Spherical and cylindrical problemshttp://calc3.askbot.com/questions/Open source question and answer forum written in Python and DjangoenCopyright Askbot, 2010-2011.Fri, 18 Jul 2014 13:17:55 -0500Finding volume using integrationhttp://calc3.askbot.com/question/135/finding-volume-using-integration/I have been thinking about this problem in section 15.1 #30: Three cylinders of radius 1 intersect at right angles at the origin, as shown in figure 15.1.6. Find the volume contained inside all three cylinders. My first thought is to find the points of intersection but other than that I'm pretty much lost. Any ideas?Fri, 18 Jul 2014 09:18:08 -0500http://calc3.askbot.com/question/135/finding-volume-using-integration/Answer by SpaceManSpiff for Finding volume using integration http://calc3.askbot.com/question/135/finding-volume-using-integration/?answer=136#post-id-136First of all. Excellent question. Getting started on this took three students, four websites, a physics professor, and a visit to the math lab. I'm still not sure if I'm right and there are some things you'll have to take at face value but I have a way to get things moving a little bit. So according to the internet, the shape this forms is called a steinmetz solid, the three cylinder steinmetz solid happens to be something called a "Rhombic Dodecahedron" (seen below). ![image description](/upfiles/calc3.askbot.com/14057061284997452.gif) This is kind of like the shape you would get if you tried to make a rubber band ball around a cube, or the one I ended up thinking of it as; "Cube with square shaped yamikas on all sides." ![image description](/upfiles/calc3.askbot.com/14057062556281134.gif) **Images courtesy of Wolfram Mathworld** The radius of the cylinders through this cube is 1 so a side length is 2. Thus the volume of the cube section is 8. Next we add the volumes of the six "square yamikas". I am totally unable to come up with a way of doing this but Wolfram has a method of parametrizing the surface area and apparently then does an iterated integral of the parametrized surface area from $-r$ to $r$ or $-1$ to $1$ in our case. And from $0$ to $\frac{\pi}{4}$. So they're finding one fourth of the volume of a yamika. Their formula (the derivation of which is not explained) implies our problem should look something like this. $$8+6 \int_{0}^{\frac{\pi}{4}} \int_{-r}^{r} (2(r^{2}-z^{2})^{\frac{1}{2}})dz$$ I don't understand their parametrization at all but if this can get anyone better than me started on this then I would greatly appreciate an explanation to this too. Mathworld really did seem to be the best resource on this even though it's really not a ton of help. Fri, 18 Jul 2014 13:17:55 -0500http://calc3.askbot.com/question/135/finding-volume-using-integration/?answer=136#post-id-136