Welcome to a lesson on line integrals
and differential form.
Line integrals are sometimes written in differential form.
So if we let our vector field equal F of r of t,
where each component of F, P, Q, and R,
is written as a function of t,
we can show where the differential form comes from
starting with the line integral of a force field
or vector field,
as discussed in the previous video,
as we see here in the first line.
So again what this is saying is
the vector field F is rewritten
as a function in terms of t,
and then differential r
is equal to r prime of t, dt.
So here's our vector field
dotted with r prime of t.
And we calculate this dot product we rewrite:
x prime of t, y prime of t, and z prime of t
as dx/dt, dy/dt, and dz/dt.
We can see that we integrate with respects to t
that these dt's would simplify out
leaving us with the differential form of a line integral.
We have the integral from a to b
of Pdx plus Qdy plus Rdz.
Let's go ahead and look at our example,
we want to evaluate this line integral
in differential form along the curve
given by r of t,
on the closed interval from zero to four.
So we know that x of t is equal to 2t,
we know that y of t is equal to t squared,
and we know that z of t
is equal to t to the 1/2.
Now let's go ahead and determine
differential x, differential y, and differential z.
Differential x is going to be equal to 2dt,
differential y is going to be equal to 2tdt,
and differential z is going to be equal to
1/2t to the negative 1/2 dt.
Let's rewrite this line integral
in terms of t now.
The integration will be from zero to four,
we're integration with respects to t.
So now we have y times dx.
Well y is equal to t squared,
and dx is equal to 2dt.
Dt is already here so we'll just multiply it by two.
Plus z squared times dy.
Well z squared would be,
t to the 1/2 squared, that's going to be t,
times differential y.
Differential y is 2tdt.
Again, the dt is already here
so we'll write down the 2t.
And lastly we have xdz.
Well x is equal to 2t.
And dz is equal to 1/2t to the negative 1/2, dt.
So we have 1/2t to the negative 1/2, dt.
So now we just need to simplify this
and then find the anti-derivative with respects to t.
So I have 2t squared,
plus another 2t squared.
Here we'll have 2 times 1/2, that's one,
and then t to the first
times t to the negative 1/2,
will be t to the 1/2.
Let's go ahead and finish this on the next page.
Let's combine our like terms.
So we have 4t squared plus t to the 1/2.
So we'll have four times t to the third over three,
or 4/3t to the third,
plus here we'll have, t to the 3/2,
divided by 3/2,
or 2/3t to the 3/2.
So when t is four,
we'll have 4/3 times four to the third, plus,
2/3 times four to the 3/2.
And then when t is zero, these are both zero.
This is going to give us 256/3,
plus four to the 3/2
is equal to eight,
eight times 2/3 will give us 16/3.
So the sum would be 272/3.
And we'll stop here for this video,
thank you for watching.