Using Greens theorm ..integration (3x+4y)dx+(2x-3y)dy for a circle
ANSWER
Let
"C= \\left \\{ \\left ( x,y \\right )\\in R^{2} :\\left ( x-a \\right )^{2}+\\left ( y-b \\right )^{2}=r^{2}\\right \\}"
be a positively oriented circle and let
"D= \\left \\{ \\left ( x,y \\right )\\in R^{2} :\\left ( x-a \\right )^{2}+\\left ( y-b \\right )^{2}\\leq r^{2}\\right \\}" .
Since the partial derivatives of the functions "L(x,y)=3x+4y," "M(x,y)=2x-3y" are continuous , then by Green's Theorem
"\\oint _{C}\\left ( Ldx+Mdy \\right )=\\iint_{D} \\left ( \\frac{\\partial M}{\\partial x}-\\frac{\\partial L}{\\partial y } \\right )dxdy"
"\\oint _{C}\\left ( (3x+4y )dx+(2x-3y)dy \\right )=\\iint_{D} \\left ( 2 -4 \\right )dxdy=-2(area\\: of \\: D)=-2\\pi r^{2}"
(because "\\frac{\\partial(2x-3y) }{\\partial x}=2, \\: \\: \\frac{\\partial(3x+4y) }{\\partial y}=4, \\: \\:" and the area of a circle "D" is equal to "\\pi r^{2}" )
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