We generalize the idea of line integrals to higher dimensions.
1 Generalizing to parametric surfaces
We’ve learned that given an explicit function \(F:\R ^2\to \R \) that graphs a surface in \(\R ^3\), we can
compute its surface area with
\[ \iint _R \d S \]
where
\[ \d S = \sqrt {1+F^{(1,0)}(x,y)^2+F^{(0,1)}(x,y)^2}\d A. \]
We will now generalize this idea to parametric
surfaces. To do this, we need to be able to compute \(\d S\) when our surface is drawn by a
parametric function. Let’s remind ourselves how we compute \(\d S\) for the surface \(z = F(x,y)\).
Consider the surface area of a “patch” of the surface, determined by \(\vec {u}\) and \(\vec {v}\) below:
In essence, we zoom in on
this portion of the surface to the extent that the tangent plane approximates the
function so well that in this figure, it is virtually indistinguishable from the surface
itself. Therefore we can approximate the surface area \(\d S\) of a “patch” of this region
of the surface with the area of the parallelogram spanned by \(\vec {u}\) and \(\vec {v}\). Here
\begin{align*} \vec {u} &= \vector {\d x,0,\pp [F]{x}\d x}\\ \vec {v} &= \vector {0,\d y,\pp [F]{y}\d y}, \end{align*}
hence
\begin{align*} \d S &= |\vec u\cross \vec v|\\ &= \left |\vector {\d x,0,\pp [F]{x}\d x}\cross \vector {0,\d y,\pp [F]{y}\d y}\right |\\ &=\sqrt {1+F^{(1,0)}(x,y)^2+F^{(0,1)}(x,y)^2}\d x\d y. \end{align*}
Now suppose we have a parametric surface:
\[ \vec {P}(u,v) = \vector {x(u,v), y(u,v), z(u,v)} \]
This case is essentially the same
as before, though now we define our patch by looking at tangent vectors
\begin{align*} \pp [\vec {P}]{u} &= \vector {\pp [x]{u},\pp [y]{u},\pp [z]{u}}\\ \pp [\vec {P}]{v} &=\vector {\pp [x]{v},\pp [y]{v},\pp [z]{v}} \end{align*}
and we may write
\[ \d S = \left |\pp [\vec {P}]{u} \cross \pp [\vec {P}]{v}\right |\d u \d v. \]
Given the following parametric formula for a sphere of radius
\(r\),
\begin{align*} x(\theta ,\varphi ) &= r\cos (\theta )\sin (\varphi )\\ y(\theta ,\varphi ) &= r\sin (\theta )\sin (\varphi )\\ z(\theta ,\varphi ) &= r\cos (\varphi ), \end{align*}
for \(0\le \theta < 2\pi \) and \(0\le \varphi \le \pi \), compute \(\d S\).
\[ \d S = \answer {r^2\sin (\varphi )} \d \theta \d \varphi \]
Use an integral of the form
\[ \iint _R \d S \]
to compute the surface area of
a sphere of radius
\(r\).
\begin{align*} S&=\int _{\answer {0}}^{\answer {\pi }} \int _{\answer {0}}^{\answer {2\pi }}\answer {r^2\sin (\varphi )} \d \theta \d \varphi \\ &=\answer {4\pi r^2} \end{align*}
2 Flux: The flow across a surface
There are many specialized applications where one is interested in the rate that a
“fluid” passes through a “surface” per unit time. We call this rate flux or the flow
across a surface. To compute the flux, we see how aligned field vectors are with
vectors normal to the surface.
- When the field vectors are going the same direction as the vectors normal
to the surface, the flux is positive.
- When the field vectors are going the opposite direction as the vectors
normal to the surface, the flux is negative.
- When the field vectors are orthogonal to the vectors normal to the surface,
the flux is zero.
Suppose you have the vector field
\[ \vec {F}(x,y,z) = \vector {0,x,0} \]
and a surface with a normal vector in the
positive direction.
Is the flux through the surface positive, zero, or negative?
positive zero negative
Since the \(x\)-values are all positive or nonnegative for points that make up our
surface, our field vectors are pointing the same direction as the normal vector.
