hello why a+b is rational ? I though the formular for contradiction is not p->q then should be like P: irrational+rational = Q:irrational? since Q remains the same in formular????
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
