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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!

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  • $\begingroup$ The origin isn't even a point on the plane... $\endgroup$ – DonAntonio Oct 15 '16 at 20:52
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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, $$ A=\int^{1}_{-1}(8t^3-12t^2+6t-1)=(2t^4-4t^3+3t^2-t)|^1_{-1}=-10 $$

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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

$W=\int_{-1}^1 (8x^3-12x^2+6x-1)dx$

$=-8-2=-10$.

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  • $\begingroup$ A stupid question: why x is from -1 to 1? $\endgroup$ – Blackgirl5 Oct 15 '16 at 21:34
  • $\begingroup$ Ok, I see. Thank you! $\endgroup$ – Blackgirl5 Oct 15 '16 at 21:39
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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 $$\int_{-1}^1 (y+xy'+zz')dx=\int_{-1}^1 ((1-x)-x+(x^2+(1-x)^2(2x-2(1-x))dx=-10.$$

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$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

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}


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

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