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we define C as the border of the area
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
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