How can I write a triple integral using different coordinate systems?

[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

 

[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.

 

[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

 

[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&#039;)^2) \frac {dz}{dz&#039;} dz&#039; <br /> = \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 <br /> = \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.

 

[blumfeld0]

Yes I do understand. Thanks for your help
blumfeld0