I saw a few posts on this sort of questions but I couldn't apply any idea on my particular question.
Question :
Calculate $\int_C (2z-x)dx+(2x-y)dy+(2y-z)dz$ when $C$ is the intersection of sphere $x^2+y^2+z^2=1$ and $x+2y+3z=0$.
My try :
I tried to find the intersection but I got complicated quality ($x^2+y^2+ xy$) and so on..
I think the center of the circle obtained by the intersection is $(0,0,0)$ ,
since the distance between $(0,0,0)$ and some point on the sphere - for example $(1,0,0)$ is $1$.
I assume the radius of the obtained circle is also $1$.
By Stoke's theorem we get -
$$\int_C (2z-x)dx+(2x-y)dy+(2y-z)dz = \int\int_S Curl(F)nds$$
The $curl(F)=(2,2,2)$ , $ n =(\frac{1}{\sqrt{14}},\frac{2}{\sqrt{14}},\frac{3}{\sqrt{14}})$
$$\int \int_SCurl(F)nds=\frac{12}{\sqrt{14}}\int\int_S dA= \frac{12}{\sqrt{14}} \pi$$
I'm really not sure this is the right answer.
Thank you in advance.
Yes you have applied the Stokes' theorem correctly and your answer is right. Note that your integration is over region $S$ and not $D$.
$ \displaystyle \iint_S( \nabla \times \vec F) \cdot \hat n ~dS=\frac{12}{\sqrt{14}}\iint_S dS$
where $S$ is a disk defined by $~S: ~x + 2y + 3z = 0, x^2 + y^2 + z^2 \leq 1$
Now as $x + 2y + 3z = 0$ is a plane passing through the center of the sphere $x^2 + y^2 + z^2 \leq 1$, the intersection forms a great circle [A great circle of a sphere is the intersection of the sphere and a plane that passes through the center point of the sphere wiki]
The area of a great circle is $\pi R^2$ where $R$ is the radius of the sphere and that leads to surface area of the disk as $\pi$ and the answer is $~\dfrac{12 \pi}{\sqrt{14}}$