I have the plane $x + y + z = 1$ and the cylinder $x^2 + y^2 = 4,$ and need to find the part of the plane which is inside the cylinder; I have the feeling it's going to be an ellipse.
I tried doing $$ x^2 + y^2 = 4x + 4y + 4z,$$ but this is a paraboloid, so it can't be what i'm looking for. What am I doing wrong and how can I find the part of the plane inside the cylinder?
Use parametrizations. The surface can be described as the image of $$\langle r\cos\theta, r\sin\theta,1-r(\cos\theta+\sin\theta)\rangle$$ for $(r,\theta)\in[0,2]\times[0,2\pi)$.
(Setting $x=r\cos\theta$ and $y=r\sin\theta$ for $(r,\theta)\in[0,2]\times[0,2\pi)$ ensures that you stay inside the cylinder $x^2+y^2=4$. Then $z$ must be $1-(x+y)$ to lie on the plane.)
EDIT
Here's a little more on cylindric sections.
If you're interested in proving that the boundary of this surface is actually an ellipse, there is a simple discussion on pages 7-8 of Hilbert's beautiful book Geometry and the Imagination. The basic idea is to embed two spheres inside the cylinder, tangent the plane on each side. The two points where the spheres touch the plane turn out to be the foci of the ellipse. To see this, take an arbitrary point on the boundary of the intersection, and draw through it the line on the cylinder parallel to the height. This line will intersect the equators of the spheres and two points of tangency, but clearly the length of this line segment is fixed as you move around the boundary. But the foci are points of tangency too; and tangents drawn from the same point must be equal in length. This shows that the sum of the distances from the two foci is fixed; so the boundary is an ellipse.
Here's the diagram Hilbert uses:
