Mumbai university > Electronics and telecommunication engineering, Electronics engineering > Sem 3 > Applied mathematics 3

Marks : 06

Years : MAY 2016

Green Theorem $\int\limits_c \bar Fd\vec r=[\int\limits_c M\space dx+ N\space dy]=\int\int\limits_s[\dfrac {dN}{dx}-\dfrac {dM}{dy}]dxdy $

$\int\limits_c\bar F dr=\int\limits_c (x^2-xy)dx + ( x^2-y^2)dy \\ =\int\limits_{OA} + \int\limits_{AO} =I_1+I_2 \\ I_1=\int\limits_{OA}(x^2-xy)dx+ (x^2-y^2)dy, \space eq^n \space of \space OA \space is \space x^2=2y\Rightarrow 2x\space dx= 2dy\Rightarrow x\space dx=dy \\ I_1=\int\limits_0^2[(x^2-\dfrac {x^3}2)dx + (x^2-\dfrac {x^4}4)x\space dx]=2 \\ I_2 =\int\limits_{AO}(x^2-xy) dx + (x^2-y^2) dy $

equation of path $x=y\\ dx=dy $

$I_2=\int\limits_2^0 [0+0]=0 \\ \int\limits_c\bar F.dr =2 ----(1) \\ \int\int\limits_s[\dfrac {dN}{dx}-\dfrac {dM}{dy}]dxdy=\int\int\limits_s(2x+x)dxdy =3\int\limits^2_{x=0}\int\limits^x_{y=\dfrac {x^2}2} x\space dx\space dy \\ = 3\int\limits^2_0[x] dx [y]^x_{x^2/2}=3\int\limits^2_0x\space dx[x-\dfrac {x^2}2] \\ =3\int\limits_0^2 (x^2-\dfrac {x^3}2 dx =3[\dfrac {x^3}3-\dfrac {x^4}8 ]^2_0 =3[\dfrac 83-\dfrac {16}8]=3[\dfrac 23]=2$

Hence Green's Theorem verified