I've been banging my head against this thing for the last 4 hours to no avail
Evaluate the integral $ \iint_S curl \vec{F}\cdot d\vec{S}$ where $S$ is the portion of the surface of the sphere defined by $x+y+z\leq 1$ and $x^2+y^2+z^2=1$ and where $F=(5x,5y,5z) \times (1,1,1)$.
Now, I've tried using the boundary of $S$, but so far, I've been unable to find a parametrization to describe it, and it's been suggested to me not to try this method. My only hope is to use the surface integral, but I can't really figure out the bounds of the integral. The only things I've managed to find out is the curl, which isn't a big deal
One of the approaches is to first use rotation of coordinate axes such that the plane is parallel to one of the coordinate axes. But we can use simple geometry instead.
The question does not state the orientation of the surface. So we assume outward normal vector.
We are to find $~\displaystyle \iint_S (\nabla \times \vec F)\cdot \hat n ~dS~$ where $S$ is the surface of the sphere $x^2 + y^2 + z^2 = 1$, below the plane $x + y + z = 1$. Applying Stokes' theorem we know that the surface integral over $S$ would be same as integral over another surface with the same boundary. So we would instead use the disc surface $~E: x + y + z = 1, x^2 + y^2 + z^2 \leq 1$
Now given the curl of $\vec F$ is a constant vector and the unit normal vector is $\hat n = \frac{1}{\sqrt3}(-1, -1, -1)$. That leads to $\vec F \cdot \hat n = c$ (a constant). So, $~\displaystyle \iint_E (\nabla \times \vec F)\cdot \hat n ~da = ac~$, where $a$ is the surface area of the disc $x + y + z = 1, x^2 + y^2 + z^2 \leq 1$.
The perpendicular distance from the origin to $x + y + z = 1$ is $\frac{1}{\sqrt3}$ and $r^2 = 1 - \frac{1}{3} =\frac 23$. Here $r$ is the radius of the disc and surface area of the disc is $a = \frac{2\pi}{3}$.
Can you take it from here?