High number curve integration problem, high number curve integration problem solving

Use the Green's formula, because the Green's formula requires a closed area, so make up first:

l1: y=1, x from 1 to 0;

l2: x=0, y from 1 to 0;

so that it becomes a positively oriented region, and then the integrals of the curve on l are s, and the integrals of the curve on l1 and l2 are s1 and s2, respectively, according to Green's formula:

s + s1 + s2 = ∫l+l1+l2[2xg(y)-y]dx+[x²g′(y)-y]dy

Double integral ( [x g (y)-y] derivative of x - 2xg(y)-y] derivative of y ) dxdy

Double integral ( 2xg (y) -2xg ('y)-1] )dxdy double integral(1) dxdy

Integral[0,1] Integral[3x -2x,1] (1) dydx integral[0,1] (1-3x +2x) dx and on l1, y=1, dy=0, so:

s1 = integral[1,0] (1) dx = 1 on l2, x=0, dx=0, so:

s2 = integral[1,0](-y)dy=1 2 sum up, s = 1 - s1 -s2 = -1 2

Hope and time to add points

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Supplement the line segment BA, into a closed curve.

i = x^2+2xy)dy = x^2+2xy)dy - x^2+2xy)dy

The former uses Green's formula, and the latter dy=0Get.

i = 2(x+y)dxdy, let x = arcost, y = brsint, dxdy = abrdrdt,i = 2ab <0, >dt <0, 1>(acost+bsint)r 2dr

2ab∫<0，π>acost+bsint)dt[r^3/3]<0, 1>

2ab/3)<0，π>acost+bsint)dt

2ab/3)[asint-bcost]<0，π>4/3)ab^2

The process of finding the integral of the curve is as follows, if you don't understand, please ask, if you are satisfied, please click to adopt.

This question: You may wish to set x=cos t and y=sin t to try.

Use the parametric equation: { x = a cost

y = a • sint

ds = x'² y')dt = a sin t + a cos t) dt = a dt, so the original formula = (0,2 )e* (cos t) 2+(sin t) 2) *r*dt=2 r*e.

If you don't understand, please ask, satisfied.

The previous one is also a quilt together.