Recipe

Choose cylindrical vs. spherical coordinates and set up a triple integral

Use when: you're asked for a triple integral over a solid described by curved surfaces, such as cylinders, cones, paraboloids, or spheres, and rectangular coordinates would need messy square-root bounds

Goal: the right coordinate system for the whole solid, and a triple integral written in it with no leftover square roots

Taught in Choosing a coordinate system

1. 1

   Look at how x, y, and z actually combine in each boundary equation, not just what the solid looks like.

   • every bound is already a constant or a plane — rectangular coordinates need no conversion at all
   • x^2+y^2 appears as a single block, as in a cylinder, cone, or paraboloid centered on the z-axis — that's the cylindrical signature
   • x^2+y^2+z^2 appears as a single block, as in a sphere centered at the origin, or a cone measured from the origin — that's the spherical signature

   Why: the combination the equation actually uses, not the solid's general roundness, is what tells you which system collapses that surface to a single variable instead of two or three.

2. 2

   Check that the SAME system simplifies every bounding surface, not just the first one you looked at.

   Example

   Find the volume of the solid inside the cylinder x^2+y^2=4, above z=0, and below the paraboloid z=x^2+y^2. Both bounds involve x^2+y^2 as a single block and nothing involves x^2+y^2+z^2: cylindrical coordinates cover the whole solid.

   Why: mixing coordinate systems piece by piece is never necessary once a single system's native surfaces already cover the whole boundary, and switching partway through only adds bookkeeping.

3. 3

   Rewrite each surface's equation using x=r\cos\theta,\ y=r\sin\theta (cylindrical) or x=\rho\sin\phi\cos\theta,\ y=\rho\sin\phi\sin\theta,\ z=\rho\cos\phi (spherical).

   Example

   x^2+y^2=4 becomes r=2. z=x^2+y^2 becomes z=r^2. z=0 stays z=0.

   Why: this is the whole payoff of Step 1's recognition: a surface that needed a square root in rectangular coordinates becomes a single native bound in the matching system.

4. 4

   Work out the range of the innermost variable in terms of the outer ones first, then the middle variable's range, then the outermost variable's constant range.

   Example

   For fixed r,\theta: z runs from 0 up to r^2. The disk x^2+y^2\le4 gives 0\le r\le2, and sweeping all the way around gives 0\le\theta\le2\pi.

   Why: a triple integral's bounds have to be found in this dependency order, because the inner variable's range is usually a function of the outer ones. A double integral's inner bound depends on its outer variable the same way.

5. 5

   Attach the system's own volume element: dV=r\,dz\,dr\,d\theta for cylindrical, or dV=\rho^2\sin\phi\,d\rho\,d\phi\,d\theta for spherical.

   Example

   V=\int_0^{2\pi}\!\int_0^2\!\int_0^{r^2} r\,dz\,dr\,d\theta = \int_0^{2\pi}\!\int_0^2 r^3\,dr\,d\theta = \int_0^{2\pi} 4\,d\theta = \boxed{8\pi}.

   Why: dropping the extra factor (r, or \rho^2\sin\phi) silently produces a wrong volume, the same failure mode as forgetting it in a polar double integral, just one dimension up.