Calculate the integral of the curve ydx xdy x 4y where L is the counterclockwise curve segment of

1. When the origin is not in the curve, p=-y (x +4y), q=x (x +4y), p and q have a first-order continuous partial derivative in l.

Calculated: p y= q x, the integral on the closed curve is easily obtained by Green's formula, and the result of this problem is = 0

2. When the origin is in the curve, p and q are not defined at (0,0), so the above method cannot be used.

Make the curve l1:x +4y = counterclockwise, small enough so that l1 and l do not intersect; Use. l1-

Indicates the inverse curve of L1 (note that the negative sign is the superscript).

Then p and q have a first-order continuous partial derivative in the region enclosed by (l+l1-), and Green's formula can be used.

l+l1-)（xdy-ydx）/（x^2+4y^2)

q/∂x-∂p/∂y)dxdy

Therefore: l(xdy-ydx) (x 2+4y 2)= l1(xdy-ydx) (x 2+4y 2).

The following is sufficient to calculate the points on L1.

l1（xdy-ydx）/（x^2+4y^2)

Note that on l1 x +4y =

1/ε²)l1

xdy-ydx).

The integrand of dxdydxdy is 1, the integral result is the area of the region, and the area of the ellipse is: ab= 2, where a= and b= 2

Complement curve l1: x +4y 0 1, take counterclockwise, the curve integral of the original L-L1 + the curve integral of L1, the former term is 0 by Green's formula, and the latter term is the curve integral of L1 (ydx-xdy).

Green's formula l1 encloses the area -2d 1 2 2

The end result is -2

Solution: Rewrite the equation of a circle x +y =1 into a parametric equation: x = cost, y = sint, dx = -sintdt, dy = costdt

s=(1/2)∮xdy-ydx

1/2)∫‹0，2π›(cos²t+sin²t)dt=(1/2)∫‹0，2π›dt

1/2)t︱‹0，2π›

Hence xdy-ydx

2 Curve integrals are divided into:

1) Curve integration of arc length (type 1 curve integration) (2) Curve integration of coordinate axes (type 2 curve integration) The integral element of curve integration of arc length is the arc length element ds; For example: the integral of the curve of l f(x,y)*ds. The integral element of the curve integration of the coordinate axis is the coordinate element dx or dy, for example:

The integral of the curve for l' p(x,y)dx+q(x,y)dy. However, the curve integral to the arc length is usually positive due to its physical significance, and the curve integral to the coordinate axis can obtain different signs according to the different paths.

From the problem, we know the p -y (x 2+y 2) of the integral of the curve

q=x/(x^2+y^2)

and they have a first-order continuous partial derivative in the region enclosed by c.

It is easy to find: partial q partial x (-x 2+y 2) (x 2+y 2) 2 , partial p partial y (-x 2+y 2) (x 2+y 2) 2 , partial q partial x - partial p partial y=0

From Green's formula, c(xdy-ydx) (x 2+y 2)= g(partial q partial x-partial p partial y)dxdy=0

A: With Green's formula.

pdx+qdy, i.e., p=-y (4x 2+y 2), q=x (4x 2+y 2).

There is p y=(-4x 2-y 2+2y 2) (4x 2+y 2) 2=(y 2-4x 2) (4x 2+y 2) 2;

q/σx=(4x^2+y^2-8x^2)/(4x^2+y^2)=(y^2-4x^2)/(4x^2+y^2)^2

This gives p y= q x, i.e., the integration result is path independent.

And the curve is not the origin, so that x=cos and y=2sin, where from 0 to 2.

Get (0 to 2 ) 2(cos ) 2+2(sin ) 2] 4[(cos ) 2+(sin) 2] d

(0 to 2) 1 2 d =

by Green's formula.

p = y²/2

q = 2xy

Then q x = 2y

p / y = y

So it can be turned into a double integral.

2y-y)dxdy = y dy dx integral interval is 0,3, calculate the curve integral y 2dx+2xydy, where c is the closed shirt song matching area surrounded by y=x and y=x, take counterclockwise.

To the final result. The process can be simplified.

Summary. dy=-dx,∫(bc)ydx-x^2dy=∫(1,0)(2-x+x^2)dx=-11/6

y=21 x2 to b(-1,0).

18.Calculate curve integrals.

i= l(x+2y 2)dx+(ey+2x)dy, l is edged by the point a(1,0).

18.Calculate curve integrals.

y=21 x2 to b(-1,0).

l is along the point a(1,0).

i=∫l(x+2y−2)dx+(ey+2x)dy,18.Calculate curve integrals.

y=21 x2 to b(-1,0).

l is along the point a(1,0).

i=∫l(x+2y−2)dx+(ey+2x)dy,18.Calculate curve integrals.

y=21 x2 to b(-1,0).

l is along the point a(1,0).

i=∫l(x+2y−2)dx+(ey+2x)dy,18.Calculate curve integrals.

y=21 x2 to b(-1,0).

l is along the point a(1,0).

i=∫l(x+2y−2)dx+(ey+2x)dy,18.Calculate curve integrals.

y=21 x2 to b(-1,0).

l is along the point a(1,0).

i=∫l(x+2y−2)dx+(ey+2x)dy,18.Calculate curve integrals.

Answer]: Using Green's formula: the singularity (0,0) is not in the integral domain.

i = l (ydx - xdy)/(x^2 + y^2)

d [(x^2 - y^2)/(x^2 + y^2)^2 - x^2 - y^2)/(x^2 + y^2)^2] dxdy

with parametric equations.

x = 1 + cost、dx = sint dt

y = 1 + sint、dy = cost dt

0 ≤ t ≤ 2π

l (ydx - xdy)/(x^2 + y^2)

Nayu(0 2 ) 1 + sint)(-sint) -1 + cost)(cost)] 1 + cost) 2 + 1 + sint) 2] dt

0→2π) sint + cost + 1)/(2sint + 2cost + 3) dt

Let u = tan(t 2), dt = 2 (1 + u 2) du, sint = 2u (1 + u 2), cost = 1 - u 2) (1 + u 2).

sint + cost + 1)/(2sint + 2cost + 3) dt

2u/(1 + u^2) +1 - u^2)/(1 + u^2) +1]/[2 * 2u/(1 + u^2) +2 * 1 - u^2)/(1 + u^2) +3] *2/(1 + u^2) du

4∫ (u + 1)/[u^2 + 1)(u^2 + 4u + 5)] du

du/(u^2 + 1) +du/(u^2 + 4u + 5)

du/(u^2 + 1) +du/[(u + 2)^2 + 1]

arctan(u) +arctan(u + 2) +c

arctan[tan(t/2)] arctan[2 + tan(t/2)] c

So i = arctan[tan(t 2)] arctan[2 + tan(t 2)]:0 2 )

The interval is divided into: 0 Changma segment, 2

i = 2 - 2) -2 - 2)

Using Green's formula: the singularity (0,0) is not in the integral domain. i = file round code l (ydx - xdy) (x 2 + y 2) = d [(x 2 - y 2) (x 2 + y 2) 2 - x 2 - y 2) line (x 2 + y 2) 2] dxdy= 0 with the parametric equation.

x = 1 + cost、dx = sint dt{ y = 1 + sint、dy = co...

(ydx+xdy)/(x|+|y|)

ydx+xdy (Because of the high value of the ears is |.)x|+|y|=1）

Bright ruler (1-1) dxdy = 0 (Green's formula).