1. ## Triple integral

Calculate the integral
$I = \int {\int\limits_D { \int {xdxdydz} } }$

where D is the region in R3 given by the inequalities $x^2+y^2+z^2<=1$ and $-x<=y<=\sqrt{3}x$

How do you deal with that last inequality?

2. Originally Posted by kloda

Calculate the integral
$I = \int {\int\limits_D { \int {xdxdydz} } }$

where D is the region in R3 given by the inequalities $x^2+y^2+z^2<=1$ and $-x<=y<=\sqrt{3}x$

How do you deal with that last inequality?
The first part is a\the inside of a sphere of radius 1

the region in the xy plane is bounded by two lines see attached graph

I would change to cylindrical coordinates from here.
If you haven't covered them yet this is it in rectangular
$
\int_{0}^{1}\int_{-x}^{\sqrt{3}x}\int_{-\sqrt{1-x^2-y^2}}^{\sqrt{1-x^2-y^2}}xdzdydx
$

cylindrical coordinates give

$\int_{-\frac{\pi}{4}}^{\frac{\pi}{3}}\int_{0}^{1}\int_{-\sqrt{1-r^2}}^{\sqrt{1-r^2}}(r\cos(\theta))rdzdrd\theta$

I hope this helps.

Good luck.

3. Ah, thank you so much. It's so obvious now.