What is the surface area of the plane $x+2y+2z=12$ cutoff by $x=0, y=0$ and $x^2+y^2=16$ 
What is the surface area of the plane $x+2y+2z=12$ cutoff by $x=0, y=0$ and $x^2+y^2=16\,?$

I am unable to solve this question since it involves $z$ variable too. 
 A: The cylinder
$$x^2+y^2=16$$
cut off an ellipse from the plane $$x+2y+2z=12$$
The projection of the ellipse in the x-y plane is given by the circle:
$$x^2+y^2\leq16$$
with radius $R=4$ and area $A=\pi R^2=16\pi$.
The ratio between the area of the circle and the area of the ellipse is given by the $\cos \alpha$, $\alpha$ being the angle between the normal to the plane and the $z$ axis.
Thus: normal to the plane: $\vec n(1,2,2) \implies \cos \alpha=\frac{2}{3}$.

The area of a quarter of ellipse is thus:
$$S=\frac32 \cdot 4\pi=6\pi$$

A: First, let me know what do you think about given equations?
here, x+2y+2z=12 is a plane cutting X,Y&Z axes at 12,6,6 respectively.
x=0 is a plane i.e. YZ-plane & y=0 is XZ-plane.
x^2+y^2=16 is a circular cylinder with radius 4 and Z-axis as its axis.
Now trace all these and you will get some part of plane x+2y+2z=12 inside the cylinder b/w axes- planes. This is the portion for which to find the surface area and take projection of this portion on xy-plane , you will get 1/4 of circle of radius 4 with centre at origin.( say this circular region R)
Now, write the equation of plane in form of Monge's equation i. e.  z=6-(x/2)-y
and find it's partial derivatives with respect to x and y both.
formula for surface area is,  S=$\iint \ ds $ over R.
here, dS =$ \sqrt( p^2 + q^2 + 1 )\ dx dy $;where p and q are partial derivatives of the Monge's equation with respect to x&y respectively.
p=-1/2  , q = -1 ; so ds=(3/2) dx dy
Now, S =3/2$\iint \ dx dy $ over R
you can see, S= (3/2)area of 1/4 of circle of radius 4 with centre at origin.
therefore S =(1/4)24$\pi$ = 6$\pi$
