Rethinking on the area division of the first type curved surface

Rethinking on the area division of the first type curved surface

@ (Calculus)

Some problems, look complex, but very good solution. In the same way, some problems look simple, but they are hard to do. Give an example of the calculation of the area of the first type of curvature.

The solution of the first type curved area based on three things to do: projection generation calculation

There is no logical order between three things, and whoever wants to do it first.

The goal is to be converted into double integrals. The curved surface is too curved, we need to be in a relatively straight scene in order to score points.

Or it can be fixed into a sequence of your own liking: one-shot, two-generation, three-calculation

Set surface s:x2+y2=a2, (0≤z≤a) x^2+y^2 = a^2, (0\leq z \leq a), ask: i=∫∫sdsx2+y2+z2 I = {\int\int}_s \frac{ds}{x^2+y^2+z^2}

Analysis: Line integral we know that we can substitute for the integrand, if we can simplify the problem, then you are not polite to enter.

I=∫∫sdsa2+z2 I = {\int\int}_s \frac{ds}{a^2+z^2}

Integral surface: Cylindrical, it is very easy to imagine.

Before you can calculate, there is one less step: projection.

Where to vote.

It is obvious that the surface is projected onto the Xoy surface as a line, and the line is not a double integral. Of course this is not the real reason to deny the projection to Xoy plane. The real reason is that the projection to the coordinate plane does not allow coincident points .

If the projection to the Xoy, compressed only left the line, you want to have how many coincident points ah, are squeezed in a line. The number is not coming up already.

It is possible to xoz, or yoz planes, but, according to the above, it is not permissible to overlap, so first you have to think about the symmetry of the surface about Xoz (now choose Projection to xoz).

Obviously about xoz symmetry, and the integrand is an even function about Y (there is no y at all in the expression, natural f (x, y, Z) = f (x,-y,z)).

So the problem is:

I=2∫∫s1dsa2+z2,y>0 I =2 {\int\int}_{s_1} \frac{ds}{a^2+z^2}, y>0

S1 S_1 is +y side of the plane.

Since it is a projection, it means that the curved part of the abruptly is flattened, is not to compensate for something.

Yes, the coefficient of compensation.

How the coefficients are compensated.

ds=1+y′2x+y′2z‾‾‾‾‾‾‾‾‾‾‾‾√dxdz</