Question

Using Greens theorm ..integration (3x+4y)dx+(2x-3y)dy for a circle

Answer 1

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 $\pi r^{2}$ )