Volume of a sphere using cartesian coordinates I'm preparing my calculus exam and I'm in doubt about how to generally compute triple integrals.
I know that the cartesian equation of a sphere is
$B_R=\{(x, y, z)|x^2+y^2+z^2=R^2\}$,
so (if I didn't want to use spherical coordinates, wich I'm aware is the best way and I already did that)
it's volume would just be $\iiint_S \mathrm{d}x\mathrm{d}y\mathrm{d}z$,
but what would the extremes be?
I know $-R \leq z \leq R$ and $-\sqrt{R^2-y^2-z^2} \leq x \leq\sqrt{R^2-y^2-z^2} $,
but what are the extremes for $y$?
I can't describe it in terms of $x$, so I have
$$\text{Vol}(B_R) = \int_{-R}^{R} \int_?^?\int_{-\sqrt{R^2-y^2-z^2}}^\sqrt{R^2-y^2-z^2} \mathrm{d}x\mathrm{d}y\mathrm{d}z.$$
What should be there instead of the '?'
 A: It should be
\begin{equation}
\text{vol}(B_R) = \int_{-R}^{R} \int_{-\sqrt{R^2-z^2}}^{\sqrt{R^2-z^2}} \int_{-\sqrt{R^2-z^2-y^2}}^{\sqrt{R^2-z^2-y^2}} \, dx \, dy \, dz
\end{equation}
The way to think about this is to successively "fix" each variable as follows: pick a variable, for example $z$. Clearly, its bounds are $-R \leq z \leq R$. Then, for a fixed $z$, we have
\begin{equation}
0 \leq x^2 + y^2 \leq R^2 - z^2.
\end{equation}
Now, pick your next variable, say $y$. From the above inequality, it follows that $y$ must satisfy $- \sqrt{R^2 - z^2} \leq y \leq \sqrt{R^2 - z^2}$ (if it was outside this interval, the inequality above would be violated). Now, keep $y$ fixed. Then, we get
\begin{equation}
0 \leq x^2 \leq R^2 - z^2 - y^2
\end{equation}
which implies $-\sqrt{R^2 - z^2 - y^2} \leq x \leq \sqrt{R^2 - z^2 - y^2}$.

So, what you're doing is you look at the constraint all $3$ variables satisfy. Then, you choose one of them (for example $z$) and ask yourself what are the bounds it can take (in this case it was obvious). Then, you think of $z$ as fixed, and ask yourself what constraint do the remaining two variables satisfy (in this case $x,y$). Then you pick one of them (for example $y$) and ask what values can it take. Now, think of $y$ as fixed ax well, so that you end up with a constraint on $x$. This is a step-by-step reduction process which should be helpful; to really solidify your understanding you should see what each step is doing geometrically as well.
A: After doing away with the integral over $x$ it remains to be integrated over $y$ and $z$, as you already know. The domain of the remaining double integral is a circle on the $YZ$ plane. So, its limits are $-\sqrt{R^2 - z^2}$ (lower) and $\sqrt{R^2 - z^2}$ (upper).
If you want to see it geometrically, think of your 2-dimensional sphere of radius $R$ in the 3-dimensional Cartesian space. Project everything onto the $YZ$ plane to get a flat 2-dimensional circle of radius $R$. You can see that circle as the union of infinitely many half-circumferences of radius $R$, for $R$ ranging from $0$ to $R$.
