we define C as the border of the area(adsbygoogle = window.adsbygoogle || []).push({});

D={(x,y)|(x^2+y^2)^2<=x^2-y^2,x>=0}

whats the value of the integral

S(x^2y^3+2y)dx+(x^3y^2+3x)dy

C

while C is against the clock direction

it's the possitive direction,

the field components are from C1 (continious and devertive continious)

the area connected and muzzled

so i tried to used grin sentence and got:

SSdQ/dx-dP/dy

D

SS1dxdy

D

trans to D*:

x=rcost

y=rsint

therefore j=r

i assigned that at to D to get new D*

r^4<=r^2(cos^2(t)-sin^2(t))

r^2<=[1+cos(2t)]/2 - [1-cos(2t)]/2

r^2<=cos(2t)

0<=r<=sqrt(cos(2t))

also

x>=0 therefore

rcost>=0

cost>=0

-pi/2<=t<=pi/2

so our integral is

SSr dr dt

D*

S [r^2/2] dt

(1/2)S cos(2t) dt

(1/2) [sin(2t)/2]

(1/4) (0+0 )

0

the right answer is 1/2

where's my mistake?

thanks

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# Green sentence

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