Suppose you have the vector field
\[ \vec {F}(x,y,z) = \vector {-z,0,0} \]
and a surface with a normal vector in the positive
direction.
Is the
flux through the surface positive, zero, or negative?
positive zero negative
Since the \(z\)-values are all positive or nonnegative for points that make up our surface,
our field vectors are pointing the the opposite direction as the normal vector.
Suppose you have the vector field
\[ \vec {F}(x,y,z) = \vector {0,0,x} \]
and a surface with a normal vector in the positive
direction.
Is the flux through the surface positive, zero, or negative?
positive zero negative
Since the \(x\)-values are all positive or nonnegative for points
that make up our surface, our field vectors are orthogonal to the normal
vector.
From our work above, we see that if we computed the flow
across a surface, we need to indicate a “positive” and “negative” direction.
These
directions are arbitrary, in the sense that they will depend on the context of the
problem.
Each point on a smooth surface has two unit normal vectors, pointing in opposite
directions. Choosing an
orientation means choosing one of these vectors to be
“positive” and the other to be “negative.”
In essence, to compute the flow across a surface, we demand that the surface has
two sides. While it might seem reasonable to assume that every surface
have two sides, in fact this is false, there are surfaces that cannot be
oriented. Consider
\[ \vec {P}(\theta ,t) = \begin{bmatrix} (1 + t \cos (\theta /2))\cos (\theta )\\ (1 + t \cos (\theta /2))\sin (\theta )\\ t \sin (\theta /2) \end{bmatrix} \]
For your viewing pleasure, we’ve included a graph:
This surface is
called a Möbius strip. It is a one-sided surface, meaning one could walk along each
side of the surface without crossing the edge! Möbius strips are really cool, and this
author invites you, the young mathematician, to explore their mysteries on your
own.
If you have a closed surface, the normal vector pointing outward indicates the
“positive” direction, and the normal vector pointing inward indicates the “negative”
direction. Moreover, given any parameterization of an orientable surface, there is a
natural orientation based on the parameterization.
Given a parameterization of an orientable surface:
\[ \vec {P}(u,v) = \vector {x(u,v), y(u,v), z(u,v)} \]
the
orientation given by the
parameterization is given by the direction of
\[ \pp [\vec {P}]{u} \cross \pp [\vec {P}]{v}. \]
Now we have all the “parts” of a surface integral, it is time to explain what they
are.
3 Surface integrals
To compute the flow across a surface, also known as flux, we’ll use a surface integral.
While line integrals allow us to integrate a vector field \(\vec {F}:\R ^2\to \R ^2\) along a curve \(C\) that is
parameterized by \(\vec {p}(t) = \vector {x(t),y(t)}\):
\[ \int _C \vec {F}\dotp \d \vec {p} \]
A surface integral allows us to integrate a vector field \(\vec {F}:\R ^3 \to \R ^3\) across a
surface \(S\) that is parameterized by
\[ \vec {P}(u,v) =\vector {x(u,v), y(u,v), z(u,v)}. \]
Consider a patch of a surface \(\d S\) along with a unit vector
normal to the surface \(\uvec {n}\):
A
surface integral will use the dot product to see how “aligned” field vectors are with
this (scaled) unit normal vector.
Let
\(\vec {F}:\R ^3\to \R ^3\) be a vector field and
\(\vec {P}:\R ^2\to \R ^3\) be a smooth
vector-valued function drawing an oriented surface
\(S\) exactly once as
\(u\) runs from
\(a\) to
\(b\)
and
\(v\) runs from
\(c\) to
\(d\):
\begin{align*} \vec {F}(x,y,z) &= \vector {L(x,y,z), M(x,y,z), N(x,y,z)}\\ \vec {P}(u,v) &= \vector {x(u,v),y(u,v),z(u,v)}. \end{align*}
A surface integral is an integral of the form:
\begin{align*} \iint _R \vec {F}\dotp \uvec {n} \d S &= \int _c^d\int _a^b \vector {L,M,N} \dotp \frac {\left (\pp [\vec {P}]{u}\cross \pp [\vec {P}]{v}\right )}{\left |\pp [\vec {P}]{u}\cross \pp [\vec {P}]{v}\right |}\left |\pp [\vec {P}]{u}\cross \pp [\vec {P}]{v}\right |\d u \d v\\ &= \int _c^d\int _a^b \vector {L,M,N} \dotp \left (\pp [\vec {P}]{u}\cross \pp [\vec {P}]{v}\right )\d u \d v \end{align*}
Consider a surface integral:
\[ \iint _R \vec {F}\dotp \uvec {n} \d S \]
Suppose that
\(\vec {P}(x,y) = \vector {x,y,G(x,y)}\) for
\(a\le x\le b\) and
\(c\le y\le d\). Simplify the integral above.
