Welcome to a lesson on line integrals
in R three.
So hopefully you already watched the video
on line integrals in R two.
This is just an extension of the same idea
into three dimensions.
So if we have a continuous function,
F of X, Y, Z,
in a region containing a smooth space curve
given by the vectored valued function as we see here
where T is on the closed interval from A to B,
then we can evaluate this function of three variables
along the space curve with respects to S
by evaluating this integral here
with respects to T.
So we first convert the function F of X, Y, Z
into a function of T
by replacing X with X of T,
Y with Y of T,
and Z with Z of T
from the components of the space curve,
and then we replace the differential S
with the magnitude of R prime of T, DT,
and as we mentioned in the last video,
replacing differential S
with the magnitude of R prime of T, DT,
should be somewhat familiar to us
because we did do this
when we determining arc length
of a space curve earlier in the semester.
Let's take a look at our first example.
We want to integrate the following line integral
from the point zero, zero, zero
to the point two, five, four.
So we first needed to find R of T
for this path,
so if we let the X component be two T,
the Y component be five T,
and the Z component be four T,
this would be the path of our line integral
as long as T is on the closed interval
from zero to one.
This also tells us that X prime of T
would be equal to two,
Y prime of T would be equal to five,
and Z prime of T would be equal to four.
Let's go ahead and set this up in terms of T.
Limits of integration would be from zero to one.
Now we're going to go ahead and rewrite the given function
in terms of T.
So we'll have two times X times Y,
so we'll have two times two T
times five T,
plus three times Z,
which would be three times four T.
We're going to multiply this by the magnitude
of R prime of T.
So we're going to have the square root of two squared
plus five squared,
plus four squared, DT.
Let's go ahead and simplify this.
So we're going to have two times two T,
times five T.
That's going to be 20 T squared.
This'll be 12 T.
This'll be the square root of four
plus 25 plus 16.
That's going to be the square root of 45.
Let's go to the next page.
Here we have a common factor of four.
Let's factor out four, square root 45.
We'll be left with five T squared
plus three T.
The square root of 45 simplifies
to three, square root five.
So we have 12, square root five.
Now we'll go ahead and integrate.
We'll have five times T to the third
over three, or 5/3 T cubed
plus three times T squared over two,
or 3/2 T squared.
So when T is one,
we'll have 5/3 plus 3/2.
When T is zero, these are both zero.
Common denominator here would be six.
So we have 12 square root of five
over one times,
this would be 19 over six,
and this simplifies here.
Now we can multiply.
We have two times 19.
That's 38, square root five,
all over one.
So the result would be 38, square root five.
Let's take a look at another example.
Here we want to integrate the following line integral,
along R of T equals cosine T,
sine T, T,
which is a helix on the closed interval
from zero to pi.
So notice in this problem,
they've already given us the vectored valued function
for the path of integration.
So here's X of T,
Y of T,
and Z of T.
So we know that X prime of T
would be equal to negative sine T,
Y prime of T is equal to cosine T,
and Z prime of T would be equal to one.
So let's go ahead and convert our function
to a function of T,
and then replace differential S
with the magnitude of R prime of T, DT.
We know the interval for T
is from zero to pi.
We're going to have two Y times sine Z.
That's going to be two times sine T,
and sine Z is equal to sine T
because Z is equal to T,
and we're going to have the magnitude
of R prime of T,
so we'll have sine squared T
plus cosine squared T
plus one square root is one.
So let's take a look at this square root here.
All of this is equal to one,
so this becomes the square root of two.
Let's go ahead and factor out
the square root of two.
We'll write this as two, sine squared T, DT,
and let's go ahead and finish this
on the next page.
Now here, you'll have to apply the power reducing formula
for sine squared T.
Here's the formula here.
Notice the twos here simplify out.
So we just have one minus cosine two T, DT.
Then of course, here we have a U substitution.
So when converting to U,
we'd have an extra factor of 1/2.
So we'll have T minus,
this'll be 1/2, sine two T.
Now, go ahead and replace T with pi
and then zero,
and when T is pi,
we'll have pi minus,
the sine of two pi is zero minus,
and then when T is zero,
these are both zero again.
So we're left with square root of two
times pi,
so the result would just be
the square root of two pi.
And that's going to do it for this video.
I hope you found these examples helpful.