Limits of integration, surface joined by paraboloid and plane I need help with part of this old exam question:
Q:
C is the intersection line between the paraboloid $z=x^2+y^2$ and the plane $2x-2y+z=1$ travelled counter-clockwise from $(0,0,10)$. Calculate $\oint\limits_C F\cdot dr$. $F(x,y,z)=(z-3y)i+(3x+z)j+e^{x-y}k$
A:
After some calculations the professor end up here:
$$\oint\limits_C F\cdot dr = \\\\
\iint\limits_{x^2+2x+y^2-2y\le1}(-e^{x-y}-1,1-e^{x-y},6)\cdot(2,-2,1)dxdy =\\\\= \iint\limits_{(x+1)^2+(y-1)^2\le3}-2e^{x-y}-2-2+2e^{x-y}+6dxdy=\\\\
=2\pi(\sqrt3)^2=6\pi$$
I understand he is exchanging the $z$ in $2x-2y+z=1$ for the one in $z=x^2+y^2$, but then what? I get to $(x+1)^2+1+(y-1)^2+1\le1$ but how to get to $..\le3$? It there perhaps a radius thing that I should know about?
My questions are:
How is he getting that final limit of integration?
How is he performing that integration to get to $2\pi(\sqrt3)^2$?
How would I solve the integral with polar coordinates? I end up at $2\int_0^{2\pi}\int_0^1rdrd\phi=2\pi$
Thank you for your time!
 A: He's completing the square. Observe:
$$\begin{align}\
x^2+2x+y^2-2y&\le1\\
(x^2+2x+1)+(y^2-2y+1)&\le1+1+1\\
(x+1)^2+(y-1)^2&\le3
\end{align}$$
I'm not sure how he's doing the integration, but the integration can be done with polar coordinates. Your region is a circle centered at $(-1,1)$.
$$\iint_D({-2e^{x-y}-2-2+2e^{x-y}+6})\,dA=\iint_D2\,dA$$
$$\int_0^{2\pi}\int_0^\sqrt{3}2r\,dr\,d\theta=6\pi$$
A: Prof. solution is easier than the following ( see below ) !!!

You can find $x$ and $y$ in terms of $z$:
$$
z - \left(1 - z \over 2\right)^{2} = \left(x^{2} + y^{2}\right) - \left(x - y\right)^{2}
=
2xy
$$
$$
2xy = {-z^{2} + 6z -1 \over 4}
$$
$$
\left(x^{2} + y^{2}\right) + 2xy
=
z + {-z^{2} + 6z -1 \over 4}
=
{-z^{2} + 10z -1 \over 4}
$$
$$
\left.%
\begin{array}{rcl}
&&\\[1mm]
x + y & = & {1 \over 2}\,\sqrt{-z^{2} + 10z -1}
\\[2mm]
x - y & = & {1 \over 2}\,\left(1 - z\right)
\\[1mm]&&
\end{array}\right\rbrace
\ \Longrightarrow\
\left\lbrace%
\begin{array}{rcl}
&&\\[1mm]
x & = & {1 \over 4}\left(\sqrt{-z^{2} + 10z -1}\ + 1 - z\right)
\\[2mm]
y & = & {1 \over 4}\left(\sqrt{-z^{2} + 10z -1}\ - 1 + z\right)
\\[1mm]&&
\end{array}\right.
$$
Roots of $\displaystyle{-z^{2} + 10z - 1 = 0}$ are $\displaystyle{z_{\pm} = 5 \pm 2\sqrt{3}}$ such that
$$
5 - 2\sqrt{3}\quad <\ z\quad <\quad 5 + 2\sqrt{3}
$$
$$
\oint_{C}\vec{\rm F}\left(\vec{r}\right)\cdot{\rm d}\vec{r}
=
\int_{z_{-}}^{z_{+}}\left\lbrack%
{\rm F}_{x}\left(x\left(z\right),y\left(z\right),z\right)\ x'\left(z\right)
+
{\rm F}_{y}\left(x\left(z\right),y\left(z\right),z\right)\ y'\left(z\right)
+
{\rm F}_{z}\left(x\left(z\right),y\left(z\right),z\right)
\right\rbrack
{\rm d}z
$$
