Evaluate Integral Using Stokes Theorem

Question

Using Stokes theorem evaluate

$$\int_\Gamma(z-y)\,dx-(x+z)\,dy-(x+y)\,dz$$

where $\Gamma$ is the intersection of $x^2+y^2+z^2=4$ with the plane $z=y$ with anticlockwise orientation when looking on $z$ from the positive side

If we take $\nabla\times F=(0,2,0)$ but how can we find $\hat{n}$

Answer 1

Since $\Gamma$ is a circle in a plane, we can apply Stokes' Theorem to the disk $D$ that it encloses to get that $\int_{\Gamma} F \cdot ds = \int_D$ curl $F \cdot dA$. At each point of $D$, the normal to the disk is the same as the normal to the plane. In general, a normal to a plane is a vector $v$ such that the equation of the plane can be written in the form $(x - x_0, y - y_0, z - z_0) \cdot v = 0$, where $(x_0, y_0, z_o)$ is any fixed point on the plane. In this case, we can take $(x_0, y_0, z_0)$ to be the origin. So you just need to find a unit vector $v$ with this property, pointing in whichever direction agrees with the orientation of $\Gamma$.

Answer 2

$\vec n$ is normal to plane. So, $ \vec n = <0,1,-1>/\sqrt 2 \\ \iint Curl (\vec F). \vec n dS = \iint <0,2,0> \cdot (<0,1,-1>/\sqrt 2) dS \\ = \iint \sqrt 2 dS = \sqrt 2 (Area \space of \space circle) = \sqrt 2 \pi(2^2) \\ = 4 \sqrt 2\pi$