Let me start from the beginning, from the basic one-dimensional integral to the second kind of surface integral.
Research on multiple integral algorithm;
Single integral (definite integral): there is only one independent variable y = f(x)
When the integrand function is 1, it is the length of a straight line (large degree of freedom).
∫(a→b) dx = L (straight line length)
When the integrand function is not 1, it is the area (rule) of the graph.
∫(a→b) f(x) dx = A (plane area)
In addition, the definite integral can also find the regular volume of the rotating body, as follows
Disc method: v = π ∫ (a → b) f? (x) dx
Shell method: v = 2π ∫ (a → b) xf (x) dx.
The calculation methods include substitution integral method, polar coordinate method and so on. I don't know if I have more contact with definite integral.
∫(α→β) ( 1/2)[A(θ)]？ Dθ = A (plane area in polar coordinates)
Double integral: there are two independent variables z = f(x, y)
When the integrand function is 1, it is the area (large degree of freedom).
∫(a→b) ∫(c→d) dxdy = A (plane area)
When the integrand function is not 1, it is the volume of the graph (rule) and the volume of the rotator.
∫(a→b) ∫(c→d) dxdy = V (volume of rotating body)
The calculation methods include rectangular coordinate method, polar coordinate method and Jacobian substitution method.
Polar transformation: {x = rcosθ
{ y = rsinθ
{α≤ θ≤ β, maximum range: 0 ≤ θ ≤ 2π.
∫(α→β) ∫(h→k) f(rcosθ，rsinθ) r drdθ
Triple integral: there are three independent variables u = f(x, y, z).
When the integrand function is 1, it is the volume and the volume of the rotator (maximum degree of freedom).
∫(a→b) ∫(c→d) ∫(e→f) dxdydz = V (volume of rotating body)
When the integrand function is not 1, it has no geometric meaning but only physical meaning.
The calculation methods include rectangular coordinate method, cylindrical coordinate slicing method, cylindrical coordinate projection method, spherical coordinate method, Jacobian replacement method and so on.
Polar coordinate transformation (slice method): {x = rcosθ
{ y = rsinθ
{ z = z
{ a ≤ z ≤ b
{ 0 ≤ r ≤ z？
{α≤ θ≤ β, maximum range: 0 ≤ θ ≤ π.
∫(a→b) ∫(α→β) ∫(0→z？ )f(rcosθ，rsinθ，z) r drdθdz
Particularly, when f(x, y, z) can be expressed as f(z),
∫∫∫∫∫Ωdxdy Dz =∫ (A→ B) F (Z) [∫∫ Ddzdxdy] Dz =∫ (A→ B) F (Z) (cross-sectional area Dz) Dz.
The expression of cross-sectional area Dz is a function of z.
Polar change (cylindrical coordinates): {x = rcosθ
{ y = rsinθ
{ z = z
{ h ≤ r ≤ k
{α≤ θ≤ β, maximum range: 0 ≤ θ ≤ 2π.
∫(α→β) ∫(h→k) ∫(z？ →z？ )f(rcosθ，rsinθ，z) r dzdrdθ
Polar coordinate change (spherical coordinate): {x = rsinφcosθ
{ y = rsinφsinθ
{ z = rcosφ
{ h ≤ r ≤ k
{a ≤ φ ≤ b, maximum range: 0 ≤ φ ≤ π.
{α≤ θ≤ β, maximum range: 0 ≤ θ ≤ 2π.
∫(α→β) ∫(a→b) ∫(h→k) f(rsinφcosθ，rsinφsinθ，rcosφ) r？ Sin? φ drdφdθ
Double integral can be simplified by symmetry:
For a single integral:
If the integrand function is symmetric about y 。
Then ∫(- a→a) f(x) dx = {0, if f(x) is the odd function of x.
{2∫(- a→a) f(x) dx, if f(x) is an even function about x.
If the integrand function is symmetric about x 。
Then ∫(- b→b) f(y) dy = {0, if f(y) is the odd function of y.
{2∫(- b→b) f(y) dy, if f(y) is an even function about y.
