Calculate line integral $\oint _C d \overrightarrow{r } \times \overrightarrow{a }$ Calculate line integral:
$$\oint_C d \overrightarrow{r}  \times \overrightarrow{a},$$
where $\overrightarrow{a} = -yz\overrightarrow{i} + xz \overrightarrow{j} +xy \overrightarrow{k}$ , and curve $C$ is intersection of surfaces given by $ x^2+y^2+z^2 = 1$, $y=x^2$, positively oriented looked from the positive part of axis $Oy$.
I first tried to calculate the intersection vector $\overrightarrow{r }$ by substituting $y=x^2$ into $ x^2+y^2+z^2 = 1$, and completing the square with $y+1/2$ and got the following : $ \overrightarrow{r}= \pm\sqrt{ \sqrt{5/4} \cos(t) - 1/2}\overrightarrow{i}, (\sqrt{5/4}\cos(t) - 1/2 )\overrightarrow{j}, (\sqrt{5/4}\sin(t) )\overrightarrow{k}$,
$  -\cos^{-1} (1/\sqrt 5)<t< + \cos^{-1}(1/\sqrt 5)$
After that i tried to vector multiply the two vectors, after differentiating $\overrightarrow{r}$ first, but I got some difficult expressions to integrate.
Is there an easier way to do this?

$x^2 + y^2 + z^2  = \frac 34\\
x^2 = y\\
y^2 + y + z^2 = 1\\
(y+\frac 12)^2 + z^2 = \frac 54$
$y = \frac {\sqrt 5}{2} \cos\theta - \frac 12\\
z = \frac {\sqrt 5}{2} \cos\theta\\
x = \sqrt {\frac {\sqrt 5}{2} \cos\theta - \frac 12}$
or we could do standard spherical:
$x = \sin\phi\cos\theta\\
y = \sin\phi\sin\theta\\
z = \cos\phi$
and then incorporate the parabola $y=x^2$
$\sin\phi\sin\theta = \sin^2\phi\cos^2\theta\\
\sin\phi = \sec\theta\tan \theta\\
\phi = \arcsin\sec\theta\tan \theta$
$x = \tan \theta\\
y = \tan^2 \theta\\
z = \sqrt{1-\sec^2\theta\tan^2 \theta}$
would be another.
But since it is a closed curve, perhaps Stokes theorem is in order
$\nabla \times a = (0,2y,2z)$
Lets use the surface of the parabolic cylinder:
$x = x\\
y = x^2\\
z = z$
$dS = (-2x,1, 0)$
$\iint 4x^2 \ dz \ dx\\
\int 4 x^2z \ dz$
I need some limits of integration.
$x^2 + x^4 + z^2 = 1\\
z = \pm\sqrt{1-x^4-x^2}\\
x^4 + x^2 - 1 = 0\\
x^2 = \frac 12(1-\sqrt5)$
$\int_{-\sqrt \phi}^{\sqrt \phi} 8 x^2\sqrt{1-x^4-x^2} \ dx$
And that is as far as I can take this.
