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!

With the surface defined by $g(x,y,z)=x+2y+3z-1=0$ over the domain $(x,y)\in C=\{(a,b):a^2+b^2\le3\}$, use the formula:
\begin{align}
\text{Surface Area} &= \int \int_C \sqrt{\frac{g_x^2+g_y^2+g_z^2}{g_z^2}}\mathrm{d}x \mathrm{d}y\\
&=\int \int_C \sqrt{\frac{1^2+2^2+3^2}{3^2}}\mathrm{d}x \mathrm{d}y\\
&=\int \int_C \frac{\sqrt{14}}{3}\mathrm{d}x \mathrm{d}y\\
&=\frac{\sqrt{14}}{3}\int \int_C \mathrm{d}x \mathrm{d}y\\
&=\frac{\sqrt{14}}{3} \pi (\sqrt{3})^2\\
&=\pi \sqrt{14}
\end{align}
Note the final expression for the double integral was simply the area of the region in the $x$-$y$ plane that we were integrating over (a circle of radius $\sqrt{3}$)
