# Homework Help: Volume of a solid with 3 boundary conditions

1. Mar 3, 2014

### Yosty22

1. The problem statement, all variables and given/known data

Find the volume of an object bounded by x2 + y2 ≤ 1, x2 + z2 ≤ 1 and y2 + z2 ≤ 1.

2. Relevant equations

3. The attempt at a solution

This stuff is very new to me (multiple integrals to find volume) so I am not entirely familiar with it. My first thought was to put each boundary condition in terms of one variable. For example, solve the first 2 boundary conditions for y and z in terms of x, but then if you substitute them in to the third boundary condition, it doesn't really get you anywhere.

Any help would be much appreciated.

Thank you.

2. Mar 3, 2014

### sa1988

Hmmm, quite hard to explain.

I tend to visualise or draw the equations out, so I can get a better feel for what I'm looking for.

Typically I would start by drawing an x-y plane and working out the 'base' of the volume in question (ok it might not be the base, but it's the part that crosses the 0 axes, which is often helpful to go from). So you can ignore all the z values/equations for the time being because you only want to know how it acts in the x-y plane. In the case of your example, you can hopefully see that the outer limits in the x-y plane form a circle.

Now, looking at the boundaries in the x-z and y-z planes, you can see that everything is defined in the same way, as a circle. Which means it must be a sphere. (You can draw all this out too to see for yourself).

Do you know how to work in spherical co-ordinates? If so, I'm fairly sure the question expects you to do this. Otherwise the hard part is actually completing the integral because anything with a circle is going to integrate into arcsines and require values of pi, which is a nightmare that is nicely evaded in a spherical system.

3. Mar 3, 2014

### Yosty22

We have done only a little bit with spherical coordinates. Sadly, my professor for the class never explicitly taught spherical coordinates, he just gave us a quiz to prove that in spherical coordinates, you have to change dx dy dz into r2sin(\theta) dr d\theta d\phi.

Other than that we haven't really done much with it besides talking about the transformations:
x = rsin(\theta)cos(\phi)
y = rsin(\theta)sin(\phi)
z = rcos(\theta)

Sorry if this isn't much background in spherical coordinates, but that is one of the reasons I came here. I knew it would be ugly in Cartesian and I know that the bounds all create circles, but due to my ignorance of spherical coordinates, myself nor any of my classmates actually know how to proceed from just simply visualizing and drawing out the bounds in Cartesian coordinate planes to try to figure anything else out.

4. Mar 3, 2014

### sa1988

Ah right ok. Well the stuff you've covered is enough, woohoo!

Now it's super easy.

Since it's a sphere, you can go on the basis of symmetry, just work out the volume of one octant then multiply it by 8.

In a single octant, you can see clearly the limits of integration for each variable.

Radius goes from 0 to 1.
Phi goes from 0 to pi/2
Theta goes from 0 to pi/2

So there we have it.

Volume = 8 0π/20π/201 [r2sin(θ)] drdθdø

The reason it's so much simpler is because the limits are now not bound by horrible looking equations of square roots, but are instead just bound as numerical limits on the axes and angles as defined by the octant you're looking at.

It's exactly the same as a Cartesian integration, where you sum over all points from 'a to b'. It just happens to be that now 'all points' is instead 'all angles and all radii'.

The r2sin(θ) thing is what maps, or 'transforms', the cartesian value into its spherical equivalent. Quite hard to explain it properly as I don't fully understand it myself, however this may help if you want to know more. Also look up 'conformal mapping'.

5. Mar 3, 2014

### Yosty22

Ahh okay! Got it! That makes a lot more sense, I really appreciate it, and thanks for the links!

6. Mar 3, 2014

### sa1988

Great stuff! Glad (and relieved) that I explained it well for you!

7. Mar 3, 2014

### LCKurtz

WHOA THERE everyone. That solid is NOT a sphere. It is the intersection of three cylinders.

8. Mar 3, 2014

### Yosty22

I was just coming back because I seemed to encounter some issues. I am not sure how to fix it at this point. My classmates have told me that the answer should be 8(2-sqrt(2)), which just so happens to be pretty much exactly 1/2 of what I got following the previous steps. How would I account for them being cylinders in order to fix it?

9. Mar 4, 2014

### sa1988

Which surely forms a sphere in this case?

The integration I pointed you towards should give the volume of a sphere of radius 1. Which is sort of close to that answer of 8(2-root2), but is also quite different.

What method did your classmates use? Maybe I'm the one who's wrong. All this talk of cylinders is scaring me... (although a cylinder volume would also include values of Pi, so... hmmmmm)

10. Mar 4, 2014

### D H

Staff Emeritus
Surely not!

