Finding surface area of part of a plane that lies inside a cylinder??? I have a question::
Let $S$ be the part of plane $x+2y+3z=1$ that lies inside cylinder $x^2 + y^2 = 3$
They want me to find the surface area of S??
This is a way harder question than all my previous ones, and I think I should start by finding the intersection point of the plane and the cylinder:
$x+2y+3z-1 = x^2 + y^2 -3$
$x^2-x + y^2 - 2y -2 = 3z$
$x(x-1)+y(y-2)-2=3z$
Now I am stuck? Help me get the equation please!

Note that the surface will be bounded by an ellipse. You're having trouble because you're trying to describe the surface in rectangular coordinates, when instead there is an obvious parametrization using polar coordinates.
Hint: parametrize the surface by $$\langle \underbrace{r\cos\theta}_{=x}, \underbrace{r\sin\theta}_{=y}, \underbrace{\frac{1}{3}\left(1-r\cos\theta-2r\sin\theta\right)}_{=z}\rangle$$ for $r\in[0,\sqrt{3}]$ and $\theta\in[0,2\pi)$. Then do the appropriate surface integral.
