Surface integral, Gauss formula Using Gauss formula to divide the area of an upper hemisphere, add a bottom surface to make it a closed surface At this time, the problem arises. After dividing the closed surface, do you want to subtract the integral of the auxiliary surface or add the integral of the auxiliary surface? One book is added and the other is subtracted. I'm confused Is there any difference between the outer side integral and the upper side integral?

Surface integral, Gauss formula Using Gauss formula to divide the area of an upper hemisphere, add a bottom surface to make it a closed surface At this time, the problem arises. After dividing the closed surface, do you want to subtract the integral of the auxiliary surface or add the integral of the auxiliary surface? One book is added and the other is subtracted. I'm confused Is there any difference between the outer side integral and the upper side integral?

To subtract, subtract is to add the second type of surface integral on the surface
When calculating the second kind of surface integral, it involves the side of the surface you added. Take the upper side to subtract the positive side to subtract, take the lower side to subtract the negative side to add
The calculation of triple integral also involves the side of closed surface
So when using Gauss formula, we should consider two positive and negative, the first is the positive and negative of Gauss formula, and the second is the positive and negative of the second kind of surface integral converted into double integral
The answers above are not accurate. They all go beyond the explanation of dividing the area of the second type of curve into double integral. This is also the reason why many people are puzzled by such questions
Supplement:
Outside, inside is the Gauss formula to consider the positive and negative judgment
The upper side and the lower side are judged when the second type of curve area differentiation is double integral

Surface integral of the second type of higher number
The integrand is xdydz + ydzdx + zdxdy, the integral surface is helicoid, x = u * cosv, y = y * SINV, z = C * V (0
I know how to do it. I just want to know what the answer is

The title is wrong
Y = y * SINV, should be y = u * SINV
The method is to transform it into the first type surface integral
Write it in the form of (pcosa + qcosb + rcosy) ds, then rewrite it with parameter equation
The key is to write the normal vector of S and the area element of DS under the parametric equation,
Please refer to the knowledge content of space surface. Related to Jacobian determinant
Specific process did not offer reward points, but also difficult to play, slightly
The reference answer is: 2C * (π ^ 2) * [root (b ^ 2 + C ^ 2) - root (a ^ 2 + C ^ 2)]
It's a bit complicated, but can you understand it

Let ∑ be the cylinder x ^ 2 + y ^ 2 = a ^ 2 at 0

Because the integral surface satisfies f (x, y) = x ^ 2 + y ^ 2 = f (y, x) = y ^ 2 + x ^ 2 = a ^ 2
So ∫ x ^ 2ds = ∫ y ^ 2ds
Then the original integral = (1 / 2) ∫ (x ^ 2 + y ^ 2) ds = (a ^ 2 / 2) ∫ DS = (a ^ 2 / 2) (2 π ah) = π a ^ 3H