Professor Dave here, I want to tell you about
line integrals. By now we are very familiar with ordinary integrals. Integrating f(x)dx lets us find the area under
a curve given by the function f(x). Integrating f(x,y)dxdy lets us find the volume
under a surface given by the function f(x,y). Now with line integrals, we can integrate
a surface f(x,y) along the path of some curve C. We will be integrating along small segments
of the curve C, which we will call ds. Just like when we initially learned integration,
these segments will be our widths, while the surface f(x,y) will be our heights. So once again, our integration will give an
area, but now we are finding the area under a surface along a particular path within that surface. The way we will write this out is the integration
along C of f(x,y)ds. Recall that when we learned how to evaluate
multiple integrals, the x and y variables were treated independently during the integration. But now with line integrals we have a single
integral with both x and y. We must keep in mind that the curve we are
integrating along constrains the relationship between the two variables. Because of this, we can no longer treat them
independently, but this also opens up the possibility of reducing things to a single
variable that is to be integrated. In order to calculate line integrals by hand,
it is often necessary to be able to express the curve C as two separate equations, one
for x and one for y, that both depend on a new variable, t. As we have learned in a previous tutorial,
x of t and y of t are referred to as parametric equations, and as t varies, the x and y coordinates
of the curve they form will be mapped out. For a quick example consider the equations
x(t) = t/2 and y(t) = t^2. We can easily see the curve this would map
out by solving for t in terms of x and substituting it into the equation for y. By multiplying the first equation by 2 we
see that t = 2x, which when plugged into the second equation gives us y = (2x)^2 or just
4x^2. We can easily plot this equation. Now let’s say t runs from 0 to 6. We could plug these values into our parametric
equations to get the starting and ending points on this curve for these values of t. For t = 0 we get x = 0 and y = 0. For t = 6 we get x = 3 and y = 36. For this range of t we are left with this
part of the curve between these two points. Now, why are parametric equations so useful
for line integrals? With parametric equations, we can write out
the ds term in the integral as something we can actually calculate by hand. So ds ends up being equal to root[(dx/dt)^2
- (dy/dt)^2]dt. This represents the infinitesimal space between
two points along the curve C with an incremental change to the parametric variable t. Now that we know this and our integration
is in terms of t, we can now express our line integral as the integral from a to b of f(x,y)
root[(dx/dt)^2 + (dy/dt)^2]dt. So given a surface f(x,y) and a curve C, we
must take the derivatives of the parametric equations with respect to t and plug everything
into this equation. Then we integrate from the lower bound of
t, which is a, to the upper bound of t, which is b. As an example, consider the surface given
by f(x,y) = 2x and our curve C being the same as the one we discussed early with parametric
equations x(t) = t/2 and y(t) = t^2, letting t run from 0 to 6. We can go ahead and plug all of this information
into our equation for line integrals, but notice that f(x,y) is still in terms of x,
whereas we want to integrate with respect to t. We must replace the x with the appropriate
expression in terms of t. Luckily the parametric equation x(t) gives
us just that, so we can replace x with t/2. Now we must get the terms dx/dt and dy/dt,
which will just be the derivatives of the parametric equations. Differentiating x(t) = t/2 we get dx/dt = 1/2,
and for y(t) = t^2 we get dy/dt = 2t. Plugging everything in, our line integral
becomes the integral from 0 to 6 of t root[(1/2)^2 + (2t)^2]dt, which simplifies to the integral
of t root(1/4 + 4t^2)dt. To solve this we must use substitution, letting
u = 1/4 + 4t^2, which when we differentiate both sides gives du = 8t dt. We have t dt in the expression, so let’s
change that to du over 8. And now can write the line integral strictly
in terms of u, which gives us 1/8 times the integral of root u du. This is simple to integrate if we think of
the root as an exponent equal to one half. We will get 1/8 times 2/3 u^(3/2), evaluated
from the lower bound of zero to the upper bound of six. Simplifying and plugging our original expression
back in, we have 1/12 times (1/4 + 4t^2) to the 3/2 from 0 to 6. Plugging in t = 6 and subtracting from that
the expression at t = 0, we get 1/12 times (1/4 + 4*36) to the 3/2 minus 1/12 times 1/4
to the 3/2. Plugging this into a calculator gives us approximately
144.36. This is the area under the surface f(x,y)
= 2x along our curve C. We have now gone over the basic approach for
line integrals, but like many concepts in math their applications and complexity can vary. A useful property for more complex cases is
the ability to split a curve up into separate pieces for integration. Given a curve like this for example, you could
consider each part of it as separate curves, and the whole line integral can be expressed
as the sum of each. It’s also possible to consider line integrals
with respect to x and y only. These are much simpler calculations as the
ds is simply replaced with dx or dy respectively. And if we want to put these in parametric
form, we get the integral from a to b of f dx/dt dt for x and the integral from a to
b of f dy/dt dt for y. One more thing to note is that line integrals
are not limited to just the dimensions we have been using here. If we wanted a line integral of a function
f(x,y,z) we would need to use a higher dimensional ds, which ends up being root[(dx/dt)^2 + (dy/dt)^2
- (dz/dt)^2]dt. Additionally, line integrals become very useful
with vector fields. Recall that vector fields have the form F
= <P(x,y,z), Q(x,y,z), R(x,y,z)>. The line integral of this vector field along
some curve C is given by the integral along C of F dot dr, where dr is the vector <dx,
dy, dz>. Going through the dot product, we see that
the line integral can be expressed as the integral over C of Pdx + Qdy + Rdz. If our curve is parameterized by t, the dx
can be replaced with dx/dt dt, the dy by dy/dt dt, and the dz by dz/dt dt. What this line integration calculates with
vector fields can be thought of as how much of the vector field is in the direction of
the curve. This makes sense as we are dotting the vector
field, which takes the tangential component of the field to the path of the curve. In physics, this calculation can be used to
find the work done by a force along a path. An important thing to note about line integrals
is that in most cases, the path taken affects the answer. That is, even with the same starting and ending
points, a and b, the line integral over different paths can yield different answers. There is an exception that will cause line
integrals to be independent of paths, and this is when the vector field F is conservative. Remember, this means that the vector field
can be written as the gradient of some function f. This is because the line integrals of gradient
vectors have a special property to them. The integral over a curve C of the gradient
of f dot dr equals the function f evaluated at the endpoint b minus f evaluated at the
starting point a. This expression is what is known as the fundamental
theorem for line integrals. Line integrals can be a bit confusing due
to the notation, and having to take the curves into account for calculations, but we have
outlined a range of uses, and this will be important in understanding a theorem which
we will be discussing soon. Let’s check comprehension.