Find the surface area of the part of the cone $z=\sqrt{x^2+y^2}$ that lies below the plane $x+2z=3$ It seems as though I'll have to use some sort of integral (surface integral?) using the distance between the plane and the bottom of the cone as limits of integration. Unfortunately I'm not familiar with surface integrals.
 A: *

*Parametrize the surface, for example:
$$
x=x, \quad y=y, \quad z=\sqrt{x^2+y^2}; \quad (x,y)\in D
$$
where $D$ is the projection of  the intersection of $x+2z=3$ and $z=\sqrt{x^2+y^2}$ in the $xy$ plane. Can you write $D$ explicitly?

*From this parametrization, it follows that the surface area equals
$$
A=\iint_D ||r_x \times r_y||\; dA
$$
where $r_x=(1,0,\frac{\partial \sqrt{x^2+y^2}}{ \partial x})$ and $r_y=(0,1,\frac{\partial \sqrt{x^2+y^2}}{ \partial y})$.


This is the general methodology. I am leaving the technical details for you to do ;)
A: To find the region over which you are to integrate, take the intersection of the two surfaces to find a curve bounding the region (this is equivalent to projecting),
$$
z=\sqrt{x^2+y^2}\;\text{and}\;z=\frac{3-x}{2}\implies \frac{(x+1)^2}{4}+\frac{y^2}{3}=1
$$
by the usual completing the square technique. 
Then, your integrand will be the norm of the vector normal to the surface 
$$
z=\sqrt{x^2+y^2}
$$
which is 
$$
\left|\left|\left(\frac{x}{\sqrt{x^2+y^2}},\frac{y}{\sqrt{x^2+y^2}},-1 \right)\right|\right|=\sqrt{2}
$$
Recall that the area of an ellipse with major axis $a$, $2$ for you, and minor axis $b$, $\sqrt{3}$ for you, is 
$\pi ab=\pi 2\sqrt{3}$, giving you the surface area,
$$
2\pi\sqrt{6}
$$
A: The projection of the intersection of the cone and plane onto the $xy$ plane is an ellipse $$\frac{(x+1)^2}{4}+\frac{y^2}{3}=1.$$(Obtained by solving equations $z=\sqrt{x^2+y^2}$ and $x+2z=3$.) Using change of variables $u=\frac{x+1}{2},\,v=\frac{y}{\sqrt{3}}$ with the Jacobian $\frac{\partial(x,y)}{\partial(x,y)}=2\sqrt{3}$. Then we have $$\int\int_{\frac{(x+1)^2}{4}+\frac{y^2}{3}\leq 1}dydx=2\sqrt{3}\int\int_{u^2+v^2\leq 1}dvdu=2\sqrt{3}\pi$$.
