This is an example in the book: finding the volume of the solid surrounded by a rotating paraboloid A: X & # 178; + Y & # 178; = R & # 178; and B: X & # 178; + Z & # 178; = R & # 178

This is an example in the book: finding the volume of the solid surrounded by a rotating paraboloid A: X & # 178; + Y & # 178; = R & # 178; and B: X & # 178; + Z & # 178; = R & # 178

Since it is an example, it has an answer. Why do you want to ask?

Calculate the second type surface integral ∫ (x ^ 3 + e ^ ysinz) dydz-3x ^ 2ydzdx + zdxdy, where s is the lower side of 1-x ^ 2-y ^ 2 in the lower hemisphere z = - radical
Detailed process ~ ~ thank you~~~

Let P = (P = (x ^ 3 + e ^ 3 + e ^ ysinz, q = (x ^ 3 + e ^ ysinz, q = -3x ^ 2Y, q = - 3x ^ 2Y, r = Z, then # ຣ∫∫∫∫∫∫∫∫∫ (3x ^ 2-3x ^ 2, surfaces': s': S': z = 0, S + s': s': s': s': s': s': s': s': s: let P = (P ∫∫∫∫∫∫∫∫∫ (3x (3x (3x ^ 2 ^ 2 (3x ^ 2 ^ 2-3 ^ 2-3 ^ 2-3 ^ ∫∫ dxdydz = 2 π / 3, (if the integrand is 1 in triple integral, the integral is equal to the volume of the integral region). For the surface s', the integral = 2 π / 3-0 = 2 π / 3 because DZ = 0, z = 0, the integral expression is replaced by integral = 0

Let the surface integral be plane X / 4 + Y / 3 + Z / 2 = 1, then ∫ (1 / 2x + 2 / 3Y + Z) ds=

Let the surface be plane x + y + Z = 1, then ∫ ∫ ∑ XDS=

It is proved that the analytic function on a certain region is a real function and it must be a constant
Can it be proved by C-R equation?

Of course
Let f = u (x, y) + IV (x, y)
V = 0
And because of the C-R equation, UX = vy (UX is the partial derivative of u to x)
So UX = 0
Similarly, uy = 0
So u = constant
So f = u = constant
Why don't you go to math and ask?

It is proved that the function f (z) is analytic in the domain D, and | f (z) | is constant in D. then f (z) is constant in D

Let f (z) = u (x, y) + I V (x, y)
If | f (z) | = 0, then: F (z) = 0
If | f (z) | ≠ 0,
And | f (z) | is constant in D, which means: {U (x, y)} ^ 2 + {V (x, y) ^ 2} = constant ≠ 0
The partial derivative of U (x, y) to X is expressed by u '(x)
There are: 2uu '(x) + 2vv' (x) = 0 (1)
2uu'(y) +2vv'(y) =0 (2)
Since f (z) is analytic, u and V satisfy the C --- r condition. U '(x) = V' (y), u '(y) = - V' (x)
Substituting (1) and (2) results in:
uu'(x) - vu'(y) = 0 (3)
uu'(y) + vu'(x) =0 (4)
Because: (* *) u ^ 2 + V ^ 2 ≠ 0, the solution from (3) (4) is: u '(x) ≡ 0, u' (y) ≡ 0
Thus we can deduce: u (x, y) ≡ C1. (constant)
In the same way: V (x, y) ≡ C2. (constant)
Thus we know: F (z) ≡ C1 + IC2
The proposition is thus proved

∫∫| XY-1 | DXDY d = 0 ≤ x ≤ 20 ≤ y ≤ 2 how to write piecewise function after opening absolute value?
The answer is: 3 / 2 + 2ln2

for xy ≤ 1
|xy-1| = -xy+1
for xy> 1
|xy-1| = xy-1
xy ≤ 1
=> x≤ 1/y for 0≤y≤2
I =∫∫ |xy-1| dxdy
= ∫∫ -xy+1 dxdy ( 0≤x≤y ) + ∫∫ xy-1 dxdy (y

The sign of a number is positive, and the absolute value is equal to - 3 and - 4. What is the product of the absolute values of two numbers

It turns out that the landlord added a comma
-The absolute value of 3 is 3, and the absolute value of - 4 is 4,
So 3 * 4 = 12

Is closed interval monotone function integrable? How to prove it?

To prove integrability is to prove that the integral is not infinite, so that a definite value can be obtained;
There must be & nbsp; Max & nbsp; and & nbsp; min for monotone functions on closed intervals
The integral theorem is: Min × [interval length] = & lt; integral value = & lt; max × [interval length]
&So: closed interval monotone function must be integrable

How to find monotone interval of function with absolute value
For example: y = 2x + 9, y = x2 + 2x + 3

The first thing to do is to turn on the absolute value
y=｜2x+9｜
x> = - 9 / 2 y = 2x + 9 y '= 2 > 0 monotone increasing
x0
y=｜x2+2x+3｜=x2+2x+3
Then we can judge the monotone interval