1: Because of its directionality, it has the property of even zero odd degree. When F(x) is an even function, if σ is symmetric about the corresponding surface, then one part takes+and the other part takes-and the result is F(x)-F (-x) = F (x)-F (x). If you take a part-the result is f (x)-f (-x) = f (x) = 2f (x), and the two parts are equal, then the formula of 2: 3 1 can be superimposed. For σ, the normal vector in the form of z = z(x, y) is n = {-z' x, -z'. 1} then ∫∫ _ (σ) pdydz+qdzdx+rdxdy = ∫∫ _ (d) {p (-z 'x)+q (-z 'y)+1} dx. P/？ x+？ Q/？ y+？ R/？ Z) dxdydz-∫ _ (σsum) pdydz+qdzdx+rdxdy (σ sum), if the original surface can't be closed in a closed space, the Gaussian formula can't be used directly, and the area needs to be closed after adding several surfaces, for example, adding several (σ sum) surfaces, then the Gaussian formula can be used. Finally, we should pay attention to reducing the integrals corresponding to those surfaces (σ and σ). 4. If there are singular points on the integrand function when digging a hole, the Gaussian formula cannot be directly applied, and a small space r=ε needs to be added, which is enough to contain all the internal singular points. Then when the radius ε tends to 0, this part of the corresponding integral should also be subtracted when using the Gaussian formula, so there is ∫∫ _ (σ) = ∫∫ _ (ω)-∫∫ _ (ε) 5: instead of if the equation of the integrand function F is on σ, then