I am trying to evaluate the following triple integral below, $$\int_{0}^{\frac{\pi}{6}}\int_{0}^{\frac{\pi}{2}}\int_{0}^{3}\rho^2\sin\phi \, \mathrm{d}\rho \, \mathrm{d}\theta \, \mathrm{d}\phi,$$
which is written in spherical coordinates. But as I am solving the integral, I noticed that $$\rho^2\sin\phi \, \mathrm{d}\rho \, \mathrm{d}\theta \, \mathrm{d}\phi=\mathrm{d}V = \mathrm{d}z \, \mathrm{d}y \, \mathrm{d}x.$$ So I thought that maybe this triple integral problem can be solved much simply when converted into rectangular coordinates, which would look similar to the mathematical form below. $$\int_{x_1}^{x_2}\int_{y_1}^{y_2}\int_{z_1}^{z_2}1 \, \mathrm{d}z \, \mathrm{d}y \, \mathrm{d}x.$$ The thing is I have trouble finding the boundaries for each $z$-axis, $y$-axis, and $x$-axis. What should be the appropriate values for $(x_1,x_2), (y_1,y_2), (z_1,z_2)$?
P.S: I am not so comfortable with LaTex yet. I would appreciate any edits to my question. Thank you.
Spherical is the easier calculation. Even if you convert to cartesian, you are going to need a trig substitution which effectively puts you back into polar coord's if not spherical...
Anyway, learning how to transform coordinates is important. I like to try to visualize the region before I start working out.
The region is one-quarter of a cone inside a sphere.
It is one quater of a cone because $0\le \theta\le \frac {\pi}{2}$
The equation for the sphere $x^2+y^2+z^2 = 3^2$
The slope of the cone associates with the angle $\frac {\pi}{6}$
$(z\tan \frac {\pi}{6})^2= x^2+y^2$ or $z^2 = 3x^2 + 3y^2$
The intersection of the sphere and the cone is $x^2+y^2 = \frac {9}{4}, z=\frac {3\sqrt 3}{2}$
Okay, that is enough information to make our change of coordinates.
$\displaystyle \int_0^{\frac 32}\int_0^{\sqrt{\frac 94 - x^2}}\int_{\sqrt {3x^2 + 3y^2}}^{\sqrt{9-x^2-y^2}}\ dz\ dy\ dx$