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Multivariable calculus
Course: Multivariable calculus > Unit 5
Lesson 5: Stokes' theorem- Stokes' theorem intuition
- Green's and Stokes' theorem relationship
- Orienting boundary with surface
- Orientation and stokes
- Orientations and boundaries
- Conditions for stokes theorem
- Stokes example part 1
- Stokes example part 2
- Stokes example part 3
- Stokes example part 4
- Stokes' theorem
- Evaluating line integral directly - part 1
- Evaluating line integral directly - part 2
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Stokes example part 3
Converting the surface integral to a double integral. Created by Sal Khan.
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At7:52, how come he integrates with respect to dr d(theta) and not r dr d(theta)?(12 votes)
Integrating with respect to r dr d(theta) is only required when changing the variables of an integral from x and y terms to polar coordinates. In this case, however, the expression for the normal vector is already in terms of r and theta so the conversion factor of r was not required.
Consider this as well though. We know that any plane with the form Ax+By+Cz+D=0 has a normal vector of Ai+Bj+Ck. Thus the plane y+z=2 (or 0x+1y+1z-2=0) will have a normal vector of 0i+1j+1k. Since the surface in question lies in that plane, it will also have a normal of 1j+1k. This vector is in terms of x y and z though, so when changing the integral to polar coordinates, r dr d(theta) is required. Therefore the double integral is of curlF(dot)(1j+1k)r dr d(theta) which is equivalent to curlF(dot)(rj+rk)dr d(theta).(22 votes)
At0:30, why is ds equal to Sr x S theta dr d theta? Thank you!(7 votes)- Here is my intuition after analyzing the problem a little, but it is by no means a rigorous proof.
S is the position vector that defines the slanted plane with respect to the origin. S can be evaluated at any (r, theta) pair to obtain a point on that plane. By picking (r, theta) as defined by the boundaries 0<r<1 and 0<theta<2*pi will limit the points to within the slanted circular area in question.
Now, if you take the derivative of S with respect r, you will obtain a vector that is tangential to the slope. This is because if you evaluate S at two points where the second r is slightly larger than the first, both of the points will be on the slanted plane. In other words, the difference between the two position vectors will be parallel to the slanted plane, in the general direction of r. Same occurs with theta. Thus, at any point on S, if you take the derivative in two orthogonal directions, you will obtain a vector normal to the slanted plane by the RHR. If you integrate using the boundaries you will obtain the area of the slanted circle.(3 votes)
- At0:59, why is the equation not rdθdr like we learned in polar coordinates?(3 votes)
- The integral areas are not the same.
An infinitesimal area on the x-y plane (or to say the z=0 plane) can be represented in cartesian coordinate as dxdy, as well as in polar coordinate as r·drdθ.
We've learned the geometric intuition behind this transformation in https://www.khanacademy.org/math/multivariable-calculus/integrating-multivariable-functions/double-integrals-a/a/double-integrals-in-polar-coordinates
In the Jacobian's point of view, if we parameterize the coordinates in the x-y plane in 3D space as [x=rcosθ, y=rsinθ, z=0], or to say V(r,θ)=[rcosθ, rsinθ, 0], the Jacobian can be calculated as the length of the cross product
V r x Vθ=
thus the length is r, and it is multiplied in the integral as r·drdθ, which is consistant with the result from the geometric intuition.i j k
cosθ sinθ 0
-rsinθ rcosθ 0
= [0, 0, r],
However in this video, we are parameterize an infinitesimal area not on the z=0 plane, but the intersection plane y+z=2, therefore it's not suitable just to multiply the r in the integral, instead, we have to evaluate the corresponding Jacobian from the cross product based on the parameterization of S(r,θ)=[rcosθ, rsinθ, 2-rsinθ].
