The curve represented by {1 - (y-4) ^} under the equation lxl-1 = root sign is two semicircles

The curve represented by {1 - (y-4) ^} under the equation lxl-1 = root sign is two semicircles

|x|-1=√[1-(y-4)²]
When | x | ≥ 1, i.e. x ≤ - 1, or X ≥ 1, the equation can be reduced to
(|x|-1)²= 1-(y-4)²
(|x|-1)²+ (y-4)²=1
When x ≤ - 1, the equation can be reduced to (- x-1) & sup2; + (y-4) & sup2; = 1, that is, (x + 1) & sup2; + (y-4) & sup2; = 1,
In this case, the equation represents the left half of the circle with (- 1,4) as the center and 1 as the radius;
When x ≥ 1, the equation can be reduced to (x-1) & sup2; + (y-4) & sup2; = 1,
In this case, the equation represents the right half of the circle with (1,4) as the center and 1 as the radius;
Therefore, the equation | x | - 1 = √ [1 - (y-4) & sup2;] denotes two semicircles

What is the image of y = 1 / x, y = - 1 / X & # 178; and so on? According to what?
Is not only the point method? Know quickly say, thank you!
Answer quickly, where are the people?

It's not only the point tracing method, but also the derivation method, inflection point method, asymptote method and so on
The point tracing method is more accurate for linear function and discrete function. For continuous function, the point tracing method is equivalent to using finite points to approximate infinite points
The image with y = 1 / X is a hyperbola in the first three quadrants
The image of y = - 1 / X & # 178; can be obtained by finding the reciprocal of parabola y = - X & # 178; and digging out the origin
You can draw it yourself

Higher number, the problem of integral of curve with arc length
∫ [l] x ^ 2ds, where l is the circle where the sphere x ^ 2 + y ^ 2 + Z ^ 2 = R ^ 2 intersects the plane x + y + Z = 0

The sphere x ^ 2 + y ^ 2 + Z ^ 2 = R ^ 2 and the plane x + y + Z = 0 are rotationally symmetric about three coordinate axes, so ∫ (L) x ^ 2ds = ∫ (L) y ^ 2ds = ∫ (L) Z ^ 2ds
So, ∫ (L) x ^ 2ds = 1 / 3 ×∫ (L) (x ^ 2 + y ^ 2 + Z ^ 2) ds = 1 / 3 ×∫ (L) R ^ 2ds
The plane passes through the center of the sphere, so the radius of circle L is r, so
∫(L)x^2ds＝1/3×R^2×2πR＝2πR^3/3

Higher number - integral of arc length curve
By using the definition of the curve of arc length, it is proved that if the curve arc L is divided into two smooth curve arcs L1 and L2, then
∫[L]f(x,y)ds=∫[L1]f(x,y)ds+∫[L2]f(x,y)ds

By definition, integral is the limit of the division and summation of curves. In this way, a division D1 of L1, together with a division D2 of L2, is a division of L. then take the limit~

How to judge the side of the surface accurately in the surface integral of coordinates

If you want to use Gauss formula, it may involve adding one side. At this time, it depends on the angle between the normal vector of the added side and the normal vector of the specified positive side. The acute angle is positive and the obtuse angle is negative

I use the surface integral of coordinates and Gauss formula to get different results
∫∫ zdxdy + xdzdy + ydxdz, s is the outside of the cylinder x ^ 2 + y ^ 2 = 1 which is cut by the plane z = 0 and z = 3. There is s under that ∫∫ and I have calculated two methods respectively. The answers are different. The Gauss method is not the correct answer. I don't know why. I hope you can tell me the correct process of solving the problem with Gauss method. Thank you

In this paper, we calculate ∫∫∑ xdydz + ydzdx + zdxdy, where ∑ is the outer rule of the part of cylindrical surface x ^ 2 + y ^ 2 cut by plane z = 0 and z = 3. We use Gauss formula: complementary surface ∑ 1: z = 3, upper side, ∑ 2: Z = 0, lower side then ∫∫ (∑ + Σ 1 + Σ 2) xdydzdx + ydzdxdy = 3 ∫∫∫∫ Ω DV = 3 * π

What do you think of the upper and lower sides when calculating the surface integral of coordinates

A memory method:
Imagine that you have a cube in front of you. You see a cube on the top, front and right
The rest are negative
The practical meaning is: the normal vectors of the three surfaces are the positive directions of the three axes
In practical application, you will compare the normal direction of the surface to be judged with the normal vectors of the three surfaces. If the direction follows the normal direction of a certain surface, it is the positive side. Otherwise, it is the negative side
If you want to judge the upper and lower sides, don't care where the surface is. First of all, look at the normal direction of the surface given by the topic. For example, the topic is given downward. You know that the top of the cube in memory is positive, while the top normal direction is upward. Therefore, the surface is the lower side
Do not you specify the normal direction of the surface? It is usually said that the normal direction of the surface points to the positive direction of the z-axis

Why can surface / curve integrals be directly brought into the integral domain ∑ equation? But double / triple integrals can't,

This is a good question. This is related to the integral region. Take the circle x ^ 2 + y ^ 2 = 1 as an example. If the circle is used as a closed curve for curve integration, then the integral curve is x ^ 2 + y ^ 2 = 1. If it is a double integral, note that the integral region is the interior of x ^ 2 + y ^ 2 = 1, not the circle itself. If it is strictly expressed by equation, it should be x ^ 2 + y ^ 2

Why can curve equation be substituted by surface integral, while double integral and triple integral can be substituted?

If you look carefully, the curvilinear equation of the curvilinear integral and the equation of the double integral are the same circle. The integrand domain equation of the curvilinear integral is X & # 178; + Y & # 178; = A & # 178; the integrand domain equation of the double integral is X & # 178; + Y & # 178; ≤ A & # 178; only the boundary domain of the circle can be used for the double integral

What are multiple integrals, curve integrals and surface integrals
What is double integral
What is triple integral
What is the solution to the curve integral of arc length
What is the curvilinear integral of coordinates
What is the division of curved area of area
What is the solution to the curvilinear area of coordinates
What is the relationship and difference between multiple integral, curve integral and surface integral

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It's really necessary to understand this. I'll tell you what I know
Double integral is the key to volume
Triple integral is the key to the quality of solid
The first kind of curvilinear integral is a method to find the mass of arc
The second kind of curvilinear integral is work seeking
The first kind of surface integral is the method to find the surface quality
The second kind of surface integral is the solution of surface flux
As for the relationship, multiple integrals are the general term. Surface integrals and curve integrals are applications of multiple integrals. To be exact, they are applications of double and triple integrals. Curve integrals and surface integrals are parallel, and their respective fields belong to multiple integrals
In physics, it is estimated that they will have other applications. These are just some aspects. I hope they can help you. Let's turn off this problem