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!
 A: 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
$W=\int_{-1}^1 (8x^3-12x^2+6x-1)dx$
$=-8-2=-10$.
A: Crossing of paraboloid $z=x^2+y^2$ and plane $x+y=1$ leads to paraabolic curve. Let us express it explicitly: $y=1-x\to z=2x^2-2x+1$. Now, introduce parameterization $x=t$, then we have parametric equation of curve: 
$$
r(t)=
\begin{cases}
x=t\\
y=1-t\\
z=2t^2-2t+1
\end{cases}\quad \text{where }t\in[-1,1]
$$
Boundaries of $t$ come from points $(-1,2,5)$ and $(1,0,1)$ declared in task.
Let us calculate the work:
$$
A=\int_C\vec{F}d\vec{r}=\int^1_{-1}\vec{F}(\vec{r}(t))\vec{r}(t)^{'}dt
$$
Where $A$-work, $C$-curve.
$$
\vec{F}=y\vec{i}+x\vec{j}+z\vec{k}\to (1-t)\vec{i}+t\vec{j}+(2t^2-2t+1)\vec{k}\\
\vec{r}(t)^{'}=\left(t\vec{i}+(1-t)\vec{j}+(2t^2-2t+1)\vec{k}\right)^{'}=(\vec{i}-\vec{j}+(4t-2)\vec{k})\\
\vec{F}(\vec{r}(t))\vec{r}(t)^{'}=(1-t)-t+(2t^2-2t+1)(4t-2)=8t^3-12t^2+6t-1
$$ 
Then,
$$
A=\int^{1}_{-1}(8t^3-12t^2+6t-1)=(2t^4-4t^3+3t^2-t)|^1_{-1}=-10
$$
A: Hint. Given $x\in[-1,1]$ then $y=1-x$ and $z=x^2+y^2$.
Are you able to compute the line integral now?
P.S. By the way, at the end, you should find
$$\int_{-1}^1 (y+xy'+zz')dx=\int_{-1}^1 ((1-x)-x+(x^2+(1-x)^2(2x-2(1-x))dx=-10.$$
A: $\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{{#1}}\,}
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Note that
  $\ds{\nabla\times\vec{F} = \vec{0}\implies\vec{F} = -\nabla\Phi}$. So, you can $\underline{'ignore'}$ the paraboloid and the plane because the integral only depends on the initial and final points. Namely,
  $$
\int_{\vec{A}}^{\vec{B}}\vec{F}\cdot\dd\vec{r} =
-\Phi\pars{\vec{B}} + \Phi\pars{\vec{A}}
$$

\begin{align}
-\,\partiald{\Phi}{x} & = y\implies\Phi = -yx + \mrm{f}\pars{y,z}
\\[5mm]
-\,\partiald{\Phi}{y} & = x\implies x - \partiald{\mrm{f\pars{y,z}}}{y} = x\implies \mrm{f}\pars{y,z} = \mrm{g}\pars{z}\implies\Phi = -xy + \mrm{g}\pars{z}
\\[5mm]
-\,\partiald{\Phi}{z} & = z\implies-\,\partiald{\mrm{g}\pars{z}}{z} = z
\implies\mrm{g}\pars{z} = -\,{1 \over 2}\,z^{2}
\\[5mm] & \implies
\bbx{\ds{\Phi = -xy - {1 \over 2}\,z^{2} + \pars{~\mbox{a constant}~}}}
\end{align}

$$
\int\vec{F}\cdot\dd\vec{r} =
\left.\vphantom{\LARGE A}-\Phi\right\vert_{\ \pars{-1,2,5}}^{\ \pars{1,0,1}} =
\bracks{-\pars{-\,{1 \over 2}}} - \bracks{-\pars{-\,{21 \over 2}}} =
\bbx{\ds{-10}}
$$
