Triple Integral Evaluation with Spherical Coordinates

[twoflower]

Hi,

I don't know whether this is the best place to ask, anyway, I would like to check my results in Maple and don't know how to evaluate something like this there:

$$\iiint_{M} z^2\ dx\ dy\ dz\mbox{ , where M = [x,y,z] \in \mathbb{R}^3, x^2+y^2+z^2 \leq R^2, x^2 + y^2 + z^2 \leq 2Rz}$$

There is function Tripleint in Maple but I don't know how to properly pass the bounds..

Thank you

 

[benorin]

What version of Maple R U using?

 

[twoflower]

What version of Maple R U using?

It's version 9.5

 

[benorin]

Me too. Have you tried using the MultiInt function call with spherical coordinates? (trying to figureit out also)

 

[twoflower]

Me too. Have you tried using the MultiInt function call with spherical coordinates? (trying to figureit out also)

I didn't know about this function, thank you.

After converting to spherical coords, I got

$$\iiint r^4\sin^2 \phi \cos \phi\ dr\ d\phi\ d\theta$$

Anyway, I'm still not sure about the bounds..

 

[benorin]

the bounds are two spheres, the first is centered at the origin with radius R, the second is $x^2+y^2+(z-R)^2\leq R^2$, which is a sphere of radius R centered at (0,0,R). These spheres overlap and have a curve of intersection which is a circle contained in some plane z=k, for some constant k. But that is not the point; the region is bounded below by the shifted sphere and bounded above by the other sphere.

 

[benorin]

I get

$$\int_{\theta =0}^{2\pi}\int_{\phi =0}^{\pi} \int_{r =2R\cos\phi}^{R} r^4\sin^2 \phi \cos \phi\ dr\ d\phi\ d\theta=\frac{4\pi R^5}{15}$$

 

[twoflower]

I get
$$\int_{\theta =0}^{2\pi}\int_{\phi =0}^{\pi} \int_{r =2R\cos\phi}^{R} r^4\sin^2 \phi \cos \phi\ dr\ d\phi\ d\theta=\frac{4\pi R^5}{15}$$

Thank you benorin. First, I didn't know how to write the bounds in spherical coordinates so I tried to compute in cartesian coordinates. But I got a different result. Here's how I approached:

When I draw it, it's clear that for $z \in [0,\frac{R}{2}]$ I have

$$A:= {[x,y] \in \mathbb{R}^2, x^2 + y^2 \leq 2Rz - z^2}$$

which I can imagine as area of circle with radius $\sqrt{2Rz - z^2}$.

For $z \in [\frac{R}{2}, R]$ I have

$$B:= {[x,y] \in \mathbb{R}^2, x^2 + y^2 \leq R^2 - z^2}$$

ie. area of circle with radius $\sqrt{R^2 - z^2}$.

So I can write the original integral as sum of these two:

$$I = \int_0^{\frac{R}{2}}\left(\iint_{A} 1\ dx\ dy \right)\ dz + \int_{\frac{R}{2}}^{R}\left(\iint_{B} 1\ dx\ dy \right)\ dz$$

$$= \int_{0}^{\frac{R}{2}} z^2\pi(2Rz - z^2)\ dz\ +\ \int_{\frac{R}{2}}^{R} z^2\pi(R^2 - z^2)\ dz = ... = \frac{59}{480} \pi R^5$$

I think it's ok, isn't it?

 

[benorin]

Better to use cylindrical coordinates, I think:

$$\int_{\theta =0}^{2\pi}\int_{r=0}^{\frac{\sqrt{3}}{2}R} \int_{z =R-\sqrt{R^2-r^2}}^{\sqrt{R^2-r^2}} z^2r dz dr d\theta=\frac{59}{480}\pi R^5$$

Yep: same as you.

 

[twoflower]

Better to use cylindrical coordinates, I think:
$$\int_{\theta =0}^{2\pi}\int_{r=0}^{\frac{\sqrt{3}}{2}R} \int_{z =R-\sqrt{R^2-r^2}}^{\sqrt{R^2-r^2}} z^2r dz dr d\theta=\frac{59}{480}\pi R^5$$
Yep: same as you.

Nice :) Anyway, could you tell me please how did you get the bounds when you were computing it with the spherical coordinates? I mean the inner integral with bounds $[2R\cos \phi, R]$...

 

[benorin]

I did it wrong. But substitute [itex]x^2+y^2+z^2=r^2\mbox{ and }z=r\cos \phi[/tex] into the equations of the spheres.