Computing cross section area between cylinder and disk

Question

I am creating a cylinder, whose center disk is placed at position vector v1 and whose orientation is given by the disk's normal vector n1. The second object is a disk, embedded in the same box, but since I don't know how to draw disks in $3D$ in Mathematica, I use a very thin cylinder to approximate the disk, which is placed at v2 and orientation n2. Both have diameter d=4. Here's the setup:

v1 = {0.5, 0.5, 0.5};
n1 = {1, 1, 1};
v2 = {1, 1.5, 0};
n2 = {1, 1, 0};
d = 4;

ef1 = 5; (*elongation factor of cylinder 1 to find endpoints to draw*)
\
ef2 = 0.00001; (*elongation factor of cylinder 2, to approximate disk*)
\
cyl1 = Cylinder[{v1 - ef1*n1, v1 + ef1*n1}, d/2];
cyl2 = Cylinder[{v2 - ef2*n2, v2 + ef2*n2}, d/2];

And drawn together: Graphics3D[{Opacity[.5], cyl1, cyl2}]:

As we see in the image, the cylinder is crossing a portion of the disk, and I'm trying to learn if:

• Is there way a to compute the area of the disk that is in intersection with the crossing cylinder? A naive approach using Area@RegionIntersection[cyl1, cyl2] seems not to work (returns Undefined).

Answer 1

You consider tw0 3D-regions, that's why its onlye possible to calculate a volume!

Try

dV=Volume@RegionIntersection[cyl1, cyl2]
(*0.000249282 *) 

To get an approximation of the disk area divison by the thickness of your pseudo-disc gives the result!

addendum The workaround cyl2 isn't necessary. Try

disk = ImplicitRegion[({x, y, z} - v2).n2 == 0&&Norm[{x, y, z} - v2] <= d/2, {x, y, z}]
dA = RegionIntersection[cyl1, disk] 
Area@dA        
(*8.81346*)