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Current time:0:00Total duration:7:44

In the last few videos, we
evaluated this line integral for this path right over here
by using Stokes' theorem, by essentially saying that
it's equivalent to a surface integral of the curl
of the vector field dotted with the surface. What I want to do
in this video is to show that we didn't have
to use Stokes' theorem, that we could have just
evaluated this line integral. And the thing to
keep in mind is, in this case, it's
kind of a toss up which one is
actually simpler to do. But Stokes' theorem is
valuable because sometimes, if you're faced with
the line integral, it's simpler to
use Stokes' theorem and evaluate the
surface integral. Or sometimes, if you're faced
with the surface integral, it's simpler to
use Stokes' theorem and evaluate the line integral. So let's try to figure
out this line integral. And hopefully, we're going
to get the same answer if we do everything correctly. So the first thing we want to
do is find a parametrization for our path right over here,
this intersection of the plane y plus z is equal to 2. And essentially, you can
imagine this hollow pipe that intersects the xy-axis
at the unit circle that goes up and down forever. And we get this path right over
here with this orientation. And since we're only
parameterizing a path, it's only going to deal
with one parameter. And so let's think
about it a little bit. We've done this
many times before, but it doesn't hurt to go
through the exercise again. So that is our y-axis. This is our x-axis. That is our x-axis. And the x and y values
are going to take on every value on
the unit circle. And then the z value
is going to tell us how far above the
unit circle we needed to be to actually
be on this path. So x and y are going to take
on all the values on the unit circle. And we've done that
many times before. The easiest way to think
about that is let's introduce a parameter called
theta that essentially measures the angle with the
positive x-axis. And theta, we're just going
to sweep it all around, all the way around
the unit circle. So theta is going to
go between 0 and 2 pi. So 0 is less than theta, which
is less than or equal to 2 pi. And in that situation,
x is just going to be-- this is just the unit
circle definition of trig functions-- it's just going to
be equal to cosine of theta. y is going to be sine of theta. And then z, how
high we have to go, we can use this constraint
to help us figure it out. y plus z is equal
to 2, or we could say that z is
equal to 2 minus y. And if y is sine
theta, then z is going to be equal to
2 minus sine theta. And so we're done. That's our parametrization. If we wanted to write it as
a position vector function, we could write r, which is
going to be a function of theta, is equal to cosine of
theta i plus sine of theta j plus 2 minus sine of theta k. And now we're ready
to at least attempt to evaluate this line integral. We need to figure
out what F dot dr is. And to do that, we need
to figure out what dr is. And we just have to
remind ourselves, dr is the same thing as
dr/d theta times d theta. So if you take the derivative
of this with respect to theta, derivative of cosine theta
is negative sine theta i. Derivative of sine
theta is cosine theta plus cosine theta j. And the derivative
of 2 minus sine theta is going to be negative
cosine theta k. And then all of that
times-- we still have this d theta
to worry about. This is dr/d theta. Let me write this down. So then we still have to write
our d theta just like that. And now we're ready to take
the dot product of F with dr. So let's think about
this a little bit. I'll write dr in this color. F dot dr is going to be
equal to-- so we look first at our i components. We have negative y squared
times negative sine of theta. That's going to be--
well, the negatives are going to cancel out. So you're going to get-- So you're going to get y
squared times sine theta, that's from the i component,
plus-- now we're going to have x
times cosine theta And then we're going
to have plus z squared times negative cosine theta. So that's going to be negative
z squared times cosine theta. And then all of
that times d theta. All of that business
times d theta. And if we're actually going
to evaluate the integral, if we're going to
evaluate it over that path that we care about, now we
have it in our theta domain. So we could just say this is
a simple integral from theta going from 0 to 2 pi. Actually, we're not fully in
our theta domain just yet. We still have it expressed
in terms of y, x's, and z. So we have to express
those in terms of theta. So let's do that. So that's going to be equal to
the integral from 0 to 2 pi. And actually, let me give
myself some more space because I have a feeling
this might take up a lot of horizontal real estate. This is going to be the
integral from 0 to 2 pi. y squared, well, y
is just sine theta. So it's going to be sine squared
theta times another sine theta. So this is going to
be sine cubed theta. Let me write it in a new color. I'll write it in blue. So this is sine squared theta
times another sine theta, so this is going to
be sine cubed theta. Let me color code it. So that's sine cubed theta. Put some parentheses here. And then x is cosine theta
times a cosine theta. So this is going to be
plus cosine squared theta. And then z squared,
this is actually going to get a
little bit involved. So let's think about
it, what z squared is. I'll do it up here. z squared is just going
to be 4 minus 4 sine theta plus sine squared theta. There's going to be negative
z squared times cosine theta. So negative z squared is going
to be equal to negative 4 plus 4 sine theta minus
sine squared theta. So that's negative
z squared, and we're going to multiply that
times cosine theta. And I'll do it in orange. So all of this business
right over here is going to be equal to cosine
theta times all of this, cosine theta times all
of this right over here. So it's going to be negative
4 cosine theta plus 4 cosine theta sine theta and
then minus cosine theta sine squared theta. And looks like we're done, at
least for this step, d theta. And so now we just have
to evaluate this integral. So we saw actually setting up
our final integral, getting to a simple one-dimensional
definite integral, is actually much
simpler in this case. But the actual integral
we have to evaluate is a little bit
more complicated. We might have to break out a
few of our trig identity tools in order to solve it
properly, but we can solve it. But I'll leave you there. In the next video,
we'll just work on actually evaluating
this integral.