Find ∫ ∫ [1 / (x ^ 2 + y ^ 2 + 1)] dxdydz, where D is a closed region bounded by cone x ^ 2 + y ^ 2 = Z ^ 2 and plane z = 1

Find ∫ ∫ [1 / (x ^ 2 + y ^ 2 + 1)] dxdydz, where D is a closed region bounded by cone x ^ 2 + y ^ 2 = Z ^ 2 and plane z = 1

In cylindrical coordinates, the variation range of Z is √ (X & # 178; + Y & # 178;) 1] RZ / (R & # 178; + 1) | [R --- > 1] Dr
=2π∫[0--->1] r(1-r)/(r²+1) dr
=2π∫[0--->1] (r-r²)/(r²+1) dr
=2π∫[0--->1] (r-r²-1+1)/(r²+1) dr
=2π∫[0--->1] r/(r²+1) dr-2π∫[0--->1] 1 dr+2π∫[0--->1] 1/(r²+1) dr
=π∫[0--->1] 1/(r²+1) d(r²)-2π+2πarctanr
=πln(r²+1)-2π+2πarctanr |[0--->1]
=πln2-2π+π²/2

Given LG2 = 0.301, find log2 * 1000

3+0.3010=3.3010

If a can be divided by B, then a divided by B can be divided completely

Yes, since a can be divided by B, a divided by B can be divided completely
If a = 6, B = 18, then 18 / 6 = 3
In fact, a can be divided by B. a divided by B has the same formula, B / A

X ^ 2 / 4-y ^ 2 = 1 p is the minimum distance from any point on the hyperbola to the fixed point m (5,0)

It is obvious that there is a minimum value from the moment P appears on the right branch of hyperbola to M,
Using the second definition of hyperbola, let the distance from m be D, and the distance from the right collimator be x, so D / X is equal to e (eccentricity)
So d = Xe when x is the smallest, D is the smallest
Obviously, when x = A-A ^ 2 / C, the minimum number of data carried in is 5 minus 2

When n is a positive integer, n (n + 1) + 1 must be a. odd B. even C. prime D. composite

Odd number

The reciprocal of loga ^ B is logb ^ a?
Is that right? If so, is loga ^ B × logb ^ C × logC ^ a = 1? How to prove it?

Yes
loga^b*logb^a
=(logc^b/logc^a)*(logc^a/logc^b)
=1
loga^b×logb^c×logc^a
=(logm^b/logm^a)*(logm^c/logm^b)*(logm^a/logm^c)
=1
Using the formula of changing bottom
loga^b
=logc^b/logc^a

How to draw the function image of y = x2-3 | x | = 2

It is divided into x > 0 and x0)
y=x2+3x+2 (x

The distance between two points a and B representing X and negative one on the number axis is (). If the absolute value of a multiplied by B is two, then x equals ()

|x+1|
±2

The difference between marginal probability density and probability density

There's no difference in essence, the edge is just because it's in a binary environment

MATLAB in the unified diagram with stem, plot drawing, double vertical axis
There are three columns of data, the first column is the horizontal axis, the second column is drawn into needle graph with stem, corresponding to Y1 axis, and the third column is made into a line with plot, corresponding to Y2 axis
xiexie

x=0:0.25:4;
y=exp(x);
plotyy(x,y,x,y,'plot','semilogy');
hold on;
Title ('title ');
Ylabel ('coordinate axis');
Xlabel ('x-axis');