Conservative vector field qu

Ok the qu is:

Write down a condition obeyed by a conservative vector field V= {P(x,y), Q(x,y)}

This must be that (dP/dy)=(dQ/dx) so that the differential equation would be exact. But for the case:

P= (x^2)y + y Q= x(y^2) + x

if using (dP/dy)=(dQ/dx) then this isnt exact but is using (dP/dx)/(dQ/dy) then it is, and as the question does not actually specify which function is dx and which is dy, then which one should i use to find a function f(x,y) such that V= (upside down triangle)f ?

Also, how would i evaluate the line integrals:

C1 : x=t, y=t (0<t<1)
C2: x=0, y=t (0<t<1); x=t, y=1 (0<t<1)

for integral of Pdx+ Qdy as stated above

I realise for this case the differential equation is not exact, so should i assume this for the beginning of the question, and assume that the vector field is not conservative?

thanks v much for ne help,

Ellie

Reply 1

Jonny W

It's confusing because Q dx + P dy is exact and P dx + Q dy is not.

(int over C1) P(x, y) dx + Q(x, y) dy
= (int from 0 to 1) P(t, t) + Q(t, t) dt . . . . . dx/dt = dy/dt = 1
= (int from 0 to 1) 2t^3 + 2t dt
= 1/2 + 1
= 3/2

Since x doesn't change on C2's first part and y doesn't change on its second,

(int over C2) P(x, y) dx + Q(x, y) dy
= (int from 0 to 1) Q(0, t) dt + (int from 0 to 1) P(t, 1) dt
= (int from 0 to 1) 0 dt + (int from 0 to 1) t^2 + 1 dt
= 1/3 + 1
= 4/3