how is $ 2\pi \int^R_{-R} \sqrt{R^2}dx = 4\pi R^2 $? (sphere area)

Question

I am trying to understand the proof of the sphere area formula.

In my math book they use the formula $y = \sqrt{R^2 - x^2}$ $-R \leq x \leq R$

They rotate the function above around the x-axis and get:

$ A = 2\pi \int^R_{-R} y \sqrt{1+ (\frac{dy}{dx})^2} dx = 2\pi \int^R_{-R} \sqrt{R^2 - x^2} \sqrt{1+ \frac{x2}{R^2-x^2}} dx = 2\pi \int^R_{-R} \sqrt{R^2}dx = 4\pi R^2 $

I understand the development until this part:

$= 2\pi \int^R_{-R} \sqrt{R^2}dx = 4\pi R^2 $

Can someone please help me with this one (and how the calculation is made)? Please also explain your approach when solving it.

Thank you!

Answer 1

\begin{align*} 2\pi \int_{-R}^R \sqrt{R^2} \, dx &= 2\pi \int_{-R}^R R \, dx \\ &=2\pi R\int_{-R}^R 1 \, dx \\ &= 2 \pi R \left( 2R \right) \\ &= 4 \pi R^2. \end{align*}

Answer 2

If $R\ge 0$, then $\sqrt{R^2}=R$, and, as the comments say, it is independent from $x$, in other words, $x\mapsto \sqrt{R^2}$ is a constant function, and if you draw its graph (over the closed interval $[-R,R]$), you will find that the integral in question is the area of a suitable rectangle which you can easily find.