Line integral around intersection of sphere and plane The unit sphere is intersected by the plane $x + y = 14$. Find the line integral of $F = \langle yz + y, xz+5x,xy+2y\rangle$ around the intersection.

$$\iint\nabla\times F\cdot\textbf{n}\ dA$$

the unit normal vector is easily found by looking at the plane equation: $\frac{1}{\sqrt2}\langle1,1,0\rangle$

Curl of $F = \langle 2,-1,4\rangle$. So the curl of $F$ dotted with that = $1/\sqrt2$, so we have 

$$\iint\ \frac{1}{\sqrt{2}}\ dA$$

It's finding an expression for the skewed circle I'm having trouble with. By some algebra, I was able to find the circles projection in the $xz$-plane:

$$2\left(\frac{x-1}{2}\right)^2 + z^2 = \frac 32$$

But I'm not sure what I'm supposed to do now...
The unit circle $x^2 + y^2 = 1$ is a great circle of the unit sphere. This unit circle also happens to intersect the circle of intersection between the unit sphere and the plane, hence the chord connecting those two points is a diameter of the skewed circle. This diameter is also of course the diagonal of the unit square in the xy-plane, so its length is $\sqrt2$.

Then the area of the skewed disk is $A = \pi (\frac{\sqrt2}{2})^2 = \frac{\pi}{2}.$
\begin{align}
1 &amp; =
x^{2} + y^{2} + z^{2}
=
x^{2} + \left(1 - x\right)^{2} + z^{2}
=
2\left(x^{2} -x\right) + 1  + z^{2}
\\[5mm] &amp; =
2\left(x - {1 \over 2}\right)^{2} + {1 \over 2}  + z^{2}
\end{align}

$$
2\left(x - {1 \over 2}\right)^{2} + z^{2} = {1 \over 2}
\quad\Longrightarrow\quad
{\left(x - 1/2\right)^{2} \over \left(\color{red}{1/2}\right)^{2}}
+
{z^{2} \over \left(\color{red}{\sqrt{2\,}/2}\right)^{2}} = 1\,,
\quad
\mbox{( Ellipse )}
$$

$S_{\rm A}$: Surface area.
$\displaystyle
S_{\rm A} = \pi\times\left(\color{red}{1 \over 2}\right)\times\left(\color{red}{\sqrt{2\,} \over 2}\right)
= {\sqrt{2\,} \over 4}\,\pi$.

Integral de linea: $\left(1/\sqrt{2\,}\right)\left(\sqrt{2\,}\,\pi/4\right)
=
\color{#ff0000}{\displaystyle{\large{\pi \over 4}}}$