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Areas versus volumes of revolution

For fun I decided to derive the surface area of a sphere of radius $1$ from the formula for the perimeter of a circle. This integral is what I came up with:

$2\pi\int_{-1}^1\sqrt{1-x^2}dx = \pi^2$

Unfortunately the desired value is $4\pi$. My rationale was simply to stack infinitely thin 'hula-hoops' whose radii followed the curvature of the sphere. I can't readily see where my conceptual misunderstandings are, can someone help elucidate them for me? Thanks.

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    Thanks for the comments and links, they've been a big help.2012-07-14

2 Answers 2

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You need to use the surface area for a surface of revolution formula $2\pi\int \rho\,ds,$ where $ds$ is an element of arc length. The smart way to go is to parametrize the semicircle as follows, $x(t) = \cos(t)$, $y(t) = \sin(t)$.

We have $ds = \sqrt{x'(t)^2 + y'(t)^2} = 1. $

The quantity $\rho$ is the radius of the surface of revolution, which, in this case, is $y = \sin(t)$.

For the circle

$\sigma = 2\pi\int_0^\pi \sin(t)\,dt = 2\pi(2) = 4\pi.$

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Check if this helps : Surface area $S$ is given by $ S= \int_{0}^{\pi}2\pi rdt$ Where $dt=Rd\theta$ and $ r=Rsin\theta$ $\theta$ being the angle of the circle in question . So we get $ S= \int_{0}^{\pi}2\pi R^2sin\theta d\theta $ $S= 4\pi$