- 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
Starting to apply Stokes theorem to solve a line integral. Created by Sal Khan.
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- x^2 + y^2 = 1 is a equation of circle isn't? how to relate it with a pole ?(4 votes)
- It is indeed the equation of a circle when restricted to the x-y plane, but when extended into the z dimension, it forms a cylinder. For any value of z, the equation gives you the same (circular) set of x-y points.(11 votes)
- Prove or disprove:
Let f be a continous increasing function on the interval (a,b). Let prime numbers p,q such that p*q is in the interval (a,b) and p < q. Then F(pq) = p*F(q), for F an antiderivative of f.(1 vote)
- Note to those who are wondering how to solve it directly it is in the next videos.
P.S sorry i knew it was question section but this was important to get attention as i had to search a whole lot of videos for its proof, hope you understand.(1 vote)
- Is is possible to use Stoke Theorem on a flat surface? For example, close curve, C integration of (x^2 + 2y + sin(x^2)dx + (x + y + cos(y^2))dy ). The C is a contour on xy plane which formed by x=0 (from 0,0 to 0,5), y= 5-x^2 (from 0,5 to 2,1) and 4y = x^2 (from 2,1 to 0,0).
I consider (x^2 + 2y + sin(x^2)) determine my vector field i component while (x + y + cos(y^2)) for my j component and k component = 0 of a vector field F.
I double integrate the (curl of F) dy from x^2/4 -> 5-x^2 then dx from 0->5.
The answer i get is 27.083 but the answer is 20/3(1 vote)
- Yes that can be done. Sal applies stokes theorem to a flat surface parallel to x-y plane in this video:
- We keep on using (curl dot normal) ds and just curl dot ds in these integrals kind of interchangably, but don't we expect normal and ds (as a vector) to be orthogonal? wouldn't that mean that these values would in general be different?(1 vote)
- the normal vector is orthogonal to the surface, so n and dS as vectors are pointing in the same direction.(0 votes)
Let's now attempt to apply Stokes' theorem And so over here we have this little diagram, and we have this path that we're calling C, and it's the intersection of the plain Y+Z=2, so that's the plain that kind of slants down like that, its the intersection of that plain and the cylinder, you know I shouldn't even call it a cylinder because if you just have x^2 plus y^2 is equal to one, it would essentially be like a pole, an infinite pole that keeps going up forever and keeps going down forever so it would never have a top or a bottom but we slice that pole, with y plus z is equal to 2, to get where they intersect we get our path C. We also have this vector field defined in this way