EDIT: Is my computation not correct, possibly because the parametrization that I used requires x,y to be on the xy-plane?
If so, can I adjust from here, and not start over? I.e., is there some sort of rotation that I can apply? Although a rotation would not help with changing the value of the answer, I think...
My work:
I computed the answer to be $3\pi$, while the answer is supposed to $\sqrt{3} \pi$.
The integral is
$$\int_C zdx + xdy + zdz$$
where C is the circle of intersection of the plane $x+y+z=0$, and the sphere $x^2+y^2+z^2=1$.
My work:
I wrote $x=rcos(\theta)$, $y=rsin(\theta)$, $z=-x-y$ $=-rcos(\theta)-rsin(\theta)$. Note that I used radius = $r=1$.
Then, I computed $d\theta$, in terms of $dx$, $dy$, and $dz$.
Finally, I computed an ordinary Riemman integral of $(1+sin^2(\theta)$)$d\theta$, from 0 to $2\pi$, and got $3\pi$ as my answer.
Have I overlooked something important? Am I off by a scaling factor? Or perhaps I paramaterized incorrectly. My computations look straightforward and ok, though.
Any hints or suggestions are welcome.
Thanks,
Your parametization of $C$ is not correct. It just does not satisfy $x^2 + y^2 + z^2 = 1$.
To find a correct one, note that $(1, 1, 1)$ is the normal vector of the plane $x+y+ z = 0$. Then one can find two orthonormal vectors $e_1, e_2$ such that both of them are prependicular to $(1, 1, 1)$, then $C$ can be parametrized by
$$\theta \mapsto \cos\theta e_1 +\sin \theta e_2.$$
I think the rest will not be difficult. (After some works from the OP, the parametrization of $C$ is given by
$$\begin{split} \theta &\mapsto \cos\theta \frac{1}{\sqrt 6}(-1,-1, 2) + \sin\theta \frac{1}{\sqrt 2} (1, -1, 0) \\ &= \left(-\frac{1}{\sqrt 6} \cos\theta +\frac{1}{\sqrt 2} \sin \theta, -\frac{1}{\sqrt 6} \cos\theta -\frac{1}{\sqrt 2} \sin \theta, \frac{2}{\sqrt 6} \cos\theta\right) \end{split}$$
Thus
$$x = -\frac{1}{\sqrt 6} \cos\theta +\frac{1}{\sqrt 2} \sin \theta,\ \ y = -\frac{1}{\sqrt 6} \cos\theta -\frac{1}{\sqrt 2} \sin \theta,\ \ z = \frac{2}{\sqrt 6} \cos\theta.$$