We know that x, y and Z are equal difference sequence. We prove that x2 [y + Z], Y2 [x + Z], Z2 [x + y] are equal difference sequence

We know that x, y and Z are equal difference sequence. We prove that x2 [y + Z], Y2 [x + Z], Z2 [x + y] are equal difference sequence

From X, y, Z into arithmetic sequence, we get x + Z = 2Y
x^2(y+z)+z^2(x+y)=y(x^2+z^2)+xz(x+z)
=y((x+z)^2-2xz)+xz(x+z)
=y(4y^2-2xz)+2xyz
=4y^3-2xyz+2xyz
=4y^3
y^2(x+z)=y^2*2y=2y^3
Therefore, x ^ 2 (y + Z) + Z ^ 2 (x + y) = 2Y ^ 2 (x + Z), that is, x ^ 2 [y + Z], y ^ 2 [x + Z], Z ^ 2 [x + y] form an arithmetic sequence

If ∑ is the upper side of the upper hemisphere z = √ (4-x ^ 2-y ^ 2), then for the surface integral of coordinates ∫ ∫ x ^ 2dxdy, the answer to this problem is 4 π,
If it's not 4 π, what's that,

The passive plane ∑ 1: z = 0, X & # 178; + Y & # 178; ≤ 4, lower side
Then ∑ and ∑ 1 form a closed surface
∫∫(Σ+Σ1) xydydz+z^2dzdx+y^2dxdy
=∫∫∫ (y+0+0)dxdydz
Only y is left in the integrand function. Because the region is symmetric with respect to the xoz plane, y is an odd function, so the result is 0
To sum up, the above integral is 0
Next, subtract the complement ∑ 1
∫∫(Σ1) xydydz+z^2dzdx+y^2dxdy
=-∫∫ y² dxdy
Using polar coordinates
=-∫∫ r³sin²θ drdθ
=-∫[0→2π]sin²θdθ∫[0→2] r³ dr
=-(1/2)∫[0→2π] (1-cos2θ) dθ∫[0→2] r³ dr
=-π(1/4)r^4 |[0→2]
=-4π
So the original integral = 0 - (- 4 π) = 4 π
Hope to help! Ha ha!

|||(x + y + Z) dxdydz = 0 integral region: x2 + Y2 + Z2 ＜ = 1

Take apart the integrand function and use symmetry, as shown in the figure below

What is a ternary function?

Ternary function can be expressed by binary function, for example, f (x, y, z) = g (x, y) + G (y, z) + G (x, z), but the binary function is expressed in the plane coordinate system, and the ternary function is the three-dimensional coordinate system. If you draw a vector in the three-dimensional coordinate system, you can decompose and project the vector to xoy, xoz, YOZ, three planes

How to draw the image of binary function by binary function

Matlab drawing software may help you
Or, if it's a simple binary function, you can draw a few points and know its rules

What does it mean that the ternary function f (x, y, z) is an odd function with respect to x? What are the characteristics of the image?
In triple integral, why is the integral region symmetric with respect to YOZ plane, the integrand function odd with respect to x, and the triple integral 0?
Thank you first!

F (x, y, z) is an odd function with respect to X
f(x,y,z)=-f(-x,y,z)
In this case, the values of the point functions with respect to YOZ plane symmetry are all opposite
So triple integral
If the integral domain YOZ faces Chen and is an odd function with respect to x
Then the integral value of the symmetric region is also the opposite
So the total triple integral is zero

Partial derivatives of higher numbers
z=(1+xy)^y
Finding the partial derivative of Z to y in the above formula
I hope the process is more elaborate,
Of course I know how to do it, but it's not right,

Take logarithms on both sides: lnz = y * ln (1 + XY) to derive Y: Z '/ z = ln (1 + XY) + YX / (1 + XY) so: Z' = Z * [ln (1 + XY) + XY / (1 + XY)] = (1 + XY) ^ y [ln (1 + XY) + XY / (1 + XY)]

How to find the volume of a body of revolution in advanced mathematics?
Make a tangent line at a point m on the curve y = x ^ 2 (x > = 0), so that the area enclosed by the tangent line, curve and X axis is 2 / 3, and calculate the volume of the body of revolution obtained by rotating the above plane figure around the X axis

Let the abscissa of the tangent point be a, then the tangent equation is y = 2ax-a ^ 2, and the intercept on the X axis is a / 2
Area 2 / 3 = ∫ (0 to a / 2) x ^ 2DX + ∫ (A / 2 to a) (x ^ 2-2ax + A ^ 2) DX = a ^ 3 / 12, so a = 2
The tangent equation is y = 4x-4
The volume of the body of revolution v = ∫ (0 to 2) π x ^ 4DX - ∫ (0 to 1) π (4x-4) ^ 2DX = 16 π / 15

Volume of high revolution
The area of the plane figure surrounded by y = x / 1, y = x, and X axis, and the volume of the body of revolution obtained by rotating the plane figure around the axis

Area = 1 / 2 AOB + integral (X: 1 - & gt; + infinity) 1 / X DX = 1 / 2 + LNX (1 - & gt; + INF) does not exist (Does X have an upper bound?)

How to solve the problem of high number rotator
Through point P (1,0), make the tangent line of X-2 under the root of parabola y. the tangent line, parabola and ox axis form a plane figure. Try to find (1) the area of the plane figure; (2) the volume of the body of revolution of the plane figure rotating around ox axis

Derivation y '= 1 / (X-2 under 2 * radical)
Set the tangent point (a, A-2 under the root)
A-2 / (A-1) = 1 / (2 * X-2)
A = 3
Tangent point (3,1)
Straight line y = 0.5x-0.5
Integral again
Cone parabola for volume integration
V cone = Π ^ 2
V parabola = 2 to 3 (Π y ^ 2) DX = 2 to 3 (Π (X-2)) DX = Π (0.5x ^ 2-2x) | 2 to 3 = 0.5 Π
The volume of the plane figure rotating around the ox axis is v = Π ^ 2-0.5 Π