For double integrals:
If the integrand function is symmetric about y 。
Then ∫∫D f(x, y) dxdy = {0, if f(x, y) is the odd function of X.
{2∫∫D？ F(x, y) dxdy, if f (x, y) is an even function about x, d? Is the first limit
If the integrand function is symmetric about x 。
Then ∫∫D f(x, y) dxdy = {0, if f(x, y) is the odd function of y.
{2∫∫D？ F(x, y) dxdy, if f (x, y) is an even function about y and d? Is the first limit
Especially when the integration region is symmetrical about two axes.
The integrand function is also an even function. And then ∫∫D x? dxdy = ∫∫D y？ dxdy = ( 1/2)∫∫D (x？ + y？ )dxdy
For triple integrals:
If the integral domain ω is symmetric about the zox plane.
∫∫∫∫∫ωf(x (x, y, z) dxdydz = {0, if f(x, y, z) is the odd function of X.
{2∫∫Ω? F(x, y, z) dxdydz, if f (x, y, z) is an even function about x, ω? Is the first limit
If the integral domain ω is symmetric about the yoz plane.
∫∫∫∫∫Ωf(x, y, z) dxdydz = {0, if f (x, y, z) is the odd function of y.
{2∫∫Ω? F(x, y, z) dxdydz, if f (x, y, z) is an even function about y, ω? Is the first limit
If the integral domain ω is symmetric about the xoy plane.
∫∫∫∫∫Ωf(x, y, z) dxdydz = {0, if f (x, y, z) is the odd function of z,
{2∫∫Ω? F(x, y, z) dxdydz, if f (x, y, z) is an even function about z, ω? Is the first limit
Especially when the integration region is symmetrical about all three coordinates.
The integrand function is also an even function. And then ∫∫∫∫ωx? dV =∫∫∫ωy？ dV =∫∫∫ωz？ dV =( 1/3)∫∫∫ω(x？ + y？ + z？ )dV
Therefore, the higher the level, the freer and wider the space available, but the more complicated and difficult it is, and
Moreover, there are fewer restrictions than the above two, and the spatial imagination is improved.
Multiple integrals can be converted into several definite integrals, and each definite integral can control different extension directions.
Another example is the area surrounded by f (x) and g(x) in a ≤ x ≤ b, where f(x)>g(x).
The formula for calculating the area by definite integral is ∫(a→b) [f(x)-g(x)] dx.
But for the upgraded double integral, the area formula is ∫(a→b) dx ∫(g(x)→f(x)) dx, and the integrand becomes 1.
Calculate the value from z = x with different integration levels? + y？ 、z = a？ Closed volume?
One-fold integral (definite integral): projected to the zox plane, z = x? Make z = a? -& gt； X = a, using the circular shell method.
V = 2πrh = 2π∫(0→a) xz dx = 2π∫(0→a) x？ dx = 2π？ ( 1/4)[ x？ ] |(0→a) = πa？ /2
Double integral: height a, z = x? + y？ X projected onto the xoy plane? + y？ = a？
So just ask ∫∫(D) (x? + y？ ) dxdy, where d is x? + y？ = a？
V = ∫∫(D) (x？ + y？ )dxdy = ∫(0→2π) dθ ∫(0→a) r？ Doctor, at this point, you will find that the steps are the same as definite integral.
= 2π ? ( 1/4)[ r？ ] |(0→a) = πa？ /2
Triple integration: the volume of a rotating body, and the integrand function is 1, which can be directly solved.
Column coordinate slicing method: dz: x? + y？ = z
v =∫∫∫(ω)dxdydz
= ∫(0→a？ )dz ∫∫Dz dxdy
= ∫(0→a？ )πz dz
= π ? 【z？ /2 ] |(0→a？ )
= πa？ /2
Column coordinate projection method: dxy: x? + y？ = a？
v =∫∫∫(ω)dxdydz
= ∫(0→2π) dθ ∫(0→a) r dr ∫(r？ →a？ )dz
= 2π ? ∫(0→a) r？ (a？ - r？ ) doctor
= 2π ? 【a？ r？ /2 - ( 1/4)r？ ] |(0→a)
= 2π ? 【a？ /2 - ( 1/4)a？ ]
= πa？ /2
There are many ways to find the volume by triple integration, which is called high degree of freedom.
