# Ask Uncle Colin: The RMS Line

Dear Uncle Colin,

My textbook claims that the RMS voltage of a sine wave is $\sqrt{\frac{1}{2}}$ because between $x=0$ and $x=\pi$, the signed area between the curves $y=\sin(x)$ and $y=\sqrt{2}$ 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 $0 \le x \le \piby 2$, and then we just need to look at the integral $\int_0^{\piby2} \left(\sin(x) - \sqrt{\frac{1}{2}}\right)\dx$.

That’s $\left[ -\cos(x) - x\sqrt{\frac{1}{2}} \right]_0^{\piby2}$, or $1-\piby2 \sqrt{\frac{1}{2}}$. 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 $\sqrt{\frac{1}{2}}$ with $k$ and solve for it.

The integral is $\left[ -\cos(x) - kx \right]_0^{\piby2}$, or $1 - k\piby 2 = 0$ – which means $k = \frac{2}{\pi}$. But that’s not the RMS.

The RMS of a function $f(x)$ over an interval $a \lt x \lt b$ is defined as $y_{RMS} = \sqrt {\frac{1}{b-a}\int_a^b [f(x)]^2 \dx}$.

I find it slightly nicer to square and cross-multiply to give $(b-a)y^2_{RMS} = \int_a^b [f(x)]^2 \dx$, which at least looks like you’ve got the same kind of thing on both sides.

In any case, we need to work out $\int_0^{\piby2} \sin^2(x) \dx$. Using the identity $\sin^2(x) \equiv \frac{1-\cos(2x)}{2}$, this integrates to $\left[ \frac{x}{2} - \frac{\sin(2x)}{4} \right]_0^{\piby2}$, which is $\piby 4$.

So we have $\left(\piby 2\right) y_{RMS}^2 = \left( \piby 4\right)^2$, which simplifies down to $y_{RMS} = \frac{1}{\sqrt{2}}$, 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