Professor Dave here, let’s take a look at surface integrals.
Surface integrals can be thought of as a higher dimensional equivalent to line integrals.
Instead of a curve C to integrate over, we will have a surface S. If we then have a function
f(x,y,z), the surface integral of f over S will be expressed as this integral. While
this expression looks fairly simple, just like constraining line integrals to the path
of a curve, the complexity here comes from constraining to the surface S. And just like
we parameterized curves for line integrals in terms of a new variable t, it proves useful
to parameterize the surface S in terms of two new variables, which we will call u and
v. The coordinates of the surface S can be expressed as a vector r = <x(u,v), y(u,v),
z(u,v)>. With this parameterization, we will be able to express the dS in the surface integral
in terms we can use. It turns out that this dS can be expressed as the magnitude of the
cross product of the r vector’s derivatives, which we will call ru and rv, du dv. Written
out, ru = and similarly rv = . So our surface
integral of f over the surface S becomes the double integral of (f of r) times the magnitude
of ru cross rv du dv. The interpretation of these surface integrals is a bit more abstract,
but if looking at the function f = 1, then the integral gives back the surface area of
S across the domain of the integration. For an example of how to set up the surface
integral, consider the surface given by x = u cos(v), y = u sin(v), and z = v. In order
to set up the integral we must take the derivatives of these expressions. The vector ru will be
given by the derivatives with respect to u, leaving us with . Meanwhile,
the vector rv will be given by the derivatives with respect to v, leaving us with <-u sin(v),
u cos(v), and 1>. Now we must take the cross product given by ru cross rv. To find the
cross product we set up the matrix shown here and take the determinant. We end up getting
sin(v)i - cos(v)j + [u cos^2(v) + u sin^2(v)]k. And since cos^2(v) + sin^2(v) = 1, we end
up with the vector . Now we need the magnitude of this vector, which
as we recall, simply means the square root of the dot product with itself. This gives
us root (sin v)^2 + (-cos v)^2 + u^2, which given the same identity we just used, becomes
root (1 + u^2). This is the magnitude of the cross product, making dS equal to root (1
+ u^2) du dv. So our surface integral of some function f over this surface will be the double
integral of f times root (1 + u^2) du dv, where f is written in terms of u and v, and
the bounds will be determined by the range of u and v.
This example illustrates how to set up the most general cases of surface integrals. However,
there is a simplified case where the surface we are integrating over is only a function
of x and y, or in other words z = g(x,y). For cases like this, x and y serve as our
parametric variables and our surface can be represented by the vector r = .
Let’s find the expression for dS in this case. First we find the vector rx by taking
the x derivative, giving us <1, 0, dz/dx>. Then we find ry by taking the y derivative,
giving us <0, 1, dz/dy>. Now taking the cross product rx cross ry, we take the determinant
of this matrix. We end up getting -dz/dxi – dz/dyj + k. Next we need the magnitude
of this cross product, which is the square root of (dz/dx)^2 + (dz/dy)^2 + 1. Now that
we have this, we see that dS in this special case is root [(dz/dx)^2 + (dz/dy)^2 + 1] dx
dy. Putting everything together, our surface integral becomes the double integral of the
function f times root [(dz/dx)^2 + (dz/dy)^2 + 1] dx dy. Calculations for this scenario
are much more simple and follow this easy to calculate expression, we only need to take
care when dealing with complex bounds for x and y.
Finally, just the way that line integrals could be used with vector fields, surface
integrals can do the same. In order to express surface integrals with vector fields, we must
first introduce the “normal vector” n. If we have a surface to integrate over, the
normal vectors at each point are unit vectors that point in perpendicular fashion away from
the surface. An easy way to visualize this would be to imagine running a flat piece of
paper across the surface as a tangent plane. The direction straight “out” of the paper
is the normal vector. Now if we have a vector field F permeating space, the surface integral
of F over the surface S will be F dot the normal vector n dS. This calculation ends
up describing the “amount” of the vector field F that passes through the surface S,
also called the “flux” of F across S. In order to go through calculations, we must
know how to express the normal vector in terms we can use. For surfaces that are given by
a parameterized r(u,v) vector like we discussed earlier, the n vector ends up being equal
to ru cross rv, divided by the magnitude of ru cross rv. So our surface integral becomes
F dot ru cross rv divided by the magnitude of that cross product, dS. But recall that
dS equals the magnitude of the cross product times du dv. Plugging this in cancels the
magnitudes out, leaving us with the double integral F dot (ru cross rv) du dv. This is
the expression we can use to calculate the flux in any general case, but let’s once
again consider a surface where z is a function of x and y. We saw in this case that rx cross
ry equals -dz/dxi - dz/dyj + k. Recall how we can write vector fields in components <P,
Q, R>, so the dot product F dot (rx cross ry) ends up being (-P dz/dx – Q dz/dy +
R). And so in this case our flux becomes the double integral of this expression, dx dy.
For a complete example, consider the vector field F = and the surface given
by z = x + y^2. Let’s find the flux of the field through this surface over the range
x from 0 to 1 and y from 0 to 1. Since the surface is given as a function of x and y,
we can jump to expressing our surface integral as the double integral of (-P dz/dx – Q
dz/dy + R) dx dy. Here P is x^2, and dz/dx is the x derivative of x + y^2, which equals
1. Q is xy and dz/dy will be the y derivative of x + y^2 so 2y. And R is z, which in terms
of x and y is x + y^2. Plugging all this in, we get the double integral of (-x^2 - 2xy^2
+ x + y^2) dx dy where both x and y range from 0 to 1. Evaluating the x integral first,
we get the antiderivative [(-x^3)/3 – (x^2)(y^2) + (x^2)/2 + x(y^2)], where x runs from 0 to
1. Evaluating at x = 1 gives (-1/3) – (y^2) + 1/2 + y^2, while evaluating at zero will
just give zero. The y squared terms cancel, leaving us with the integral from 0 to 1 of
1/6 dy. This gives us y/6 evaluated from 0 to 1, which is just 1/6. This will be our
flux through this surface. Surface integrals prove useful in many fields,
such as when using Gauss’s Law in physics, which relates the amount of charge to the
flux of an electric field through a closed surface. They are also used in some theorems
which we will be discussing next, so with that in mind, let’s check comprehension.