On the algorithm of curve integral and surface integral;
If you continue to study, you will find it easier to find the formula of (plane) area and volume than (surface) area.
After learning the formula for finding the volume, there will be a formula for finding the surface.
There are "curve integral" and "surface integral", which are divided into "the first category" and "the second category"
When the integrand function is 1, the first kind of curve integral is to find the arc length, and the contrast definite integral can only find the length of the straight line.
∫(C) ds = L (curve length)
When the integrand function is not 1, it is to find the area of the surface with the arc as the bottom line.
∫(C) f(x, y) ds = A (surface area)
The application of the second kind of curve integral includes the work done along the curve L in the force field and so on.
The first algorithm of arc length curve integration;
If the integrand function is a parametric equation x = x(t), y = y(t)
Then ∫ (l) f (x, y) ds = ∫ (a→ b) f [x (t), y (t)] √ [x' (t)? + y'(t)？ ] dt
If the integrand function is y = y(x)
Then ∫ (l) f (x, y) ds = ∫ (a→ b) f [x, y (x)] √ [1+y' (x)? ] dx
If the integrand function is r = r(θ)
Then ∫ (l) f (x, y) ds = ∫ (α→β) f (rcos θ, rsin θ) √ [r? (θ) + r'(θ)？ ] dθ
If the integral region is symmetrical about y axis.
Then ∫(L) f(x, y) ds = {0, if f(x, y) is the odd function of x 。
{2∫(L？ ) f(x, y) ds, if f(x, y) is an even function about x, l? Is the first limit
If the integral region is symmetric about x 。
Then ∫(L) f(x, y) ds = {0, if f(x, y) is the odd function of y 。
{2∫(L？ ) f(x, y) ds, if f(x, y) is an even function about y, l? Is the first limit
If the integration region is symmetric about y = x:
Have ∫(L) x? ds = ∫(L) y？ Repeat from the mark.
There is ∫(L) x ds = ∫(L) y ds.
If the integral region is symmetric about the y = x plane: (rotationally symmetric)
Have ∫(L) x? ds = ∫(L) y？ ds = ∫(L) z？ Repeat from the mark.
Have ∫(L) x ds = ∫(L) y ds = ∫(L) z ds.
The second algorithm of coordinate curve integration;
If the integrand function is a parametric equation x = x(t), y = y(t)
Then ∫ (l) p (x, y) dx+q (x, y) dy = ∫ (a → b) {p [x (t), y (t)] x' (t)+q [x (t), y (t)]
If the integrand function is y = f(x)
Then ∫ (l) p (x, y) dx+q (x, y) dy = ∫ (a → b) {p [x, f (x)]+q [x, f (x)] f' (x)} dx.
If the curve l can surround the closed area d, Green's formula is used:
Then ∮(L) P(x, y)dx+Q(x, y)dy = ∫∫D (? Q/？ x -？ P/？ y ) dxdy
If curve L cannot close a closed area, you can add a line segment to close a closed area D, and then use Green's formula:
Then ∫ (l)+∫ (l1)+∫ (L2)+...+∫ (ln) = σ (k =1→ n) ∫ (l _ k) =.
Counter-clockwise+clockwise-
If you want to use Green's formula, and the integral domain D contains singularities, you should add the excavated singularities, and then use Green's formula:
The excavated L 1 part is usually round or oval.
That is, ∫(L)+∫(L 1 clockwise) = ∮(L+L 1)
= => ∫ (l) = ∮ (l+l1)-∫ (l1clockwise)
= = & gt∫(L) = ∫(L 1 counterclockwise), if the value of the previous double integral is 0.
If the integrand function is three-dimensional, Stokes formula can be used.
∮(c)pdx+qdy+rdz =∫∫σrota * n ds
= ∫∫Σ (? R/？ You-? Q/？ z)dydz +(？ P/？ z -？ R/？ x)dzdx +(？ Q/？ x -？ P/？ y)dxdy
When the integrand function is 1, the first kind of surface integral is to find the surface area, while the contrast double integral can only find the plane area.
