WEBVTT
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CHRISTINE BREINER: Welcome
back to recitation.
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In this video, I'd like
to do two problems that
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ask us to determine the flux of
a vector field along a surface.
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So the first one is
I'd like you to find
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the outward flux of the
vector z comma x comma
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y through the piece of
the cylinder that's shown.
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So it's just shading
the cylindrical part.
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So it's a cylinder of
radius a, and you're
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taking the piece in the
first octant up to height h.
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So that's the first question
I'd like you to answer,
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is the flux of this vector
field through that piece
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of the cylinder,
the outward flux.
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And then I'd also like you to
find the outward flux of this
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vector field x*z comma y*z comma
z squared through the piece
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of the sphere of radius
a in the first octant.
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So again, it will be
in the first octant,
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like this one is in
the first octant.
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It'll be the piece
of the sphere that
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sits in the same part of
three-dimensional space
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as the piece of the cylinder
we're looking at here.
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So what I'd like
you to do, again,
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is just find the
flux-- in both cases,
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the outward flux-- of the vector
listed through the surface that
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is listed.
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And then when you feel
comfortable and confident
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with your answer, you can
bring the video back up
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and I'll show you how I did it.
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OK, welcome back.
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So again, what
we're trying to do
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is determine flux of a vector
field through a surface.
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So I am going to determine
first what the normal is here,
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and then what F dot
n is here, and then
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using the fact
that I know F dot n
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and I know dS in a
good parametrization,
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I'm going to be
able to calculate
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the integral quite simply.
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So let's point out
first, that the normal
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at any point on the
surface is going
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to be equal to x comma
y comma 0 divided by a,
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where a is, again, the radius.
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Because in the normal, I
know there's no z-component.
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And I actually know it's
the same as the normal would
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be on a circle of radius a.
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And so this vector x comma
y has length a obviously,
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so that's why I'm dividing by a.
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So when I look at the
normal and I dot it with F,
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let's see what I get.
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F dotted with the normal
is going to be x*z plus x*y
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divided by a, right?
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So I just took x dotted
with x times z, y times z,
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and 0 times y, I add
those up, and I still
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have to divide by my a.
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So that's actually the vector
dotted with the normal.
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Now, what's a natural
parametrization to use here
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is obviously the
cylindrical coordinates,
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because I'm on a cylinder.
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The radius is fixed.
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But what I'm interested in is
changes in theta and changes
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in the height.
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So I'm going to be
interested in d theta and dz,
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and I need to understand
what dS in this case is.
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And it's just a d theta dz.
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So let me write
down what I'm going
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to need to completely determine
the rest of number one
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is I'm integrating
over the surface.
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And I'll put the
bounds momentarily.
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Actually, I should
also point out,
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I'm going to write x
and y in terms of theta,
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because now I know
what they are.
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x in terms of theta is a cosine
theta, and y in terms of theta
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is a sine theta.
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So when I simplify the
expression F dot n,
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I get z times cosine theta
for my first component.
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And I get a times
cosine theta sine
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theta for my second component.
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So let me maybe point
out again how I got that.
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Let me come over here.
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x is a cosine theta.
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So I get z times a cosine
theta divided by a.
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So I get z times cosine theta.
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x again is a cosine
theta. y is a sine theta.
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So I get a squared
cosine theta sine theta.
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And I divide by a, so I get
a single a cosine theta sine
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theta.
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And then my dS, as I mentioned
before, is a d theta dz, right?
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And so now I just have to
figure out the bounds in theta
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and the bounds in z.
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Well, the bounds
in z are very easy.
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The bounds in z
are simply 0 to h,
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and so I'm going to put
those here on the outside.
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And then the bounds in theta.
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Maybe it's helpful to come
over and look at my picture.
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This direction, in the
x-direction, is theta equals 0.
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When I swing around
to the y-direction,
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I'm at theta equals pi over 2.
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So I need to go from
0 to pi over 2, right?
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So let me come back over here.
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OK, so now, really,
I have a couple
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of constants that are letters,
but everything now, I'm
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ready to integrate.
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So a is just a constant,
and h is a constant,
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and z and theta
are my variables,
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and I can actually do this
integration quite simply.
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I'm not going to do
it because I know
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this is at this point something
we already know how to do,
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but I'll give you the answer.
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So let me write down the answer.
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You get a h squared over
2, plus a squared h over 2.
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So that's the solution
you actually get.
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So again, I mean, you have to
integrate in d theta first.
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So you have to deal with this,
you have to deal with this,
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and this uses a
good trig identity.
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You can use the fact
that 2 sine theta cosine
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theta is sine 2 theta.
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I'll give you that
little hint, and then you
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can figure it out from there.
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And then you integrate in z,
and you evaluate from 0 to h.
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So that's the solution
to number one.
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So now, let's look
at number two.
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In number two, we
have to again figure
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out-- we have our vector field
and we know our surface is
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a piece of a sphere.
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And so if we're
going to parametrize
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the surface of a
sphere, we know we want
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to use d theta and d phi, OK?
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And let me point out first,
that again, the normal is going
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to be in some ways similar to
what we saw on the cylinder,
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but now instead of x comma y
comma 0, because it's a sphere,
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it's going to be x comma
y comma z divided by a.
