Looking for someone who has done a square: Use Stokes' theorem to find the circulation of F (yz, xy, xz) on the boundary of the square with vertices (0,0,2), (1,0, 2), (1,1,2), and (0,1,2), oriented counter-clockwise as viewed from above.
Since the z coordinate is constant for all of the points, you know that all coordinates are coplanar to the xy plane. So the normal vector to the surface is n = (0,0,1) The surface that you're integrating over is analagous to the area of that square. I hope this helps! - From a fellow Calc 3 student
Very nice explanation, thanks from Tanzania (East 🌍)
Thank you so much, sir
Looking for someone who has done a square: Use Stokes' theorem to find the circulation of F (yz, xy, xz) on the boundary of the square with vertices (0,0,2), (1,0, 2), (1,1,2), and (0,1,2), oriented counter-clockwise as viewed from above.
Since the z coordinate is constant for all of the points, you know that all coordinates are coplanar to the xy plane. So the normal vector to the surface is n = (0,0,1)
The surface that you're integrating over is analagous to the area of that square. I hope this helps!
- From a fellow Calc 3 student
Was there a way to directly calculate that integral without using Stokes' theorem?
no
@@yvonneho876 there is you can use a line intergral
@@lucasl4644 i got C 🥲
Y limits how it is?
Brilliant, thank you!
Great thanks!
In duration of 4:10 ---4:20 , why we take the projection of the surface only on xy plane
Not on x-z or y-z plane
We can but if we project it in xz the we have to find parametric form of y and substituite to find the answer
From india that's a nice question
Horse shit, a length is not equal to a surface area. Why do you omit the units of the integrals?