Calculate line integral $\oint _C d \overrightarrow{r } \times \overrightarrow{a }$

Question

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?

Answer 1

$x^2 + y^2 + z^2 = \frac 34\\ x^2 = y\\ y^2 + y + z^2 = 1\\ (y+\frac 12)^2 + z^2 = \frac 54$

$y = \frac {\sqrt 5}{2} \cos\theta - \frac 12\\ z = \frac {\sqrt 5}{2} \cos\theta\\ x = \sqrt {\frac {\sqrt 5}{2} \cos\theta - \frac 12}$

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.