# Triple Integral Problem

1. Dec 26, 2005

### blumfeld0

Hello. This is my first of many posts at this forum. For fun recently I came across a triple integral that i would really like to know how to do.
basically i have two equations
x^2+y^2+z^2=1
x^2+y^2+(z-1)^2=2
if you plot these you will see that they intercept.
i need to find the area under the graph where they intercept.
I would like to write this as a SINGLE integral using cartesian, cylindrical and sphereical coordinates.
anyone know how to write it any or all of these three ways. thank you very much
khurram

2. Dec 26, 2005

### maverick6664

Is integral necessary? (or am I making a mistake?)
subtract these equations side by side, and you'll get
z=0.
That's where these spheres intercept.
So put z=0 into the first equation (or the second one), you'll have:
$$x^2 + y^2 = 1 (z = 0)$$
and this is the graph of interception and its area is evidently $$\pi.$$

I hope it helps

If you want to calculate the "volume" of the common volume region instead of area, it's quite different. (but I guess this is not what you want.)

The volume is calculated as

$$\int_0^1 \pi (\sqrt{(1-z^2)})^2 dz + \int _{-\sqrt{2}}^0 \pi (\sqrt{(2-z^2)})^2 dz = \pi \frac {2+4 \sqrt{2}}{3}$$

It's not single integral or cylindrical or spherical, though...

I think simple is best.

Last edited: Dec 26, 2005
3. Dec 26, 2005

### blumfeld0

Triple Integral

Hello. You are absolutely right sir. I meant to ask for the volume not the area.

How do i write that sum of those two integrals as one integral.
Also How would I go about converting that into cylidrican and spherical coordinates. I mean I know the volume elements for spherical and cylindrical
spherical
dV==r^2 sinphi dphi dtheta dr.
cylindrical
dV==r dr dtheta dz
thank you for all your help

blumfeld0

4. Dec 26, 2005

### maverick6664

In Cartesian coordinates, the latter integral

$$\int _{-\sqrt{2}}^0 \pi (\sqrt{(2-z^2)})^2 dz = \int _ 0 ^ {\sqrt{2}} \pi (2-z^2)dz$$

can be put using

$$\sqrt{2} z' = z$$

into

$$\int _0^1 \pi (2- (\sqrt{2} z')^2) \frac {dz}{dz'} dz' = \int _0 ^ 1 \pi 2 \sqrt{2}(1-z^2) dz$$

So the total volume can be written as a single integral

$$\int _0^1 \pi ((1-z^2) + 2 \sqrt{2} (1-z^2)) dz = \int _0^1 \pi (1+ 2 \sqrt{2}) (1-z^2) dz$$

Does it make sense? It looks like just a trick of calculation to me.

As for polar and cylindrical coordinates, I wrote spheres as:

polar:
$$r^2-2r\cos(\theta)=1$$ and
$$r^2=1$$

cylindrical:
$$r^2+z^2-2z=1$$ and
$$r^2+z^2 = 1$$.

But I haven't make integral fomulae yet.

Last edited: Dec 26, 2005
5. Dec 26, 2005

### blumfeld0

Yes I do understand. Thanks for your help
blumfeld0

Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook