Let the tangent plane of the point (x, y, z) on the surface ∑: x ^ 2 / A ^ 2 + y ^ 2 / b ^ 2 + Z ^ 2 / C ^ 2 = 1 be π, and calculate the surface integral ∫ ∫ ∑ 1 / λ DS, where λ is the distance from the coordinate origin to π

Let the tangent plane of the point (x, y, z) on the surface ∑: x ^ 2 / A ^ 2 + y ^ 2 / b ^ 2 + Z ^ 2 / C ^ 2 = 1 be π, and calculate the surface integral ∫ ∫ ∑ 1 / λ DS, where λ is the distance from the coordinate origin to π

Step 1, find the equation of the tangent plane π at the point (x, y, z) on ∑:
The equation of ∑: F (x, y, z) = x ^ 2 / A ^ 2 + y ^ 2 / b ^ 2 + Z ^ 2 / C ^ 2-1 = 0
The normal vector of π n = {f'x, f'y, f'z} = {2x / A ^ 2,2y / b ^ 2,2z / C ^ 2}
The equation of π is: 2x / A ^ 2 * (x-x) + 2Y / b ^ 2 * (Y-Y) + 2Z / C ^ 2 * (Z-Z) = 0
The results show that XX / A ^ 2 + YY / b ^ 2 + ZZ / C ^ 2-1 = 0
Step 2, find the distance from the coordinate origin to π
Using the distance formula from point to plane, we can get λ = 1 / √ (x ^ 2 / A ^ 4 + y ^ 2 / b ^ 4 + Z ^ 2 / C ^ 4)
Step 3, find DS in the surface integral ∫ ∫ ∑ 1 / λ ds:
The equation of integral surface: F (x, y, z) = x ^ 2 / A ^ 2 + y ^ 2 / b ^ 2 + Z ^ 2 / C ^ 2-1 = 0
That is Z = ± C √ (1-x ^ 2 / A ^ 2-y ^ 2 / b ^ 2) ★
Using the derivative formula of implicit function, z'x = - f'x / f'z = - x C ^ 2 / Za ^ 2
z’y=- F’y/F’z=-y c^2/ zb^2
Then DS = √ 1 + (z'x) ^ 2 + (z'y) ^ 2) DXDY can be obtained
=c^2/┃z┃*√(x^2/a^4+y^2/b^4+z^2/c^4) dxdy▲
Step 4, find the surface integral ∫ ∫ ∑ 1 / λ ds:
First, we have to divide the integral surface of this integral into two parts
Then the two curved areas are divided into two double integrals by the formula
Note that the integral regions of the two double integrals are the same elliptic regions, that is, the upper and lower ellipsoids are projected onto the xoy surface respectively
However, the difference between the integrand of the two double integrals is only one sign, as shown in {}
So the surface integral is equal to zero