# Improper triple integral bounds of integration in cylindrical coordinates

Regrettably I don't understand the process of change of coordinates and change of bounds very well.

$$x = r \, \cos \, \theta \\ y = r \, \sin \, \theta \\ z=z$$

where

$$\displaystyle {r \ge 0,\;\;\;}\kern-0.3pt {0 \le \theta \le 2\pi ,\;\;\;}\kern-0.3pt {- \infty \lt z \lt \infty .}$$

In case we have a triple integral of the form

$$\displaystyle \int _0^{\infty }\int _0^{\infty }\int _0^{\infty }f(x,y,z)dxdydz$$

the corresponding integral in cylindrical coordinates will be

$$\displaystyle \int_c^d \int_{\beta}^{\alpha} \int_a^b f(r, \theta, z) r \, dr \, d\theta \, dz.$$

For the bounds of integration of $$(x,y,z)$$ going from zero to infinity, what would be the values of $$a,b,c,d,\alpha,\beta$$ ?

The variable $$z$$ has not changed, so that one is easy: $$c=0, d = \infty$$.
For the rest, use this approach: sketch the region that you are integrating over. It's as simple as representing that region in terms of your new coordinates. In this case, the region is just the octant of the $$3$$-dimensional plane with positive $$x$$, $$y$$, and $$z$$ coordinates. Since we already took care of $$z$$, let's look at where $$x$$ and $$y$$ can lie. In the two-dimensional $$xy$$ plane, they must lie in the first quadrant. How can we represent this in terms of a radius ($$r$$), and a central angle ($$\theta$$)?
Since we can have a point in the first quadrant arbitrarily far from the origin, we know $$r$$ goes from $$0$$ to $$\infty$$. But what about $$\theta$$? Well, the first quadrant only has $$0 \leq \theta \leq \pi/2$$, so we have our answer.