Solid of revolution -- General question

[0kelvin]

There are two ways to revolve, around Y or X and the formulas are different.

If I have something bounded by $$f(x) = x^2 + 1$$. I can write $$x = \sqrt{y - 1}$$. But, is it wrong to swap axis to show that I'm integrating dy, not dx?

 

[mfb]

The boundaries matter, and different integration directions lead to different formulas (but the same result).

Consider a simpler two-dimensional example. What is the area between your f(x), the x and y-axis and x=1? Well, it is $\int_0^1 f(x) dx = \frac{3}{2}$. What if we want to integrate over y? f(x) changes from 1 to 2 but our area starts at y=0, so we set up $\int_0^2 ? dy$. The function we need to integrate over is the length between (f(x) OR x=0) and x=1 in the horizontal axis, the length is 1 for y<1 and $1-x(y) = 1-\sqrt{y-1}$ for y between 1 and 2. To avoid using a case structure let's split the integral in two parts: $\int_0^1 1 dy + \int_1^2 (1-\sqrt{y-1}) dy$. If you calculate that you should get the same result as before, $1+\frac 1 2 = \frac 3 2$ - but it was way more messy.

The same applies to the solids of revolution. You can integrate over the volume in any order you want, but some are more convenient.