Triple integral $\iiint x^2 \,dx\,dy\,dz.$

Here's the question $$\iiint x^2 \,dx\,dy\,dz.$$ I am asked to evaluate this integral over the region $$D:=\left \{ (x,y,z) \in\mathbb{R}^3 :x^2 \leq y^2+z^2 \leq 4\right \}.$$

I've shown that :

• $$2 \leq x \leq -2$$

then:

$$\iiint x^2 \,dx\,dy\,dz= \int_{-2}^{2}dx \int\int x^2 dydz$$

From here,do i have to use the cylindrical coordinates?

Any support for this question would be appreciated.

• ok @Arthur i try to do it – andrew Jan 23 at 18:21
• The region is a (sort-of) hollowed-out cylinder centered around the $x$-axis. Maybe cylindrical coordinates makes this easier? (Sorry about saying spherical, I misread the inequalities.) And if you want to do it in Cartesian coordinates, I suspect that you want $x$ to be the innermost integral. – Arthur Jan 23 at 18:22
• yeah, you're right – andrew Jan 23 at 18:23
• i obtained that the solution is $\frac{64*\pi}{3}$ – andrew Jan 23 at 18:36

I use cylindrical coordinates $$y^2+z^2=r^2$$ and $$x=h$$. Then $$h$$ varies between $$-r$$ and $$r$$. Then my integral is $$\iiint_D rdr d\theta h^2dh=\int_0^{2\pi}d\theta\int_0^2dr\ r\int_{-r}^rh^2dh$$