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### 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 4

Finding the curl of the vector field and then evaluating the double integral in the parameter domain. Created by Sal Khan.

## Want to join the conversation?

- When he integrated the area at then end, because it is now in polar coordinates, shouldn't the respect to area change from "dxdy" to "r dr d(theta)" ?(18 votes)
- The normal vector n that he calculated actually comes out equal to the Jacobian that you're trying to include(as you should) in the change of variables. In surface integration, the norm of the normal evaluated by the cross product of S(r)×S(thita) is equal to the Jacobian(4 votes)

- What is a good application of this theorem?(12 votes)
- when using cylindrical coordinates to evaluate the integral shouldn't the differentials be r*dr*d(theta)?(6 votes)
- The first response (by RyanMillerM) is correct- usually when changing from rectangular to polar coordinates we multiply the function we are integrating by r- this is called a Jacobian (not a Jacobin, the guys who started the French Revolution :P ). The basic concept of a Jacobian is that it is the determinant of the matrix of the partial derivatives of the new function's two variables with respect to the function's old variables. There will always be 4, 9, etc partials to calculate, so a square invertible matrix is possible as a way to represent these partials. When you make a transformation from rectangular to polar coordinates, the Jacobian comes out to be r. This is a very simplistic explanation, but the point of all this was: Sal chose to represent this surface with a parametrization that happened to use r and theta and look eerily similar to a polar transform. However, as RyanMillerM stated, this parametrization was not an actual transform between coordinate systems, so the parametrization does not require a Jacobian.(8 votes)

- How is the partial derivative of z^2 with respect to y = 0 ? Isn't z a function of y?(4 votes)
- We parameterized z as a function of y (2-rsin(theta)) for the unit circle, which is another field. When taking the cross product, we are taking it in terms of F which is -y^2 i + x j + z^2 k. Therefore we never parameterized the z as a function of y for F.(9 votes)

- I did not understand how we can use just one surface of the figure and evaluate it instead of considering the whole figure(2 votes)
- Actually, what we cared at the beginning was only the line integral of a closed loop (any closed loop), and the Stokes theorem allows us to instead evaluate the surface (any surface) enclosed by that loop.

This works because in both cases you are measuring similar things. In the case of the line integral, you are measuring how the field does "work" over the loop, and since it's a closed loop, you are measuring a kind of rotational work (it has to rotate for the loop to close on itself).

In the case of the surface, you are measuring the rotational in the surface enclosed by the loop. And the rotational itself already has the information about how much the field wants to exert "rotational work". So you end up measuring the same thing from 2 different perspectives.(3 votes)

- at 1.26, why does Sal does "-j" instead of just "j"?(2 votes)
- How come most of the comments in this playlist are 5 years old? Is no one watching this anymore?(1 vote)
- at the end, didnt he forget that cos 0=1 thus giving the integral 2r^2 in addition to the 2(pi)(r^2) he got?(1 vote)
- but it cancels out when you evaluate 2r^2cos(theta) in theta=2*pi(1 vote)

- Still confused as to what the answer tells us...the total mass leaving the surface? And if we wanted to find the total mass leaving by using stokes theorem shouldnt we do three separate integrals because there are three separate parts?(1 vote)
- The result is pi, a positive scalar which means you get an overall positive curl for the vector field F along the boundary curve C.(1 vote)

- So this video describes how stokes' thm converts the integral of how much a vector field curls in a surface by seeing how much the curl vector is parallel to the surface normal vector. Fine.

In my maths book however there is another application of this where stokes is used twice in a row to convert

- first from a surface integral of this kind of the dot product of a vector field and the normal vector of a surface to a line integral of the surface boundary, i.e the opposite of what we see here,

- secondly from the line integral of the surface boundary back to a surface integral but this time the curl of the vector field is dotted with just the ^k unit vector and dA, an area segment of the surface's projection onto the xy-plane.

Does anyone reqognize this?

Does this mean that all information about how a vector field rotates in a surface is contained in the boundary of it's projection on the xy-plane?

