Volume and triple integral

Question

I've been studying some triple integrals and their relation to volumes in 3 dimensions and got a little confused by the following problem:

Let $$U=\{(x,y,z) \in \mathbb{R}^3; \; x^2+y^2+z^2\leq4 \; ; 0\leq z \leq5 \; ; x \leq y \leq2x\;; x\geq0 \}$$ find the integral that represents the volume of the set $U$

The solution is provided and is shown to be:

$$\int^{\frac{2}{\sqrt{5}}}_{0}\int^{2x}_{x}\int^{\sqrt{4-x^2-y^2}}_{0}\,dx\,dy\,dz+ \int^{\sqrt{2}}_{\frac{2}{\sqrt{5}}}\int^{\sqrt{4-x^2}}_{x}\int^{\sqrt{4-x^2-y^2}}_{0}\,dx\,dy\,dz$$

I just don't understand why he splits the integral into two parts and the limits of the $x$ and $y$ variables.

Answer 1

As @Arthur states: "It's rarely a bad idea to make a drawing." Does this help?

Answer 2

That the solution wrote $dxdydz$ instead of $dzdydx$is unfortunate, but hopefully you worked out which variable's integration range each $\int$ denoted. Let's flesh out @sudeep5221's explanation of the middle $\int$ in each piece. That's the hard one, the one that explains why we have to split into two integrals in the first place.

The main thing we need to understand is why $x$'s range is $[0,\,\sqrt{2}]$, and worth splitting at $x=\tfrac{2}{\sqrt{5}}$. Since $0\le x\le y$, $2x^2\le x^2+y^2\le4-z^2\le4$, so that answers the first part. But the case $y=2x$ is consistent with $x^2+y^2\le4$ iff $x^2\le\tfrac{4}{1+2^2}=\left(\tfrac{2}{\sqrt{5}}\right)^2$, so in the first integral $y$ is capped at $2x$, whereas in the second it's capped at $\sqrt{4-x^2}<2x$.

There's also a bit of a trick question here: clearly $z^2\le4$, so $z$'s true cap is $2$ (well, $\sqrt{4-x^2-y^2}$ given how the triple integrals are nested), not the more generous-sounding $5$.

Personally, I'd rather compute $U$'s volume in cylindrical coordinates as$$\int_0^2dz\int_0^2\rho d\rho\int_{\pi/4}^{\arctan2}d\varphi=2\cdot2\cdot(\arctan2-\pi/4)=4\arctan\tfrac13.$$