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Find the volume common to two circular cylinders, each with radius r, if the axes of the cylinders intersect at right angles. (using disk/washer)

I saw no example of this problem anywhere.. I saw an example how to solve it without calculus but I'm afraid that'll be deductions on my test.. I have the volume of a cylinder and the method of area between two curves but this is out of my reach.... I don't know what I can try...

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  • $\begingroup$ "...each with radius..." ?? $\endgroup$ – DonAntonio Sep 15 '13 at 10:22
  • $\begingroup$ ______________r $\endgroup$ – J L Sep 15 '13 at 10:29
  • $\begingroup$ This post is chosen to be the target for duplicates. Although neither the content of the question post itself or the answers is outstanding, there are 5 existing duplicate links. There are only two other posts with merely one existing dup-link. $\endgroup$ – Lee David Chung Lin Jan 22 at 10:06
  • $\begingroup$ can also be quickly computed without calculus: re. to this related post $\endgroup$ – G Cab Jan 22 at 11:18
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The main challenge in this problem is to depict the solid itself. Take a look at the image below

enter image description here

As we increase the hight $z$ of the intersecting plane the section of the solid is a square with a side $a$. Since the solid is created with two cylinders of radius $r$, $a$ and $z$ are related by $$a^2+z^2=r^2$$

The area of the square is $A=a^2$ or in terms of $z$: $$A(z)=a^2=r^2-z^2$$ The method of washers tells us $$V=8\int_0^rA(z)dz=8\int_0^r(r^2-z^2)dz=8\left(r^3-\frac{r^3}{3}\right)=\frac{16}{3}r^3$$

Note that the coefficient $8$ is required since we are considering a $1/8$th part of the solid.

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http://www.math.tamu.edu/~Tom.Kiffe/calc3/newcylinder/2cylinder.html this link gives proper visualisation of intersection of 2 cylinders having same radii. Volume is calculated by 2 different methods.

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    $\begingroup$ It is always good to enter the gist of the answer while including links, since at some later point in time, the links may actually disappear. BTW, Welcome to Math.SE $\endgroup$ – Shailesh Feb 6 '16 at 12:20
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The derivation above looks good. This volume is known as a "Steinmetz Solid." There is a page discussing its properties at Mathworld. That page shows another way to get the volume by integration.

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These are the standard hints I was given in high school: Consider the two following cylinders: $y^2+z^2=1$ and $x^2+z^2=1$. Let $\mathcal{C}$ be the region common to both:

a) Can you explain why the volume is between $\frac{4}{3}\pi$ and $8$?

b)What section is obtained when $\mathcal{C}$ is sliced by a plane that is parallel to the $xy$-plane?

c)Integrate to find the volume.

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Let r bet the radius of both cylinders and their axes are both perpendicular and intersecting. There is a common sphere enveloped by the 2 intersecting cylinders. If we look at the planar section which passes through both cylinder axes we can see a square area common to both cylinders. This square has an area of (2.r.2.r). We can also see a circle which is produced as a result of the plane passing through the enveloped common sphere. This circle has an area of (Pi.r.r).

If the plane is moved away from the cylinders axes, the intersection surface will reduce until the plane exits the cylinders completely. During this transition an ever changing common surface area is produced, but it will always consist of a square with a circle inside it which is tangential to that square at the middle of each side of the square. The ratio of the square to the circle is (4.r.r)/(pi.r.r)= 4/pi.

We know the volume of the common sphere is (4.Pi.r.r.r/3),So by using a similar shape approach, we can deduce that the common volume of the intersecting cylinders will be (4.Pi.r.r.r/3).(4/Pi)= 16.r.r.r/3 There is no Pi in the answer. hope that's useful, and no Calculus either. Lol.

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The volume of a Steinmetz solid $(SS)$, formed by the intersection of $2$ cylinders (radii $r$) at right angles is:

$$V = 4r \cdot r \cdot r \left(\frac{\pi}{2} - \frac{1}{3}\right) = 4r^3\left(\frac{\pi}{2} - \frac{1}{3}\right)$$

As the volume of $SS$ is formed by sections of a cylinder, the answer must contain $\pi$.

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    $\begingroup$ With no derivation and no references, it's hard to guess what object this formula describes. The last sentence of the answer is simply false. $\endgroup$ – David K Jul 2 '15 at 12:51
  • $\begingroup$ "The last sentence of the answer is simply false." Do you have any references to support that statement? No way can an expression, which gives the volume of part of a cylinder could be without pi. The Steinmetz solid is made of 4 sections of a cylinder and thus the expression for its volume must contain pi. There are countless websites on the Steinmetz solid showing the volume of 16r^3/3. Some of the sources dated back 50 years. It was wrong then and it is still wrong. My viewing of these sites (and some of them have good credentials), spurred me to make my entry on this site. $\endgroup$ – Bill Crean Jul 3 '15 at 14:14
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    $\begingroup$ As you yourself say, there are sources with good credentials that give $16r^3/3$ as the volume. I think the onus is on you to give a rigorous proof showing why they are wrong. "You can't have part of a cylinder without pi" is nonsense--start with an infinite cylinder and just cut off pieces at will, and you can make the volume anything you like. Show us exactly why this particular cut leaves any $\pi$ in the simplified expression of the volume. $\endgroup$ – David K Jul 6 '15 at 3:16
  • $\begingroup$ Are there references for -- or a derivation of -- your formula? $\endgroup$ – DBS Mar 26 '17 at 1:18

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