Fortunately F is a conservative field because the vector product of
F and the operator is zero . Therefore, W = F(a,b,c) - F(p,q,r)
I started this question;
Find the work done by F=e^(yz)i + (xze^(yz) + zcosy)j + (xye^(yz) + siny)k over the following paths from (1,0,1) to (1,pi/2, 0)
a) The line segment x=1 , y=(pi/2)t , z=1-t t between 0 and 1
b) The line segment from (1,0,1) to the origin followed byt he line segment from the origin to (1,pi/2,0)
c) The line segment from (1,0,1) to (1,0,0) followed by the x-axis from (1,0,0) to the origin, followed by the parabola y=pix^2/2 , z=0 from there to (1,pi/2,0)
***I started by solving the potential function which I got to be
f(x,y,z)=xye^(yz) + siny + C ----> F=gradient(xye^(yz) + siny)
I think dr/dt for part A is : pi/2j - k
This is where I'm stuck...
Help!!!
since curl F is zero
in part b , the total work done is
work done over the first segment from (1,0,1) to (0,0,0) +
work done over the second segment from (0,0,0) to (1,pi/2,0)
= [F(0,0,0) - F(1,0,1)] + [F(1,pi/2,0) - F(0,0,0)]
= F(1,pi/2,0) - F(1,0,1)
we can see that the shape of the path is not important unless curl F is
not zero