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So again, as before, let me
just point out the normal
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that I'm going to be using
is x comma y comma z,
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all divided by a.
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So I'm going to dot F with n
and look at the surface integral
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with respect to dS--
d capital S there--
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and I'll see what I get.
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So again, I know how I'm going
to parametrize this sphere.
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I already mentioned it,
but let me say it again.
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It's going to be in
theta and phi, right?
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Because we have a
constant radius.
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We're on a sphere of radius a.
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So I don't need to change rho.
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It's a two-dimensional thing.
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So theta is varying
and phi is varying.
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So let's see what we get first.
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Let me do a little work
and see what we get when
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we look at F dotted with n.
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So let me first point out
that F dotted with n looks
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like it's x squared
z plus y squared
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z plus-- just to
make this obvious-- z
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squared z, all divided by a.
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x squared plus y squared plus
z squared is a squared, right?
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So it's actually a squared
times z divided by a.
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So it's just a times z, right?
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So far, all I've done was
I dotted F with the normal,
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and I knew the fact that x
squared plus y squared plus z
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squared was a squared, because
I was on a sphere of radius a.
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So a squared times z
divided by a is a times z.
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And now, if I want to use
the right coordinates,
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think about theta and phi.
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z is a cosine phi.
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So F dot n in the
coordinates I'm interested in
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is going to be a
squared cosine phi.
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So let me get out of the
way so you can see that.
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So that's what our
F dot n will be.
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And now, if we're going to
figure out the flux, of course,
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it's the integral of F dot n dS.
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And let's remind
ourselves what dS is.
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dS is going to be a squared
sine phi d theta d phi.
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You saw this in
lecture, actually, also.
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So this should look familiar.
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And so now I just have
to integrate F dot n
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dS over the right bounds
for theta and phi.
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So let's determine
what those are.
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I'll put everything
together, and we'll
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determine what those are.
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So I've got F dot n dS.
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That's going to be a to the
fourth sine phi cosine phi
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d theta d phi.
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And let's think about
what is the picture
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that I need in terms of the
first octant of a sphere.
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Maybe I should draw a
quick picture over here
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so we can remember
what that looks like.
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So it's going to
be-- this is not
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going to be the greatest
drawing ever-- but it's
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something like this.
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And so I've got pieces of
a circle at each level.
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I've got a piece
of a circle, right?
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If this is the x-direction,
this is the y-direction,
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and this is the z-direction,
theta is going from 0-- again--
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to pi over 2, and phi is going
from 0 to pi over 2, right?
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So they're both going
from 0 to pi over 2.
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So hopefully, you were able
to get this far at least,
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in terms of figuring
out the flux of F
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through that piece
of the sphere.
00:09:54.490 --> 00:09:57.100
And I'm, again, just going
to write down the solution,
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and then you can check your
answer against the solution.
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And I got a to the
fourth over 4 times pi.
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So this whole solution
is a to the fourth
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divided by 4 times pi.
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So you can check
your solution there.
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Again, I want to
point out, what we
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did in both of these
problems is we were trying
00:10:17.110 --> 00:10:21.070
to compute the flux of a certain
vector field through a surface.
00:10:21.070 --> 00:10:24.860
And if you'll notice, what
I actually did in this case
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is I kept the parametrization in
terms of the x, y, z variables
00:10:30.090 --> 00:10:33.040
first, and then I put it
in the parametrization
00:10:33.040 --> 00:10:34.550
of theta and phi after.
00:10:34.550 --> 00:10:37.540
And that made it a little easier
to hang on to and figure out
00:10:37.540 --> 00:10:38.040
what it was.
00:10:38.040 --> 00:10:41.470
Because notice, that x, y,
and z become very complicated
00:10:41.470 --> 00:10:43.070
in theta and phi.
00:10:43.070 --> 00:10:44.780
I have to write a lot
more down, I guess.
00:10:44.780 --> 00:10:47.590
And then simplify
things more carefully.
00:10:47.590 --> 00:10:50.760
This way, it was very obvious
that I got an a squared times
00:10:50.760 --> 00:10:52.030
z in the numerator.
00:10:52.030 --> 00:10:54.030
So sometimes it's
a little easier
00:10:54.030 --> 00:10:57.650
to compute the F dot n in the
initial x, y, z variables,
00:10:57.650 --> 00:11:00.320
and then change it to the
appropriate parametrization
00:11:00.320 --> 00:11:01.887
for the surface
you're looking at.
00:11:01.887 --> 00:11:03.220
So that's what we did, actually.
00:11:03.220 --> 00:11:06.140
In both cases we
computed F dot n,
00:11:06.140 --> 00:11:08.550
we put it in the
right parametrization,
00:11:08.550 --> 00:11:10.340
and then we had to
figure out what dS was.
00:11:10.340 --> 00:11:12.220
We had to make sure
we knew dS, and then
00:11:12.220 --> 00:11:14.440
we just had to integrate
over the appropriate bounds
00:11:14.440 --> 00:11:15.670
for our parameters.
00:11:15.670 --> 00:11:18.230
And that's giving us the
flux across the surface.
00:11:18.230 --> 00:11:20.915
So I think that's
where I'll stop.