That's dope in that case but does anyone have an intuitive explanation?(1 vote)

## Video transcript

- [Instructor] We're
now in the home stretch. We just have to evaluate the curl of f and then this dot product and then evaluate this double integral. So let's work on the curl of F. So the curl of f is going to be equal to, and I just remember it as the determinant, so we have our i, j, k components, and it's really you could imagine it's the del operator crossed
with the actual vector. So the del operator, I'll
write this in a different color just to ease the monotony,
so this is partial with respect to x,
partial with respect to y, partial with respect to z, and then our vector field, I copied and pasted it right over here. It is just equal to negative
y squared, is our i component, x is our j component, and Z
squared is our k component. And so this is going to be equal to, this is going to be equal to i, is going to be equal to i times the partial of Z squared
with respect to y. Well, there's the Z
squared is just a constant with respect to y so
the partial of Z squared with respect to y is
just going to be zero, so this is going to be zero. Minus the partial of x with respect to z. Well, once again this is just a constant when you think in terms of z, so that's just going to be zero. So that's nice simplification, and then we're gonna have minus j, we need our little checkerboard patterns, we put a negative in front of the j, minus j and so we'll
have the partial of x, the partial of z squared
with respect to x, that's zero again, and then minus the partial of negative y
squared with respect to z, well that's zero again,
and then finally we have our k component, k, so plus, plus k, and k, we're gonna have the
partial of x with respect to x, well that actually gives us a value that's just gonna be one minus the partial of negative y
squared with respect to y. So the partial of negative y
squared with respect to y is negative two y and we're subtracting that, so it's going to be plus, plus, two y. So curl of f simplifies to just, all of this is just zero up here, is just one plus two y times
k or k times one plus two y. And so if we go back
to this right up here, if we go back up to
that, we're going to get let me re-write the
integral so zero to one and that's our r, our
r parameter is gonna go from zero to one, theta is
gonna go from zero to two pi. And now curl of f has simplified to, and I won't skip any steps
although it's tempting, it's one plus two y, and actually
instead of writing two y, let me write it in
terms of the parameters. We saw it up here, y was r sine theta, if I remember correctly,
right, y was r sine theta. So let me write y that way. Two times r sine theta k. And we're gonna dot this, we're gonna take the dot product of that
with this right over here, with r times j plus r
times k, d theto d r. And so we take the dot product, this thing only has a k component, the j component is zero, so
when you take the dot product with this j component
you're gonna get zero. And neither of them you actually
even have an i component. And so the inside is
just going to simplify to this piece right over
here is going to simplify to, we're just gonna have to
think about the k components, cause everything else is
zero, so it's gonna be r times this and we're
done! So it's gonna be r plus two r squared sine
theta, d theta d r, d theta d r and, once again, theta
goes from zero to two pi and r goes from zero to one. And now this is just a
straight-up double integral. We just have to evaluate this thing. And so, first we take the antiderivative with respect to theta,
so the antiderivative with respect to theta is going to give us, so this is going to be giving, so we're going to focus on theta first, so the antiderivative of r
with respect to theta is just r theta, you can just do r as a constant, and then the antiderivative of this, antiderivative of sine of theta
is negative cosine of theta. So this is gonna be negative
two r squared cosine of theta. And we're gonna evaluate
it from zero to two pi. And then we have the outside integral, which I will, I'll re-color
in yellow, re-color in yellow, so we'll still have to
integrate with respect to r and r's gonna go from zero to one. But inside right over
here, if we evaluate all of this business right
over here at two pi, we get two pi r, two pi r,
that's that right over there, minus... Cosine of two pi is just one. So it's minus two r
squared and then from that, we're going to subtract from that, we're gonna subtract this evaluated zero. Well r times zero is just zero, and then cosine of zero is one. So it's just minus two r squared, or negative two r squared, negative two r squared. And at this negative and this negative, you get a positive, and but then you have a negative two r squared and
then a plus two r squared it's just going to cancel out, that and that cancel out, and so this whole thing
has simplified quite nicely to a simple definite integral,
zero to one of two pi, two pi r dr, and the
antiderivative of this is just going to be pi r squared, so we're just gonna evaluate pi r squared from zero to one, when you
evaluate it at one, you get pi; when you evaluate it
at zero, you just get zero, so you get pi minus
zero, which is equal to, and now we deserve a drumroll
'cause we've been doing a lot of work over many
videos, this is equal to pi. So just to remind ourselves
what we've done over the last few videos, we
had this line integral that we were trying to figure out, and instead of directly
evaluating the line integral, which we could do and I
encourage you to do so, and if I have time, I might
do it in the next video, instead of directly
evaluating that line integral, we used Stokes theorem to say, oh we could actually instead
say that that's the same thing as a surface integral over
a piecewise-smooth boundary over piecewise-smooth surface that this path is the boundary of,
and so we evaluated this surface intergal and
eventually, with a good bit of, little bit of calculation,
we got to evaluating it to be equal to pi.