Some thoughts on the integral of line surface

Some thoughts on the integral of line surface

@ (Calculus)
For line area points, consider the following path:
1. The first curve integral: Direct calculation, the emphasis is on the conversion of DS, with partial derivative formation coefficient compensation. such as X=x (t), y=y (t), →x′2 (t) +y′2 (t) ‾‾‾‾‾‾‾‾‾‾‾‾‾√dt x = x (t), y = y (t), \rightarrow \sqrt{x ' ^2 (t) +y ' ^2 (t)}dt
2. The spatial curve is also the same, the direct calculation, is also the coefficient compensation, x=x (t), y=y (t), z=z (t) →x′2 (t) +y′2 (t) +z′2 (t) ‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾√dt x = x (t), y = y (t), z = z (t) \ri Ghtarrow \sqrt{x ' ^2 (t) +y ' ^2 (t) +z ' ^2 (t)}dt
3. The second type of curve integral is the integral to the coordinates. The plane can be projected to the axis of the calculation, into a unary integral. But the topic often makes a dollar difficult to seek, that is, the integrand complex, by seeking partial derivative rather concise, closed curve with green formula. Note that as long as the closed curve can be used, not to say that the path is independent, path-independent, with the green formula, the result is 0 good felling. Note that the denominator is not zero, otherwise with the method of digging to do, in general this case, the first independent of the path, but the middle hole can be solved. So, dig a hole, assuming the outer curve is reversed. Some books on the solution is to dig the hole set to the clockwise direction, so that the outer layers into the inner layer, so that the formation of the region (the left side, the right is outside) in the integration of 0, equivalent to the original based on the addition of a reverse, now to subtract the reverse, so the total result is the inverse of the internal curve. Privately think this is a little slow, faster, directly think of the outer curve ring up the area, cut off the small area with a hole, the other is the green formula can be solved, get 0.The area that is crossed out is restrained, and nature is directly the inverse of the curve. This small area circle has the technique, namely: can and the problem is the accumulation function relations, brings in can simplify, removes the denominator is 0 of that point, directly to the axis integral can, this very simple, is two one dollar integral.
4. Closed curves in space, the second curve integral is also interesting, Stokes grand appearance. The surface area of the curve is divided. There are two kinds of methods, Dydz,dxdz,dxdy, or Cosα,cosβ,cosγcos\alpha,cos\beta,cos\gamma. correspond to two kinds of curved area respectively. Pay special attention to the plane cut a curved surface, the first type curved area is more simple, because the plane of the Cosα,cosβ,cosγcos\alpha,cos\beta,cos\gamma is constant, very good, and then projected to the coordinate plane, only once.
5. First type surface area: This is what I call the coefficient of compensation. But this is not the same as the coefficient compensation. Think it is a curved surface, is curved, so the area is larger than the projection surface, we want to, where to project it. Assuming that the Xoy face is cast, then Z disappears and disappears. In X, Y. Z=z (x, y), the coefficient of compensation is when projecting toward the Xoy face: 1+z′2x+z′2y‾‾‾‾‾‾‾‾‾‾‾