I have the cylinder $x^2+y^2=R^2$ and the plane $z=ax+b$ ($R,a,b$ are constants). I need to find the area of the ellipse that is the region of the plane enclosed by the cylinder.
I've done it with a double integral in cylindrical coordinates and I obtained $A=\pi R^2 \sqrt{1+a^2}$. But now I would like to do it using the formula of the area of an ellipse $A= \pi AB$ where A and B are the semi-axes.
The problem is I don't know how to find the size of the axes of the ellipse.
If it helps, I have the intuition that the minor semi-axe is R.
Since the plane only tilts in the $x$ direction, by symmetry we have that the vertices of the major and minor axes lie above the points $(\pm R, 0)$ and $(0,\pm R)$ in the $xy$ plane, respectively. Plugging in the equation of the plane, the distance between the points is:
$\textbf{Minor Axis}$
$$|(0,R,b)-(0,-R,b)| = 2R$$
$\textbf{Major Axis}$
$$|(R,0,aR+b) - (-R,0,-aR+b)| = \sqrt{(2R)^2+(2aR)^2} = 2R\sqrt{1+a^2}$$
This means the area of the ellipse is
$$\text{Area} = \pi a b = \pi\left(R\sqrt{1+a^2}\right)(R) = \pi R^2\sqrt{1+a^2}$$
which incidentally means your initial calculation was incorrect.