How to prove the symmetry theorem of double integral This theorem is very good, but I want to know the principle, which is the process of proving this theorem

Symmetry theorem of double integral: if the integral domain D is symmetric about the origin and f (x, y) is an odd or even function of X, y, then ∫ ∫ f (x, y) DXDY (integral on domain D) = 0 (when f is an odd function of X, y, that is, f (- x, - y) = - f (x, y)) DXDY (integral on domain D) = 2 ∫ ∫ f (x, y) DXDY (...)

A circle passing through the origin is connected with the origin by the two ends of one diameter. Are these two lines perpendicular? Is there any theorem?

Theorem of circle angle. The circle angle of an arc is equal to half of its center angle
The diameter is a straight line, and the angle with the center of the circle as the vertex is 180 degrees
However, it has nothing to do with whether it passes through the origin. The two ends of any diameter passing through a circle are connected with another point on the circle. The triangle formed by three straight lines is a right triangle, that is, the two straight lines beyond the diameter are perpendicular to each other. This is the deformation application of the circular angle theorem

What is the rotational symmetry of integral variables?

If the letters in an algebraic formula rotate in a certain order, and the resulting algebraic formula is identical with the original algebraic formula, then the algebraic formula is called the rotation symmetry formula about these letters

How to understand the symmetry theorem of double integral fundamentally

If the integral domain is symmetric about the X axis, the integrand is an odd function about y, which is equal to 0
The even function of the integrand with respect to y is equal to twice
If the integral domain is symmetric about the Y axis, the integrand is an odd function about X, which is equal to 0
The even function of the integrand with respect to X is equal to twice
If the integral domain is symmetric about the X, Y axis, the integrand is an odd function about X, y, which is equal to 0
The even function of integrand with respect to X and Y is equal to twice
It should be easy for you to remember like this. That's how I remember

Double integral (square of X - 2x + 3Y + 2) DXDY D: square of X + square of Y

Trigonometric substitution x = RCOs α y = rsin α 0 ≤ R ≤ a, 0 ≤ α ≤ 2 π, | J | = R, then I = ∫ [0,2 π] d α∫ [0, a] (R ^ 2 * cos ^ 2 (α) - 2rcos α + 3rsin α + 2) RDR = ∫ [0,2 π] [a ^ 4cos ^ 2 (α) - (2 / 3) a ^ 3 * cos α + A ^ 2 * sin α + A ^ 2] d α = a ^ 2 * ∫ [0,2 π] d α = 2 π a ^ 2

Finding the minimum value of y = SiNx / 2 + 2 / SiNx (0 〈 x 〈π)
1. The minimum value of y = SiNx / 2 + 2 / SiNx (0 〈 x 〈π)
2. When x > 0, f (x) = 2x / x ^ 2 + 1
Two questions,

Y=sinx/2+2/sinx (0
Please check the function

00,(2/sinx)>0
Y=(sinx/2)+(2/sinx)
=(sin^2x+4)/(2sinx)
2sinx≤2
SiNx = 1,2sinx max = 2
Y = (SiNx / 2) + (2 / SiNx) min = 2.5

Given that the coordinates of any point P (x, y) on the circle C: X & # 178; + Y & # 178; - 2y-1 = 0 make the inequality x + y + m ≥ 0 tenable, find the value range of M?
It's better to have a picture.

Trigonometric substitution
x²+y²-2y-1=0
That is X & # 178; + (Y-1) &# 178; = 2
Let x = √ 2cosa, y = 1 + √ 2sina
Then x + y + M = √ 2 (Sina + COSA) + 1 + M = 2Sin (a + π / 4) + m + 1
The minimum is M-1
∴ m-1≥0
∴ m≥1

Inequality (2Y + 3) & #178;

4y²+12y+9

How to prove the symmetry theorem of double integral This theorem is very good, but I want to know the principle, which is the process of proving this theorem