So far we've defined work in one dimension.
But let's consider the path of a particle
in more than one dimension.
And on this screen it looks like a two-dimensional path,
and here's our particle.
And what we'd like to do is define the concept of work
for this type of path.
Now let's say that at this instant
our particle is experiencing a force F.
And the particle is being displaced by a distance, ds,
which is always tangent to the path.
And what we'd like to do now is talk about the work
done for this particle.
Now what we'd like to first say is
what we're going to do is imagine that we're
going to break the path down into a bunch
of small intervals.
And let's call this the j-th point
and the j-th plus 1 point.
Then our particle will start at j
and we'll displace it delta sj to the j-th
plus 1 position on the path.
And here, let's denote the force by Fj.
And now what we'd like to do is define the work
for this particular section.
Now our concept of work is the force
in the direction of the displacement.
So what I'd like to do is I'd like to define the quantity Fj,
and I put a little parentheses upstairs-- parallel lines
upstairs.
This is the component of the force Fj
in the direction of delta sj.
Now the way we denote that is we can then
say that if I take the dot product of Fj dot delta sj,
this is the component of the force
in the direction of the motion.
Because remember, our dot product,
if we call this angle theta j, is
taking how much of one vector is in the direction
of the other times the magnitude of that displacement.
And again, when we write that component in our notation,
if we call this angle theta j, we
know that that's the magnitude of Fj times sine of-- cosine
of theta j.
Now because this is the amount of force
in the direction of the motion, that's
how we've defined work done.
And we'll symbolize the work done
in taking the particle from j to j plus 1 by the symbol Wj.
And now, this is a scalar quantity.
And when we want to find the total work, what
we have to add up is from some initial point
to some final point, how much work
is being done in taking our particle
from the initial position to the final position.
And it's just the sum for all of these j's, and let's say
we have n of them of Wj.
And right now we're going to call this W,
it's just that scalar sum, And, that's
equal to the sum j goes from 1 to n of Fj dot delta sj.
So now we'll use the vector dot product
because we already know that this is how much of the force
is in the direction of the motion.
However, this answer depends on how fine we broke up this path.
And what we've seen many times now
is that the actual work that we want, let's write this as n
because it depends on the number of paths,
is the limit as n goes to infinity of this sum of scalar
amounts of work.
Now when we take that limit, as n goes to infinity,
of this sum, j equals 1 to n of Fj dot delta sj,
this quantity is formerly what we mean by a line integral.
And we denote that with a new notation.
We'll denote a line from our initial to final
and it's a dot product of the force dotted
into the small displacement.
Now whenever we see the dot product in an integral
that distinguishes it from a normal integral,
and this is what we call a line integral.
Now again, why did we switch from our delta s's to ds's?
Because when we take the limit, as n goes to infinity,
our grid becomes finer, and finer,
and finer until we're shrinking down to a point
where we can just talk about the differential displacement, ds,
and the force acting on a point.
And that's what our line integral is.
Now, what we'll learn next is how
to calculate this line integral for a number
of important physical examples that occur
in our interest in mechanics.