# More Triple Integrals!!!!

• Aug 6th 2008, 02:59 PM
Dubulus
More Triple Integrals!!!!
i have an idea on how to solve this problem, but it seems like itd be way too much work to even be possible...

triple integral over R of dV / (x^2 + y^2 + z^2)
where R is the region between:
x^2 + y^2 + z^2 = 1
x^2 + y^2 + z^2 = 4

am i correct in thinking that first i have put convert to spherical coordinates and then insert pdpdthetadphi?

ps how do you insert the integral symbol and all that fun stuff?
• Aug 6th 2008, 03:10 PM
Chris L T521
Quote:

Originally Posted by Dubulus
i have an idea on how to solve this problem, but it seems like itd be way too much work to even be possible...

triple integral over R of dV / (x^2 + y^2 + z^2)
where R is the region between:
x^2 + y^2 + z^2 = 1
x^2 + y^2 + z^2 = 4

am i correct in thinking that first i have put convert to spherical coordinates and then insert pdpdthetadphi?

ps how do you insert the integral symbol and all that fun stuff?

I'll answer your last question first. We use LaTeX here at the forums.

As I answer you're problem, I'll show what code was used.

So we are to integrate $\iiint\limits_G x^2+y^2+z^2 \,dV$?

Code:

$$\iiint\limits_G x^2+y^2+z^2 \,dV$$
Note that the region is defined as the area between two circles [when looking down at the xy plane]

So our limits for the radius would then be from 1 to 2.

The best coordinate system to use would be the spherical coordinate system. We see that $\varrho$ is defined between 1 and 2.

$\varphi$ would be defined between 0 and $\pi$

$\vartheta$ would be defined between 0 and $2\pi$

So our triple integral would be $\int_0^{2\pi}\int_0^{\pi}\int_1^2 \varrho^4\sin\varphi \,d\varrho\,d\varphi\,d\vartheta$

Code:

$$\int_0^{2\pi}\int_0^{\pi}\int_1^2 \varrho^4\sin\varphi \,d\varrho\,d\varphi\,d\vartheta$$
Does this make sense?

--Chris
• Aug 6th 2008, 03:16 PM
Dubulus
most of it makes sense yes but where does the
$\varrho^4\sin\varphi\$ come from? is that from pluggin in the spherical coordinates into f(x,y,z)?
• Aug 6th 2008, 03:33 PM
wingless
Quote:

Originally Posted by Chris L T521
So we are to integrate $\iiint\limits_G x^2+y^2+z^2 \,dV$?

..
Quote:

Originally Posted by Dubulus
triple integral over R of dV / (x^2 + y^2 + z^2)

• Aug 6th 2008, 03:37 PM
Dubulus
im afraid i dont know where to begin with taht sum. the point of the problem is using spherical coordinates.
• Aug 7th 2008, 12:03 AM
Chris L T521
Quote:

Originally Posted by wingless
..

Shoot...

$\iiint\limits_G\frac{\,dv}{x^2+y^2+z^2}$??

Thus, $\iiint\limits_G\frac{\,dv}{x^2+y^2+z^2}\implies\ii int\limits_{G_{\varrho\varphi\vartheta}}\frac{1}{\ varrho^2}\varrho^2\sin\varphi\,dV_{\varrho\varphi\ vartheta}\implies\iiint\limits_{G_{\varrho\varphi\ vartheta}}\sin\varphi\,dV_{\varrho\varphi\vartheta }$

To clear things up a little bit, this is the integral setup when we go from rectangular to spherical:

$\int_k^l\int_c^d\int_a^b f(x,y,z)\,dx\,dy\,dz\implies\int_{\vartheta_0}^{\v artheta}\int_{\varphi_0}^{\varphi}\int_{\varrho_0} ^{\varrho}f(\varrho,\varphi,\vartheta)\varrho^2\si n\varphi\,d\varrho\,d\varphi\,d\vartheta$

The $\varrho^2\sin\varphi$ term comes out of the conversion to spherical coordinates.

I hope this clarifies things for now...for I'm dead tired. I'll post more on this tomorrow...I mean today... (Sleepy)

--Chris