- 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
Evaluating line integral directly – part 1
Showing that we didn’t need to use Stokes’ Theorem to evaluate this line integral. Created by Sal Khan.
Want to join the conversation?
- why didnt we have to use another parameter, ‘r’ in this?(7 votes)
- Because for this side of Stokes’ theorem we are only taking a line integral and not a surface integral. A line integral only requires a parametrization in one variable since it is the integral across a curve and not a surface, which requires two variables for its parametrization. The curve parametrization and the surface parametrization are actually related- the curve parametrization (with respect to f instead of theta) is (cos f)*i + (sin f)*j +(2-sin f)*k, whereas the surface parametrization (with respect to r and f instead of r and theta) is (r*cos f)*i + (r*sin f)*j + (2-r*sin f)*k. The surface parametrization is different from the curve parametrization in that it multiplies all theta terms by r. Doing so creates the surface out of the curve- as Sal explained in previous videos, for any value of theta we can take different values of r and include every point on the interior of the curve by adding r as a parameter.(7 votes)
- At, If a question provides us with a surface integral, we could therefore evaluate the surface integral by calculating the line integral? Is this the concept of stokes theorem? What would be the equation in this case? 0:35(3 votes)
- Atsir said only one parameter has to be used.What if the circle is not essentially a unit circle (eg x^2+ y^2=4)?. Then r can’t be considered as 1. Thanks in advance!! 1:07(1 vote)
- if it wrong? there should be a minus sign before y^2 sinTheta(1 vote)
- Look closely, you can see negative times negative equals positive.(1 vote)
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.