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

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:

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

Thank you!

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Note that the integrand doesn't depend on $x$. –  littleO Nov 16 '12 at 10:40
Also, note that $\sqrt{R^2} = |R|$. –  beauby Nov 16 '12 at 10:43
@beauby no need to assume that, $R$ - is a sphere radius so it's always positive. –  Kaster Nov 16 '12 at 23:43

\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*}

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Thank you! This was exactly the calculation I was looking for. –  Lukas Arvidsson Nov 16 '12 at 11:34
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.