# Why doesn't $2\pi\int_{-1}^1\sqrt{1-x^2}dx$ give the surface area of a sphere of radius $1$? [duplicate]

Possible Duplicate:
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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## marked as duplicate by Jason DeVito, Rahul, Guess who it is., t.b., Nate EldredgeAug 1 '12 at 21:31

These hula hoops will be parallel to the $x$ axis I presume. They need to be slanted! Imagine measuring a the length of a graph on the inverval $[a,b]$ by adding little line segments to the curve parallel to the $x$-axis: the sum of these lengths will simply be $[a,b]$. –  anon Jul 14 '12 at 21:52
Thanks for the comments and links, they've been a big help. –  Thoth Jul 14 '12 at 22:10

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 $$S= \int_{0}^{\pi}2\pi R^2sin\theta d\theta$$ $$S= 4\pi$$

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