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 = .
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 .
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.