Subsection Surfaces in \(\RR^3\)
A typical surface in \(\RR^3\) can be parameterized by a vector function of the form
\begin{equation*}
\rr(u, v) = \mqty[x(u, v) \amp y(u, v) \amp z(u, v)]\text{,}
\end{equation*}
where \((u,v)\in R\) and \(R\) is some region of the \(uv\)-plane. If a surface \(D\) is parameterized by \(\rr(u, v), (u,v)\in R\text{,}\) we will usually indicate such by writing \(D : \rr(u, v)\text{.}\)
Example 4.4.1. Parameterizing a Surface in \(\RR^3\).
What surface in \(\RR^3\) is represented by
\begin{equation*}
\rr(u, v) = \mqty[\cos(u)\sin(v) \amp \sin(u)\sin(v) \amp \cos(v)]
\end{equation*}
where \(0\leq u\leq2\pi\) and \(0\leq v\leq \pi\text{?}\)
Solution.
If we let
\begin{align*}
x(u, v) \amp = \cos(u)\sin(v) \\
y(u, v) \amp = \sin(u)\sin(v) \\
z(u, v) \amp = \cos(v) \text{,}
\end{align*}
then we see that \(x^2 + y^2 + z^2 = 1\text{.}\) This suggests that the surface \(D:\rr(u, v)\) is contained within the unit sphere in \(\RR^3\text{.}\) In fact, it is the entire unit sphere! We can verify this using Sage:
We will primarily be interested in using surface integrals to represent and calculate the flux of a vector field
\(\FF\) over a surface
\(D:\rr(u, v)\text{.}\) As part of this process, we will need to find normal vectors to surfaces (just as we did when finding
flux integrals across curves). If
\(D\) is a smooth surface, or equivalently if
\(\rr(u, v)\) is differentiable, then finding the normal vector is just a matter of computing
\(\mathbf{n} = \rr_u\times\rr_v\text{.}\) This is because both
\(\rr_u\) and
\(\rr_v\) are tangent to the surface
\(D\text{.}\)
Example 4.4.2. Finding a Normal Vector.
Find a normal vector to the surface \(y = x^2\) at the point \((2, 4, -3)\text{.}\)
Solution.
First, we parameterize our surface by using
\begin{equation*}
\rr(u, v) = \mqty[u \amp u^2 \amp v]\text{.}
\end{equation*}
Then
\begin{equation*}
\rr_u = \mqty[1 \amp 2u \amp 0]\text{ and }\rr_v = \mqty[0 \amp 0 \amp 1]\text{.}
\end{equation*}
Now we can find the normal vector by computing the cross product of these two vectors, either by hand or by using Sage as below:
Either way, we get the normal vector \(\mathbf{n}(u, v) = \mqty[2u \amp -1 \amp 0]\text{.}\) At the point \((2, 4, -3)\text{,}\) this becomes \(\mqty[4 \amp -1 \amp 0]\text{.}\)
If a surface is given by a Cartesian equation of the form \(F(x, y, z) = 0\) as in the last example, then we can also find normal vectors by computing gradients. To see why, suppose that \(\rr(t)\) traces out a curve on the surface \(D : F(x, y, z) = 0\text{.}\) Then \(F(\rr(t)) = 0\text{.}\) However, by the chain rule we then get
\begin{equation*}
0 = \dv{t}(F(\rr(t))) = \del F(\rr(t))\cdot \rrp(t)\text{.}
\end{equation*}
Since \(\rrp\) is tangent to \(D\text{,}\) this shows that \(\del F\) must be perpendicular to \(D\text{.}\)
In the last example, we could choose \(F(x, y, z) = y - x^2 = 0\) giving the gradient
\begin{equation*}
\del F = \mqty[-2x \amp 1 \amp 0]\text{.}
\end{equation*}
At
\((2, 4, -3)\text{,}\) this becomes
\(\mqty[-4 \amp 1 \amp 0]\text{.}\) Note that this is the
negative of the answer we obtained in
Example 4.4.2. We will always have two choices for the direction of our normal vector, so going forward we will need to specify which direction we will take.
Subsection Surface Integrals
Now we are ready to define surface integrals of vector fields. In all that follows, we will assume that \(D:\rr(u, v), (u,v)\in R\) is a piecewise-smooth surface (i.e., a surface with finitely many cusps or edges). Unless specifically mentioned otherwise, we will assume that our corresponding normal vector is \(\mathbf{n}(u, v) = \rr_u\times\rr_v\text{.}\)
Definition 4.4.3. Surface Integrals of Vector Fields.
Let \(\FF\) be a continuous vector field defined over \(D\text{.}\) The (flux) surface integral of \(\FF\) over \(D\) is the quantity \(\iint_D \FF\cdot\dd{\SS}\) given by
\begin{equation*}
\iint_D \FF\cdot\dd{\SS} = \iint_{R} \FF(\rr(u, v))\cdot\nn(u, v)\dd{A}\text{.}
\end{equation*}
The flux integral \(\iint_D\FF\cdot\dd{\SS}\) represents the flow of \(\FF\) across the surface \(D\) parallel to the normal vector \(\nn\text{.}\) We will often pick \(\nn\) to be the “outward” direction relative to \(D\text{,}\) and so the flux integral will measure the flux of \(\FF\) out of \(D\text{.}\)
Example 4.4.4. Computing a Surface Integral.
Let \(\FF = e^y\ii + e^x\jj + \kk\) and let \(D\) denote the part of the plane \(x + y + z = 1\) restricted to the first octant. Find \(\iint_D\FF\cdot\dd{\SS}\text{.}\)
Solution.