∫∫ (σ) ds = a (surface area), and the degree of freedom is greater than the first kind of curve integral.
∫∫ (σ) f (x, y, z) ds, physical applications, such as the mass, center of gravity and moment of inertia of a curved surface, the velocity of the velocity field flowing through a curved surface, etc.
The application of the second kind of surface integral includes the flow of six-country surface σ in unit time and so on.
Algorithm of surface integral of the first kind;
For xoy surface, surface σ: z = z (x, y).
∫∫σf(x，y，z) dS = ∫∫D f[x，y，z(x，y)]√[ 1 +(？ z/？ x)？ + (? z/？ y)？ ] dxdy
For the yoz plane, the surface σ: x = x (y, z)
∫∫σf(x，y，z) dS = ∫∫D f[x(y，z)，y，z]√[ 1 +(？ x/？ y)？ + (? x/？ z)？ ] dydz
For the zox plane, the surface σ: y = y (x, z).
∫∫σf(x，y，z) dS = ∫∫D f[x，y(x，z)，z]√[ 1 +(？ y/？ z)？ + (? y/？ x)？ ] dzdx
If the integral domain σ is symmetric about the zox plane.
Then ∫∫Σf(x, y, z) dS = {0, if f (x, y, z) is the odd function of X.
{2∫∫Σ? F(x, y, z) dS, if f (x, y, z) is an even function about x, σ? Is the first limit
If the integral domain σ is symmetric about the yoz plane.
Then ∫∫Σf(x, y, z) dS = {0, if f (x, y, z) is the odd function of y.
{2∫∫Σ? F(x, y, z) dS, if f (x, y, z) is an even function about y, σ? Is the first limit
If the integral domain σ is symmetric about the xoy plane.
Then ∫∫Σf(x, y, z) dS = {0, if f (x, y, z) is the odd function of z.
{2∫∫Σ? F(x, y, z) dS, if f (x, y, z) is an even function about z, σ? Is the first limit
If the integrand function is symmetric about three coordinate planes: (rotational symmetry)
And ∫∫σx? dS =∫∫σy？ dS =∫∫σz？ dS =( 1/3)∫∫σ(x？ + y？ + z？ )dS
Algorithm of surface integral of the second kind;
For xoy surface, surface σ: z = z (x, y).
∫∫σf(x，y，z) dxdy = ∫∫ d f [x，y，z (x，y)] dxdy。 Top+bottom-
For the yoz plane, the surface σ: x = x (y, z)
∫∫σf(x，y，z) dydz = ∫∫ d f [x (y，z)，y，z] dydz。 Front side+rear side-
For the zox plane, the surface σ: y = y (x, z).
∫∫ σ f (x，y，z)dzdx =∫d f[x，y (x，z)，z] dzdx。 Right+left-
Or use the three-in-one formula:
∫∫σPdydz+qd zdx+Rdxdy =∫∫D[-P *？ z/？ x - Q *？ z/？ Y+R] dxdy。 Top+bottom-
Conversion between two kinds of surface integrals;
∫∫σPdydz+Qdzdx+Rdxdy =∫∫σ(PCOSα+qcosβ+rcosγ)dS
Gauss formula: If σ is the outside of a closed surface.
∫∫σpdy dz+Qdzdx+Rdxdy =∫∫∫ω(？ P/？ x +？ Q/？ y +？ R/？ z ) dxdydz
If there are singularities in the integral domain ω, it is necessary to sum up the integrals of the excavated singularities (take internal measurements), and then use the Gaussian formula:
That is, ∫ ∫ σ+∫ ∫ σ1= ∫ ∫ (σ+σ1) = ∫ ∫ ω.
∫∫ σ = ∫∫∫Ω-∫∫ σ 1
Outside+inside-
The second kind of curve integral/the second kind of surface integral is mainly used in physics, which is directional and involves vector range.
These two are more complicated and have deeper concepts.