\[ \iint _R \vec {F}\dotp \vector {-\pp [\answer {G}]{\answer {x}}, -\pp [\answer {G}]{\answer {y}},\answer {1}}\d A \]
Now let’s work some examples.
Consider the vector field
\(\vec {F}\) and the surface
\(\vec {P}\) \begin{align*} \vec {F}(x,y,z) &= \vector {y-z,x-z,x-y}\\ \vec {P}(s,t) &= \vector {s-t,t-s,2t+s} \end{align*}
for \(0\le s\le 1\) and \(1-s\le t\le 1\). Using the orientation given by the parameterization of \(\vec {P}\), compute:
\[ \iint _R \vec {F}\dotp \uvec {n}\d S \]
First
we’ll compute
\(\uvec {n}\d S\). We know that
\begin{align*} \uvec {n}\d S &= \left (\pp [\vec {P}]{\answer [given]{s}}\cross \pp [\vec {P}]{\answer [given]{t}}\right )\d t \d s\\ &=\left ( \vector {\answer [given]{1},\answer [given]{-1},\answer [given]{1}} \cross \vector {\answer [given]{-1},\answer [given]{1},\answer [given]{2}} \right )\d t \d s\\ &= \vector {\answer [given]{-3},\answer [given]{-3},\answer [given]{0}} \d t \d s. \end{align*}
Now write:
\[ \vec {F}(\vec {P}(s,t)) = \vector {\answer [given]{-2s-t},\answer [given]{-3t},\answer [given]{2s-2t}} \]
So, write with me,
\begin{align*} \iint _R \vec {F}\dotp \uvec {n}\d S &= \int _{\answer [given]{0}}^{\answer [given]{1}}\int _{\answer [given]{1-s}}^{\answer [given]{1}} \left (\answer [given]{6s + 12 t}\right )\d t \d s\\ &= \int _{\answer [given]{0}}^{\answer [given]{1}}\answer [given]{12s} \d s\\ &=\answer [given]{6}. \end{align*}
Consider the vector field
\(\vec {F}\) and the surface
\(\vec {P}\) \begin{align*} \vec {F}(x,y,z) &= \vector {x,y,z}\\ \vec {P}(\theta ,\phi ) &= \vector {\cos (\theta )\sin (\phi ),\sin (\theta )\sin (\phi ),\cos (\phi )} \end{align*}
for \(0\le \theta <2\pi \) and \(0\le \phi \le \pi \). Using the outward pointing orientation, compute:
\[ \iint _R \vec {F}\dotp \uvec {n}\d S \]
First we’ll compute
\(\uvec {n}\d S\),
paying attention to the orientation. We know that
\begin{align*} \uvec {n}\d S &= \left (\pp [\vec {P}]{\answer [given]{\theta }}\cross \pp [\vec {P}]{\answer [given]{\phi }}\right )\d \phi \d \theta \\ &=\begin{bmatrix} \answer [given]{-\sin (\theta )\sin (\phi )}\\ \answer [given]{\cos (\theta )\sin (\phi )}\\ \answer [given]{0} \end{bmatrix} \cross \begin{bmatrix} \answer [given]{\cos (\theta )\cos (\phi )}\\ \answer [given]{\sin (\theta )\cos (\phi )}\\ \answer [given]{-\sin (\phi )} \end{bmatrix} \d \phi \d \theta \\ &= \begin{bmatrix} \answer [given]{-\cos (\theta )\sin ^2(\phi )}\\ \answer [given]{-\sin (\theta )\sin ^2(\phi )}\\ \answer [given]{-\cos (\phi )\sin (\phi )} \end{bmatrix} \d \phi \d \theta . \end{align*}
Let’s now examine our normal vector. If we check at the angles \((\theta ,\phi ) = (0,0)\) or the angles \((\theta ,\phi ) = (0,\pi )\)
something strange happens, we find the vector \(\vector {\answer [given]{0},\answer [given]{0},\answer [given]{0}}\). This is an unavoidable consequence
