∫∫ x ^ 2dydz + y ^ 2dzdx + Z ^ 2dxdy, where the surface is outside the upper part of x ^ 2 + y ^ 2 + Z ^ 2 = 1

∫∫ x ^ 2dydz + y ^ 2dzdx + Z ^ 2dxdy, where the surface is outside the upper part of x ^ 2 + y ^ 2 + Z ^ 2 = 1

If the complement plane ∑ 1: z = 0, X & # 178; + Y & # 178; ≤ 1, the plane and the original surface form a closed surface. The Gauss formula ∫ ∫ x ^ 2dydz + y ^ 2dzdx + Z ^ 2dxdy = 2 ∫ ∫ ∫ (x + y + Z) dxdyz can be used. Because the integral region is symmetrical about xoy plane and xoz plane, the integrals of X and y are all 0, and the integrand function is only z = 2

∫∫ x ^ 2dydz + y ^ 2dzdx + Z ^ 2dxdy, where s is the outer side of spherical (x-a) ^ 2 + (y-b) ^ 2 + (z-c) ^ 2 = R ^ 2

If the variables X and y satisfy the inequality constraints {x-2y + 1 ≤ 0, 2x-y ≥ 0, X ≤ 1,}, then the point P (2x-y, x + y) denotes the area of the region
A,3/4 B,4/3 C,1/2 D,1
Because I don't have the answer to this question in my hand, and the result of my calculation is 2, please help me to verify whether my conclusion is correct? Is my calculation wrong, or is the option wrong?

The point satisfying the constraint condition is inside the triangle ABC (including the boundary),
Among them, a (1,1), B (1,2), C (1 / 3,2 / 3), a (1,1), B (1,2), C (1 / 3,2 / 3),
Therefore, the region corresponding to point P is the triangle a1b1c1, where A1 (1,2), B1 (0,3), C1 (0,1), A1 (1,2), B1 (0,3), C1 (0,1), A1 (1,2), B1 (0,2), B1 (0,3), C1 (0,1), C1 (0,1), A1 (,
The area is 1 / 2 * 2 * 1 = 1
Choose D

Given the function f (x) = x2-2x, the area of the region formed by the point (x, y) satisfying the condition f (x) + F (y) ≤ 0f (x) − f (y) ≥ 0 is ()
A. 4πB. 2πC. 3π2D. π

∵ f (x) = x2-2x ∵ the constraint condition f (x) + F (y) ≤ 0f (x) − f (y) ≥ 0 can be transformed into (x − 1) 2 + (Y − 1) 2 ≤ 2 (x − 1) 2 − (Y − 1) 2 ≥ 0, and its corresponding feasible region is shown as follows: its area is: 12 ·π· (2) 2 = π, so D is selected

Given that the image of the function y = (K-3) x2 + 2x + 1 has an intersection with the X axis, then the value range of K is

When △ = B ^ 2-4ac = 4-4 (K-3) ≥ 0, the image of function y = (K-3) x ^ 2 + 2x + 1 has intersection with X axis
So, K ≤ 4

Find the maximum or minimum value of the function y = x2-2x-3 in the following range
（1）0

Draw the graph of the function with the opening upward and the intersection points with the X axis are - 1 and 3,
The X coordinate of the lowest point is 1, and the Y coordinate is - 4,
(1) At 0

If the function f (x) = (2a-1) ^ x is a decreasing function on R, then the value range of a is

0

Let f (x) = (2a-1) x + B be a decreasing function on R, then the range of a is______ ．

∵ f (x) = (2a-1) x + B is a decreasing function on R, ∵ 2a-1 < 0, the solution is a < 12

If the function f (x) = (2a + 1) ^ x is a decreasing function, then the value range of a is

The exponential function has 0

If the function f (x) = x ^ 2 + (3a + 1) x + 2a is a decreasing function on (negative infinity, 4), then the value range of A

Because the opening of the function is upward, we can see that the left side of the axis of symmetry is decreasing by associating with the image. Therefore, a decreasing function on (negative infinity, 4) means that the axis of symmetry is greater than or equal to 4, that is - (3a + 1) / 2 ≥ 4, and a ≤ - 3 is obtained
Remember: the function of the number of lines is a very easy way to use, that is, when the problem combined with the function of the image, this is conducive to the solution of the problem. Come on!