Any point inside the unit sphere is inside this intersection of three cylinders, but the converse is not true. This intersection of three cylinders circumscribes the unit sphere and is in turn circumscribed by a sphere with radius ½√3.

11. Mar 4, 2014

### sa1988

Oh dear oh dear oh dear oh dear oh dear!

Not gonna lie - I can't visualise this at all now. To me it seemed like a blatant sphere. I've just done a bit of a model on Mathematica and yeah it looks like the intersection of three cylinders really doesn't form a sphere.

OP... I'm really sorry for misleading you. At least I taught you something about spherical co-ordinates...? Heh.

Please do let us know when you find or are given the solution. I'm very curious now!

12. Mar 4, 2014

### Yosty22

A classmate of mine showed me the integral that they set up to solve the problem to get the result of 8(2-sqrt(2)) but I cannot quite figure out why.

They set up a triple integral as follows:
16 0∫π/40∫10∫(sqrt(1-r^2cos^(θ))) [r] dzdrdθ

Any ideas where this comes from?

13. Mar 4, 2014

### Yosty22

edit: the integral should be read as follows:

the bounds on the first integral go from 0 to pi quarters. The bounds on the second integral go from 0 to 1. The bounds on the third and final integral go from 0 to sqrt(1-r2cos2(theta))

14. Mar 4, 2014

### sa1988

Hmm they've used cylindrical co-ordinates, which is basically polar co-ordinates for the 'base' and cartesian for the z-axis, usually given as (ρ,θ,z), where ρ is the radius and (ρ,θ) are confined to what would be the x-y plane. For example (1,π/4,1) has cartesian co-ordinates of (1/√2, 1/√2, 1)

It's multiplied by 16 because it's easier to break it into 16 sections, work out the volume of just one of the sections, then multiply to get the full volume again. If you look at this image, you can see how the cylinders cross each other. Draw a circle onto the centre of that shape, to represent the other central cylinder, and you'll see it forms 8 similar segments. So if you keep in mind that the shape has a bottom half to it as well, that makes a total of 16 segments.

Looks like the first integral, from 0 to root(1-r2cos2theta, defines the height of the thing from 0 to its maximum point, for all points. I don't know how they came up with this, but it's presumably taken from something very similar to the spherical co-ordinate properties you gave me before, where you said:

x = rsin(\theta)cos(\phi)
y = rsin(\theta)sin(\phi)
z = rcos(\theta)

You can find out more here - http://en.wikipedia.org/wiki/Cylindrical_coordinate_system

So there'll be something that defines the maximum height of the volume for any point. In the case of this question, it happens to be that the volume can never exceed a height of root(1-r2cos2theta), which I guess makes sense if you try to picture it.

And then the other integral values are the nice polar versions, which just go from 0 to r, and 0 to Pi/4, where the Pi/4 represents the fact that you're dealing with just that 1/8th of the top half of the volume.

That's about it really. The integral certainly does give a value of 8(2-root2), which is damn close to that of a unit circle, which makes sense because of the limits set out in the question, and so I'm now willing to concede that that surely is the right answer and everything I said before was totally wrong, hah.

15. Mar 4, 2014

### D H

Staff Emeritus
That's $16\int_0^{\pi/4}\int_0^1\int_0^{\sqrt{1-r^2\cos^2\theta}} r\,dz\,dr\,d\theta$.

So where does that come from? The answer is symmetry (the factor of 16) and use of cylindrical coordinates (integrating in terms z, r, and θ). The second part is easy. When you are using cylindrical coordinates to find the volume of some object, the integral is $\iiint_V r\,dz\,dr\,d\theta$, where the integral is over the volume V. That the Cartesian $dx\,dy\,dz$ becomes $r\,dz\,dr\,d\theta$ under transformation to cylindrical coordinates is derivable, but this is one of those key things you should just know.

The factor of 16 and the integration limits -- that's the tricky part. The basic idea is to use symmetry to divide that complex shape into something that is more readily integrable. A common approach is to see if there's some way to divide the volume in half. In this case, it should be obvious that the portion of the volume about the z=0 plane is exactly equal to that below z=0. Restraining z helps simplify things. In particular, if z is positive, √z2=z. Similarly, it often simplifies things if one can use symmetry arguments to keep x≥0 and y≥0. Combining these three, if you can keep each of x, y, and z positive (i.e., the first octant), you don't have to worry about nasty problems such as √z2=-z (and the same goes for x and y).

So does that kind of symmetry exist with this problem? It certainly does. In fact, there's yet one more symmetry of which one can take advantage. xy when the cylindrical coordinate θ is between 0 and pi/4. That means that the cylinder that constrains the z coordinate is x2+z2≤1, or z2≤1-x2. Since both x and z are positive in this region, this is equivalent to $z \le \sqrt{1-x^2}$ -- and that's where that final integration limit comes from.