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
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
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

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.

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.

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.

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.

For some interesting extra reading check out: