# Thread: Home work check please - Triple integrals and centre of gravity

1. ## Home work check please - Triple integrals and centre of gravity

I need to find the volume of region bounded above by sphere $x^2 +y^2 +z^2=4$ and below by the inverted cone $z=-\sqrt{x^2 +y^2} , -\sqrt{2}\leqslant z \leqslant 0$ and also find the centre of gravity with density equalling constant.

I set up my bound and triple integral as

$\int_{0}^{(3\pi /4)}\int_{0}^{(2\pi)}\int _{0}^{\2} r^2\sin \varphi drd\theta d\varphi$

and after solving i got approx 28.60

using symetry, the centre of gravity is on the z axis and using the formula for centre of mass:

$1/m\iiint z dV$

and subsituting above result and

$z = r\cos\varphi$

i obtained
$\iiint r\cos \varphi r^2\sin \varphi drd\theta d\varphi$

solving with bounds stated in first part i got
$\bar{z} = 0.22$

does this look correct? thanks for any help.

2. Originally Posted by olski1
I need to find the volume of region bounded above by sphere $x^2 +y^2 +z^2=4$ and below by the inverted cone $z=-\sqrt{x^2 +y^2} , -\sqrt{2}\leqslant z \leqslant 0$ and also find the centre of gravity with density equalling constant.

I set up my bound and triple integral as

$\int_{0}^{(3\pi /4)}\int_{0}^{(2\pi)}\int _{0}^{\2} r^2\sin \varphi drd\theta d\varphi$

and after solving i got approx 28.60

using symetry, the centre of gravity is on the z axis and using the formula for centre of mass:

$1/m\iiint z dV$

and subsituting above result and

$z = r\cos\varphi$

i obtained
$\iiint r\cos \varphi r^2\sin \varphi drd\theta d\varphi$

solving with bounds stated in first part i got
$\bar{z} = 0.22$

does this look correct? thanks for any help.
Without going through your working we can still check the numerical answers: Crude Monte-Carlo gives results close to yours, so yes they look OK

(28.577 +/- 0.085 and 0.2186 +/- 0.0023 both 2 sigma error estimates)

CB

Worried now that people in my course will just copy my working! any way i can hide it?

4. Originally Posted by olski1