Question

Let F = (P,Q) be the vector field defined by -x+y . P(x,y) = 22, (x, y) + (0,0) 0, (x, y) = (0,0) Q(x,y) = (x, y) + (0,0) x2+y2; 10,(x, y) = (0,0). (a) Show that F is a gradient vector field in R2 \ {y = 0}. (b) Letting D = {2:2020 + y2020 < 1}, compute the line integral Sap P dx + Q dy in the counter-clockwise direction. (c) Does your calculation in part (b) violate Green's theorem? Explain.

Answer 1

ANSWERR :

Let be the vector field defined by

F = (P,Q)

P(x,y) = 1+y 12+y? 0 (2, 4) + (0,0) (x,y) = (0,0)

Q(x,y) = -T+ 12+y2 0 (2,4) + (0,0) (2,y) = (0,0)

a)

Show that $\vec{F}$ is a gradient vector field in $\mathbb{R}^{2} | \left \{ y=0 \right \}$

F = (P,Q)

F is gradient vector field if

$\frac{\partial Q}{\partial x}=\frac{\partial P}{\partial y}$

$\frac{\partial Q}{\partial x}=\frac{\partial }{\partial x}\left [ \frac{-x+y}{x^{2}+y^{2}} \right ]$

$=\frac{(x^{2}+y^{2})(-1)-(-x+y)-2x}{(x^{2}+y^{2})^{2}}$

$=\frac{-x^{2}-y^{2}+2x^{2}-2xy }{(x^{2}+y^{2})^{2}}$

$=\frac{x^{2}-y^{2}-2xy }{(x^{2}+y^{2})^{2}}$

$\frac{\partial P}{\partial y}=\frac{(x^{2}+y^{2})(1)-(x+y)-2y}{(x^{2}+y^{2})^{2}}$

$=\frac{x^{2}+y^{2}-2xy-2y^{2} }{(x^{2}+y^{2})^{2}}$

$=\frac{x^{2}-y^{2}-2xy }{(x^{2}+y^{2})^{2}}$

$so, \frac{\partial Q}{\partial x}=\frac{\partial P}{\partial y}$

Therefore F is gradient vector field.

b)

$D=\left \{ x^{2020}+y^{2020}\leq 13 \right \}$

$\int _{2D}Pdx+Qdy=0$

$\therefore$ For closed boundary F is gradient vector field So,$\exists$ a fun G such that .

$F=\Delta g$

So integration with respect to closed boundary will be zero. ( For same intial and ending point on boundary ).

c)

Does your calculation in part (b) Violate Green's Theorem

$I = \int _{\partial D}Pdx+Qdy$

By using greem tjepre,

$I = \int _{D}\int \left ( \frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y} \right )dxdy$

$=\iint _D\left(\frac {\partial p}{\partial y}-\frac{\partial p}{\partial y}\right) dx dy\, \, \, By (a)part$

$=0$

Now calculation is a pant (b).do not violate green's theorem.