Dear Uncle Colin,

My textbook claims that the RMS voltage of a sine wave is 12 because between x=0 and x=π, the signed area between the curves y=sin(x) and y=2 is 0 – but I checked and it isn’t. What’s going on?

Really A Disappointing Instructional Overview

Hi, RADIO, and thanks for your message!

You’re quite right – we can use the symmetry of the graph to restrict ourselves to 0xπ2, and then we just need to look at the integral 0π2(sin(x)12)dx.

That’s [cos(x)x12]0π2, or 1π212. That’s clearly not 0 (it’s about -0.11).

So, two questions: one, where should the equal-area line lie; and two, what should the RMS explanation have said?

We’ve done most of the work for the first one already: we just need to replace 12 with k and solve for it.

The integral is [cos(x)kx]0π2, or 1kπ2=0 – which means k=2π. But that’s not the RMS.


The RMS of a function f(x) over an interval a<x<b is defined as yRMS=1baab[f(x)]2dx.

I find it slightly nicer to square and cross-multiply to give (ba)yRMS2=ab[f(x)]2dx, which at least looks like you’ve got the same kind of thing on both sides.

In any case, we need to work out 0π2sin2(x)dx. Using the identity sin2(x)1cos(2x)2, this integrates to [x2sin(2x)4]0π2, which is π4.

So we have (π2)yRMS2=(π4)2, which simplifies down to yRMS=12, which is a relief.

It turns out that it’s not that the curves have the same area beneath them – it’s that the volumes of revolution they define (about the x-axis) have the same volume.

Hope that helps!

- Uncle Colin