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!
 A: 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 $\sqrt{3}$)
A: 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.
A: The orthogonal cross section of the cylinder is a circular disc $D$ with known radius, and the given plane has an  inclination with respect to $D$ that can be computed from the data. Using this the area of $S$ can be computed without reference to any integrals.