We can use Sage to produce a quick plot of \(x + y + z = 1\) as below. However, we must remember that we're only considering the part of this plane in the first octant, and therefore must restrict \(x, y\) and \(z\) to be nonnegative.
We can parameterize this surface using
\begin{align*}
\rr(u, v) \amp = \mqty[u \amp v \amp 1 - u - v] \\
0 \amp \leq v\leq 1 - u \\
0 \amp \leq u\leq 1 \text{.}
\end{align*}
This gives us the corresponding normal vector \(\nn(u, v) = \mqty[1 \amp 1\amp 1]\text{,}\) and so
\begin{align*}
\iint_D\FF\cdot\dd{\SS} \amp = \int_0^1\int_0^{1-u}\mqty[e^v \amp e^y \amp 1]\cdot\mqty[1 \amp 1\amp 1]\dd{v}\dd{u} \\
\amp = \int_0^1\int_0^{1-u}(e^v + e^u + 1)\dd{v}\dd{u} \\
\amp = \int_0^1 (e^{1-u} + (1 - u)e^u - u)\dd{u} \\
\amp = 2e - \frac{7}{2} \text{.}
\end{align*}
As this is positive, we view the vector field \(\FF\) as flowing out of \(D\) relative to the normal vector \(\nn\text{.}\)
In
Example 4.4.4, we could have set up the surface integral by using the gradient to find our normal vector instead. One possibility is to choose
\(F(x, y, z) = 1 - x - y - z = 0\) to give our surface, which produces
\(\nn = \mqty[-1 \amp -1 \amp -1]\text{.}\) Therefore we would have
\begin{equation*}
\iint_{D}\FF\cdot\dd{\SS} = \int_0^1\int_0^{1-x}(-e^y - e^x - 1)\dd{y}\dd{x} = \frac{7}{2} - 2e\text{.}
\end{equation*}
What's happened is that we've inadvertently chosen the opposite direction for our normal vector \(\nn\) which flips the sign of our result from the example. Mathematically, there's no reason to prefer \(\mqty[1 \amp 1\amp 1]\) over \(-\mqty[1 \amp 1\amp 1]\text{.}\) This is why it's important to specify an orientation of our surface by choosing the direction of the normal vector.
In general, we say that a surface is orientable if it has a well-defined “inside” and “outside” as determined by selecting some consistent direction for its normal vectors. To illustrate this concept, it's helpful to consider a surface that fails to be orientable. The fundamental example here is the Möbius strip, which is parameterized by
\begin{equation*}
\rr(u,v) = \mqty[(1 + u\cos(\frac{v}{2}))\cos(v) \amp (1 + u\cos(\frac{v}{2}))\sin(v) \amp u\sin(\frac{v}{2})]
\end{equation*}
for \(-\frac{1}{2}\leq u\leq \frac{1}{2}, 0\leq v\leq 2\pi\text{.}\)
The Möbius strip is non-orientable because we can't choose an unambiguous orientation for normal vectors to this surface. This is because any normal vector on the surface can be moved around over the surface in a continuous fashion until it's back at the same position it started but is now in the opposite direction. To see this more precisely, consider the point \(\rr(0,0) = \mqty[1 \amp 0 \amp 0]\) on the strip. Then the normal vector can be found by computing \(\nn(u, v) = \rr_u\times\rr_v\text{,}\) and it can be checked that \(\nn(0,0) = \mqty[0 \amp 0\amp 1]\text{.}\) On the other hand, it's also true that \(\rr(0,0) = \rr(0, 2\pi)\text{.}\) However, \(\nn(0,2\pi) = \mqty[0 \amp 0\amp -1]\text{.}\) Therefore it's impossible for us to select a consistent orientation for this surface.
We can also compute surface integrals of scalar fields. To do so, we replace \(\nn(u, v)\) in the integral with \(\norm{\nn(u,v)}\text{:}\)
\begin{equation*}
\iint_D G(x, y, z)\dd{S} = \iint_{R} G(\rr(u, v))\norm{\rr_u\times\rr_v}\dd{A}\text{.}
\end{equation*}
If \(D: z = f(x, y)\text{,}\) then this expression simplifies down a bit since we can choose \(\rr(u, v) = \mqty[u \amp v\amp f(u, v)]\) and therefore
\begin{equation*}
\nn(u, v) = \mqty[-\pdv{f}{u} \amp -\pdv{f}{v} \amp 1]\text{.}
\end{equation*}
Hence
\begin{equation*}
\iint_D G(x, y, z)\dd{S} = \iint_{D} G(x, y, f(x, y)) \sqrt{1 + (f_x)^2 + (f_y)^2}\dd{A}\text{.}
\end{equation*}
Example 4.4.5. Computing a Scalar Surface Integral.
Find the mass of a thin hemisphere \(D\) centered above the origin in \(\RR^3\) of radius \(1\) and density \(\rho(x, y, z) = z\text{.}\)
Solution.
Since mass is just the integral of density, we can compute the mass by finding \(\iint_D \rho(x, y, z)\dd{S}\text{.}\) The surface itself is given by the equation \(z = \sqrt{1 - x^2 - y^2}\text{,}\) and so
\begin{align*}
\iint_D z\dd{S} \amp = \iint_{D}\sqrt{1 - x^2 - y^2} \sqrt{1 + \frac{x}{1 - x^2 - y^2} + \frac{y^2}{1 - x^2 - y^2}}\dd{A} \\
\amp = \iint_D \sqrt{1 - x^2 - y^2 + x^2 + y^2}\dd{A} \\
\amp = \iint_D \dd{A} \\
\amp = 2\pi \text{.}
\end{align*}
