $\mathbf{r}_{u}\times\mathbf{r}_{v}$ before deciding on the sign
• sqrt(cross.dot(cross)) — the magnitude of a vector is the square root of its dot product with itself
• When the surface lies in a plane $z=g(x,y)$, substitute $z=g$ into $\mathbf{F}$ before taking the dot product
• For a closed surface, the outward normal points away from the enclosed volume on every face