Expert: Alon Mandes - 1/2/2009

Question
QUESTION: evaluate the line intergralof  integration over c  ((xy+x+y)dx+(xy+x-y)dy) ,when c is the circle x^2+y^2=ax by green's theorm.

ANSWER: According to the green's theorm : ∫Pdx+Qdy=∫∫(∂Q/∂x)-(∂P/∂y)dA.
In our case : P=xy+x+y -> ∂P/∂y=x+1 &  Q=xy+x-y -> ∂Q/∂x=y+1.
The closed curve is x^2+y^2=ax Let's perform polar paramitrization:
x=a*r*cos(θ) & y=a*r*sin(θ). So now the 1-D integral becomes 2-D :
∫(xy+x+y)dx+(xy+x-y)dy=∫∫(y-x)rdrdθ. Lets calculate the limits of
r & θ : Our circle is x^2+y^2=ax ->
(a^2)(r^2)[cos(θ)]^2+(a^2)(r^2)[sin(θ)]^2=a*r*cos(θ) ->
(a^2)(r^2)=a*r*cos(θ) -> a*r=cos(θ) -> r=cos(θ)/a. So:
0<θ<2Π & 0<r<cos(θ)/a. Now our 2D integral is :
∫∫[a*r*sin(θ)]^2-[a*r*cos(θ)]^2 rdrdθ.
I will leave to you to proceed.

Alon.


---------- FOLLOW-UP ----------

QUESTION: evaluate the line intergral of  integration over c  ((xy+x+y)dx+(xy+x-y)dy) ,when c is the circle x^2+y^2=ax by line integral.


Answer
Line Integral Li=∫M(x,y)dx+N(x,y)dy = ∫M(t)(dx/dt)+N(t)(dy/dt) dt
where (x-a)=(a/2)cos(t) & y=(a/2)sin(t). Beuase x^2+y^2=ax is the same circle as (x-a/2)^2+y^2=a^2/4. Therefore :
dx/dt=(a/2)cos(t)
dy/dt=(a/2)sin(t)
M(t)=[a+(a/2)cos(t)][(a/2)sin(t)]+a+(a/2)cos(t)+(a/2)sin(t)
N(t)=[a+(a/2)cos(t)][(a/2)sin(t)]+a+(a/2)cos(t)-(a/2)sin(t)

I'll leave it for you to proceed from here.

Alon.