# triple integrals and cylindrical coordinates

my question is can cylindrical coordinates be somehow changed in order? I mean the standard cylindrical coordinates suppose a line of length $r$ that makes an angle $\theta$ with the $x$ axis and let $z$ vary and then $x=rcos\theta$ and $y=rsin\theta$ and the line $r$ (at least initially) is in the $xy$ plane

now can we suppose the the line of length $r$ makes an angle with the $x$ axis but instead is in the $xz$ plane and let $y$ vary and then $x=rcos\theta$ and $z=rsin\theta$ and if this can be done then what is wrong with this?

the problem is to find the volume of the solid resulting from the intersection of two cylinders one with equation $x^2+y^2=r^2$ and the other $z^2+y^2=r^2$

to be clear I assumed that the solid is highly symmetrical so I divided it into eight parts and tried to find the volume of one of them and I chose the one in the first octant

now surely $r$ ranges from 0 to r and there is $y^2=r^2 -x^2$ and $y^2=r^2-z^2$ so $y^2$ $\leq$ $r^2 -y^2$ and at the same time $y^2$ $\leq$ $r^2 -z^2$ now converting to cylindrical coordinates with $x=rcos\theta$ and $z=rsin\theta$ and taking the positive square root (recall that it's in the first octant) we get $y$ $\leq$ $rsin\theta$ and $y$ $\leq$ $rcos\theta$

so all I need is the set of all points that are less than both $rsin\theta$ and $rcos\theta$ now if we take $\theta$ to be $\pi/4$ then from the graphs of $sin\theta$ and $cos\theta$ we see that that sin is always less the cosine so we can safely assume that from $0$ to $\pi/4$ ,,, $y$ $\leq$ $rsin\theta$ so the integral becomes

$$\int_0^{\frac\pi4} \int_0^{rsin\theta} \int_ 0^r r \,dr\,dyd\theta$$

after evaluating you multiply the result by 2 to get the volume of the part of the solid in the first octant and then by 8 to get the full solid and I get in the end this result which is wrong $V=4r^3({2-\sqrt2})$

and I'm fully aware of the other methods to get the right answer which is $\frac{16}{3}\ r^3$ but I wanted to try out this solution

my only guess is that it's the symmetry supposition but then I would have no idea why it's not symmetrical since it's very hard to visualize and not easy to examine even if I'm looking at a graph of it

• It is very symmetrical about the y axis, in fact the cross sections perpendicular to the y axis are squares centered on the axis. – Triatticus Aug 3 '18 at 4:12
• what's wrong then? – Km356 Aug 3 '18 at 4:15
• Im not entirely sure since I cannot work it out on some paper at the moment, is there some reason you want to use cylindrical coordinates even though Cartesian is easier? – Triatticus Aug 3 '18 at 4:19
• yes I want to try changing the the variable from the ordinary z to y or maybe x.. it's not about this certain problem it's about testing a method on a complicated enough example – Km356 Aug 3 '18 at 4:21

Yes, you're free to decide along which coordinate axis to point the symmetry axis of your cylindrical coordinates. I don't see a reason not to choose the $z$ axis for this problem as usual, but you certainly can; it just might make the integrals more difficult to evaluate.

I haven't checked all of what you've done, but there's an error right at the beginning -- you're not distinguishing between the radius in the limits and the radius as one of the cylindrical coordinates. If we denote these by $r$ and $\rho$, respectively, the limits are

$$x^2+y^2\le r^2\;,\\ z^2+y^2\le r^2\;,$$

and the coordinates you chose are

$$x=\rho\cos\theta\;,\\ z=\rho\sin\theta\;\hphantom,$$

and $y$, so the limits in these coordinates are

$$\rho^2\cos^2\theta+y^2\le r^2\;, \\ \rho^2\sin^2\theta+y^2\le r^2\;.$$