Volume of the Intersection of Ten Cylinders I'm in Calculus 2, and we were first given the problem to find the intersection of two perpendicular cylinders of equal radius.

This breaks down into eight times the volume of a quarter circle (with radius r) with perpendicular square cross sections.
$$V=8\int_0^r \sqrt{r^2-x^2}^2dx=8\int(r^2-x^2)dx=8\left[ r^2x - \frac{1}{3}x^3 \right]^{r}_{0}=\frac{16}{3}r^3$$
After this question on the problem set, my teacher has written  "Aren't you glad I didn't have you find the intersection of ten cylinders?"
Assuming the ten cylinders intersect in an equal way, like the faces of an icosahedron, I assume this would make some sort of curvy-face icosahedron. 
My question is two parts


*

*Can I find the volume using a Calculus II base of knowledge (including a bit of multivar)?




  
*What is the volume of the intersection of ten cylinders of equally radius equally spaced?
  

Edit: The question should be so that the axis of each cylinder is perpendicular to the face of an icosahedron- because this is 10 pairs of parallel sides, that should be ten cylinders.
Edit 2:
Question 1 is answered: No, but maybe. (That wasn't the important part anyway)
Question 2 is still hanging, as I'd like to see the methodology involved, I'll restate the problem with my current understanding of it.

Ten cylinders, each of radius r intersect along the lines that are perpendicular to the faces of a regular icosahedron at the center of each face. What is the volume of the intersection?

I have created rather crude pictures with my limited Geogebra knowledge:


 A: Question Two
Consider the example you gave: orienting the poles of the cylinders in a perpendicular fashion must give the smallest possible intersection volume. If the cylinders point in the same direction, you'd have the largest intersection, because the cylinders would fully intersect, and as you rotate away from this extreme, you would decrease the volume of intersection.
You can imagine a skewer going through the middle of a sphere (which you can imagine as the long side of a cylinder), how would you insert another skewer so that the points which the second skewer enter and exit the sphere are as far away from the others as possible? You should see that they would be perpendicular there.
If you had ten cylinders, that would be trying to solve the problem where you have ten skewers (cylinders) to poke into the sphere, now the problem of arranging them has become far more difficult :)
But you can convince yourself (or calculate), the fact that these skewer entry and exit points are as far away from each other as possible, when they are equally spaced by symmetry; since the surface of a sphere is regular.
Question One
So since orienting the cylinders is a problem in itself, I don't believe there will be an easy way to solve this with Calculus II knowledge. But in principle it could be done. You'd just take the integral of the cylinders one at a time.