of the so-called “ Hairy ball theorem.” Regardless, since there are only a finite
number of isolated “bad points” our computation is unaffected. Checking the
orientation at the angles \((\theta ,\phi ) = (0,\pi /2)\), we find the normal vector
\[ \vector {-1,0,0} \]
meaning that these normal
vectors are oriented
inwardoutward. Hence we will now set
\[ \uvec {n}\d S = \vector {\answer [given]{\cos (\theta )\sin ^2(\phi )},\answer [given]{\sin (\theta )\sin ^2(\phi )},\answer [given]{\cos (\phi )\sin (\phi )}} \d \phi \d \theta . \]
Now write:
\[ \vec {F}(\vec {P}(\theta ,\phi )) = \vector {\answer [given]{\cos (\theta )\sin (\phi )},\answer [given]{\sin (\theta )\sin (\phi )},\answer [given]{\cos (\phi )}} \]
So, write
with me,
\begin{align*} \iint _R \vec {F}\dotp \uvec {n}\d S &= \int _{\answer [given]{0}}^{\answer [given]{2\pi }}\int _{\answer [given]{0}}^{\answer [given]{\pi }} \left (\answer [given]{\sin (\phi )}\right )\d \phi \d \theta \\ &= \int _{\answer [given]{0}}^{\answer [given]{2\pi }}\answer [given]{2} \d \theta \\ &=\answer [given]{4\pi }. \end{align*}
In the first episode of the science fiction series
The Expanse, the
Canterbury (a
space ship hulling water) is destroyed. Several “lucky” crew members escape
destruction in a small shuttle
\(2.6\cdot 10^7\) meters away, only later to be pummeled by
debris.
Setting
\begin{align*} \vec {F}(x,y,z) &= \frac {2\cdot 10^5 \vec {x}}{\pi |\vec {x}|^3}\\ \vec {P}(\theta ,\phi ) &= 2.6\cdot 10^7\vector {\cos (\theta )\sin (\phi ),\sin (\theta )\sin (\phi ),\cos (\phi )} \end{align*}
where \(\vec {x} = \vector {x,y,z}\), \(0\le \theta <2\pi \), and \(0\le \phi \le 4\cdot 10^{-7}\) the surface integral
\[ \iint _R \vec {F}\dotp \uvec {n}\d S \]
will compute the momentum (in kilograms-per-second)
that will impact the shuttle. Find this value.
From our work above we know that
\[ \uvec {n}\d S = 6.76\cdot 10^{14} \vector {\answer [given]{\cos (\theta )\sin ^2(\phi )},\answer [given]{\sin (\theta )\sin ^2(\phi )},\answer [given]{\cos (\phi )\sin (\phi )}} \d \phi \d \theta . \]
Now write:
\[ \vec {F}(\vec {P}(\theta ,\phi )) = \answer [given]{\frac {1}{3.38\cdot 10^{9}\pi }}\vector {\cos (\theta )\sin (\phi ),\sin (\theta )\sin (\phi ),\cos (\phi )} \]
So, write with me,
\begin{align*} \iint _R \vec {F}\dotp \uvec {n}\d S &= \frac {2\cdot 10^5}{\pi }\int _{\answer [given]{0}}^{\answer [given]{2\pi }}\int _{\answer [given]{0}}^{\answer [given]{4\cdot 10^{-7}}} \left (\answer [given]{\sin (\phi )}\right )\d \phi \d \theta \\ &=4\cdot 10^5 \left (\cos \left (\frac {1}{2.5\cdot 10^6}\right )-1\right )\\ &\approx 3.2\cdot 10^{-8}. \end{align*}
Hence it is likely no harm would come from the debris.
For some interesting extra reading check out:
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