And the result of the cross product is
S r x Sθ=
thus the length is √2 r.i j k
cosθ sinθ -sinθ
-rsinθ rcosθ -rcosθ
= [0, r, r],
Then, why don't we multiply it into the integral? Because it will be cancelled out from the denominator of n̂ as we've learned in https://www.khanacademy.org/math/multivariable-calculus/integrating-multivariable-functions/flux-in-3d-articles/a/unit-normal-vector-of-a-surface
Finally, only the numerator of n̂ remains, and it is the cross product of S r x Sθ. Therefore the whole double integral becomes F · (S r x Sθ) dθdr. Hope it will help.(5 votes)
Why is the "cross product" be thought of as two hands with fingers pointing in perpendicular directions with the thumb facing up? Is the "thumb" vector thought of in respect to the other two vectors, or one in particular?(3 votes)- shouldn't you have divide the cross product with its magnitude to get the unit vector ?(3 votes)
- i wonder do we have to divide r/2^0.5 for the rj+rk cause it needs to be unit vector?(4 votes)
- I guess that if you use n times dS (scalar), you do need to divide it, but since you're using n times dS (vector), you do not need to do it. Go to4:56, it says: n.dS (scalar) = dS (vector) = the cross product we're using (hopefully when you see it on the vid you'll see what I'm talking about).(0 votes)
- should n-hat not be a unit vector? in this it essentially came out to (0, j-hat, k-hat) but this is great than 1 magnitude. Why is this not divided by the square rout of 2?(1 vote)
- Because the length will be cancelled out from the denominator of n̂ as we've learned in https://www.khanacademy.org/math/multivariable-calculus/integrating-multivariable-functions/flux-in-3d-articles/a/unit-normal-vector-of-a-surface
Finally, only the numerator of n̂ remains, and it is the cross product of S r x Sθ. Therefore the whole double integral becomes F · (S r x Sθ) dθdr.(2 votes)
- I guess ds is the differential area element of the projected circle on the plane and from the plane equation the normal vector to the plane is (0,1,1) if we think in that fashion then in the polar coordinates rdrd(theta) is the differential area in the xy plane but our interest is the differential area on the surface of the projected circle on the plane which would be different but the end expression which sal came with is rdrd(theta) how do we justify ?(1 vote)
He didn't quite end up with this as the cross product was still present before the drd(theta). You are correct, though, in that the differential area rdrd(theta) represents a differential are upon the input space and that the differential area he showed using the cross product is the one we should be interested in.(1 vote)
- why isn't the normal vector a unit normal vector?(1 vote)
- After dot product of (curl f) in strokes theorem
If i get sign on 'z' is negative and given normal is positive 'z' then should i negative the dot produc(1 vote)
7:52
Video transcript
We're now ready to get into the
meat of evaluating this surface integral. And we really need
to re-express it in terms of a double integral
in the domain of the parameters. And the first thing
I'm going to do is rewrite this part right
over here using our parameters. And we already know that n, our
normal vector times our surface differential, can
also be written as kind of a vector version
of our surface differential that points in the same
direction as our normal vector. And this is going to
be the same thing. And we need to make
sure that we get the order on the
cross product right as the partial
derivative-- and I'm going to confirm
this in a second. The partial derivative
of the parametrization with respect to one
of the parameters crossed with the
partial derivative of the parametrization with
respect to the other parameter. And then that whole
thing-- I'm not going to take the absolute
value because I need a vector right over here--
times differentials of the parameters-- d theta
dr. And we can swap these two things around depending
on what will make our eventual double
integral easier. But we can't swap
these two things around because this would
actually change the direction of the vector. So we need to make sure
that this is popping us out in the right direction. So let's think
about the direction that the partial with
respect to r will take us. So as r increases, we're
going to be moving radially outward from the
center of our surface. Let me do this in
a different color. As r increases, we'll be
moving radially outwards. So this quantity
will be a vector that looks something like that. And then as theta
increases, we'll be going roughly
in that direction. And so if we take the cross
product of those two things-- and we could use
the right-hand rule. Take your right hand,
point your index finger in the direction of
that yellow vector. Let me make it clear-- this
is the orange vector right over here. Take your index finger
in the direction of that yellow vector. So that's my index finger,
my shakily drawn yellow index finger. Put your middle finger in the
direction of the orange vector. So my middle, you
bend it and put it in the direction of
the orange vector. And we don't care what
the other two fingers do. And then, your thumb
