Given that $\textbf{F} = \langle z,x,y \rangle$,
The plane $ z=2x+2y-1$ and the paraboloid $ z= x^2 +y^2$ intersect in a closed curve.
I'm trying to use stokes theorem to find the line integral.
Attempt:
We know that Stokes Theorem is given by: $\iint_S (\nabla \times \textbf{F}) \cdot \textbf{N}\: d\textbf{S} $
$\nabla \times \textbf{F} = \vec{\imath} + \vec{\jmath} + \vec{k} = \langle 1 , 1, 1\rangle$
Now, the problem I have is parameterizing. This is what I did.
$$\vec{r} = \langle x,y,2x+2y-1\rangle$$ $R_x \times R_y = 2\vec{\imath} - 2\vec{\jmath} + 0\vec{k} $
Now, $\iint_S \langle 1 , 1, 1\rangle \cdot \langle 2 , -2, 0\rangle \: d\textbf{S} = 0 $. But this is not right!
Perhaps what I'm struggling most is getting the parameterization right for the surface. Also would I have to convert to polar coordinates to complete the integral?
Any help would be appreciated! Thank you
parameterizing the surface.
$R_x = (1,0,2)\\ R_y = (0,1,2)\\ R_x \times R_y = (-2,-2,1)$
Don't know were you may have lost it.
As a general rule:
$S = (x,y,f(x,y))\\ dS = (-\frac {dz}{dx}, -\frac {dz}{dy}, 1)$
Also consider $dS$ will always be normal to the surface, and the normal line of a plane is just the coefficients of $x,y,z$ in the equation of a plane (when in standard form).
I guess you need to know to scale the $z$ coordinate in $dS$ so that it equals $1.$