Find the work done by the force on a curve The question is: The work done by the force 
$
\vec{F}=y\vec{i}+x\vec{j}+z\vec{k}
$
in moving from (-1, 2, 5) to (1, 0, 1) on C the curve of intersection of the paraboloid 
$
\ z = x^2+y^2\
$and the plane $ x+y=1 $.
From what I understand, the intersection of the paraboloid and the plane is just a point which is the origin. Then, how can the curve be formed?
Any suggestion, please? Thank you!

we have
$z=x^2+y^2$
$dz=2xdx+2ydy$
$y=1-x$
$dy=-dx$
$\vec{dl}=dx\vec{i}+dy\vec{j}+dz\vec{k}$
$\vec{F}.\vec{dl}=ydx+xdy+zdz$
$=(1-x)dx-xdx+2(x^2+(1-x)^2)(xdx-(1-x)dx)$
$=(   1-2x+2(2x^2+1-2x)(2x-1)      )dx$
which we integrate between $x_1=-1 $ and $x_2=1  $ to find
$W=\int_{-1}^1 (8x^3-12x^2+6x-1)dx$
$=-8-2=-10$.