will be in the direction of the cross product. So your thumb will
point outward like that. That's my best
attempt at drawing it. Which is exactly the direction
we needed to point it. We need it to point
upward here in order to be oriented properly
with the direction that we are actually
traversing the path. So this is actually
the right order. If when we did this we
got the thumb pointing into it or below
the plane, then we would actually have
to swap these orders. So with that out
of the way, let's actually evaluate
this cross product. So the cross product of the
partial of our parametrization with respect to r
crossed with the partial of our parametrization
with respect to theta. I like to set up the matrix
to take the cross product. So we'll put i, j, and k
components, < like that. And then, first, I will write
the partial with respect to r. So the i-component, if you
take the derivative of this with respect to r, is just
going to be cosine theta. The derivative of
this with respect to r is just sine theta. And the derivative of
this with respect to are r is just going to be a
negative sine theta. And then we're going
to cross it with this, so the derivative of this
with respect to theta is going to be negative
r sine of theta. The derivative of
our j-component with respect to theta
will be r cosine theta. And the derivative
or our k-component-- or our z-component
with respect to theta-- is going to be-- let's see--
negative r cosine theta. Is that right? Yeah. Derivative of sine
theta's cosine theta. Yeah. So it's negative r cosine theta. And now we just evaluate
this determinant over here. Our i-component is
going to be-- so ignore that row and this column. And we get sine times
negative r cosine theta. I'll just do this
in a new color. So we're going to get negative
r-- that wasn't a new color. I'll do it in purple. We will get negative r cosine
theta sine theta minus-- well, this is going to be
a negative number. So when you subtract a negative
number, it's going to be plus. So it's going to be plus
r cosine theta sine theta. It's always nice when
things cancel out like this. This plus this is just 0. Negative of it plus
the positive of it. So that all canceled out to 0. We don't have an i-component. Now, let's go to
the j-component. And remember, we need to do our
little checkerboard pattern. So it's going to be minus j. And it's going to be-- ignore
this column, that row-- cosine theta times negative r. Cosine theta is negative
r cosine squared theta. I just multiplied those two. And then, from that I'm going
to subtract this times that. And so this times that,
the negatives cancel out. We get r sine squared theta. So this is minus--
let me make sure. Yeah, I'm going to subtract
the product of these two. The product is positive. It's going to be r
sine squared theta. This is always a hard part. You can make a lot of
careless mistakes here. And this looks like we
might be able to simplify this in a second, but I'll wait. Actually, I'll just distribute
this negative sign just for fun. So if we distribute
the negative sign, these all become positive. Helps simplify
things a little bit. And now let's worry
about the k-component. The k-component I will be
doing in purple-- well, I'll do it in blue. The k-component. Ignore this row,
ignore this column. So plus k times cosine
theta times r cosine theta is r cosine squared of theta. And then from that, I'm going
to subtract negative r sine theta times sine theta. So that's going to be
negative r sine squared theta, but I'm subtracting it. So it's going to be plus
r sine squared theta. And this looks like we're
going to simplify it as well. And so this piece
right over here, we can factor out--
let me just rewrite it. This can be rewritten as r
times cosine squared theta plus sine squared theta. Basic trig identity. That just evaluates to 1. So this is just r times j. And this over here simplifies
for the exact same reason. This is r times cosine squared
theta plus sine squared theta. This also is just 1. So this just simplifies
to r times k. This whole cross product--
all of this business right over here-- simplified
to, quite luckily, is equal to r times
our j unit vector plus r times our k unit vector. And so now we can write
our surface integral. Our original surface
integral we can write as the double integral. Or we might want to change the
order in which we integrate, but we'll give ourselves that
option a little bit later. Double integral. Now, it's going to be over our
parameter domain or the r theta domain. So it's the double integral of--
we still have the curl of f. And we're going to have
to evaluate the curl of f. So I'll just write "curl
of our vector field f" dotted with this
business, rj plus rk. And then we have
our two parameters. And we might want
to switch the order. So maybe we could
write d theta dr. And then if we do
it in this order, theta goes from 0 to 2 pi,
and r is going from 0 to 1. But if we swap these
two, then obviously, we're going to have to
swap these two as well. I'll leave you there. In the next video, we will
evaluate the curl of f. And maybe in that
video, if we have time, we will get to the finish line.