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When solving some exercises, I forgot the formula for the surface area of the unit sphere. However, I remember that the length of the perimeter of a circle of radius $r$ is $2 \pi r$. So I figured I'll just compute

$$2 \int_0^1 2 \pi r \, \mathrm dr,$$

starting to integrate from the north pole of a unit sphere up to its center and as that area occurs twice, I added a factor of $2$. However, this integral yields $2 \pi$ and not the desired $4 \pi$. Where did I forget another factor of $2$?

I tried doing the same computation with the volume of the sphere since its surface area will just be the derivative of the volume. However, again

$$2 \int_0^1 \pi r^2 \, \mathrm dr = \frac{2}{3} \pi,$$

missing a factor of $2$. What am I doing wrong?

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    $\begingroup$ Your first integral gives you twice the area of a circle. I'm not sure why you think that integrating the equation of a circle will give you the surface area of a sphere. If you want to see how to go about calculating the surface area, click here. $\endgroup$ – Rocket Man Nov 18 '13 at 12:49
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Hint about the surface:

Consider the surface defined by $$r=\cos u\sin v~\mathbf{i}+\sin u\cos v~\mathbf{j}+\cos v~\mathbf{k},~~0\leq u\leq 2\pi,~0\leq v\leq \pi/2$$ so if we set $x=\cos u\sin v,~y=\sin u\cos v,~z=\cos v$ then $x^2+y^2+z^2=1$ and this is clearly your unit sphere. Now we can use the following well-known formula. I hope you don't mind I am giving you this not pretty one:

$$S=\iint_S ds,~ds=\sqrt{\left(\frac{\partial(y,z)}{\partial(u,v)}\right)^2+\left(\frac{\partial(z,x)}{\partial(u,v)}\right)^2+\left(\frac{\partial(x,y)}{\partial(u,v)}\right)^2}$$

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    $\begingroup$ $\ddot\smile +1$ $\endgroup$ – amWhy Nov 18 '13 at 14:34
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Let $V_n(R)$ be the ($n$)-volume of the Euclidean ball $\mathbb B^n_R$ with radius $R$ in $\mathbb R^n$and let $ S_n(R)$ be the $n$ ("area") of the sphere of radius $R$ in $\mathbb R^{n+1}$. We have $$ \frac{d}{dr} V_n(r)=S_{n-1}(r), e.g. \text{for $n=2, V_2(r)=π r^2, S_1(r)=2π r$,} $$ and for $n=3, V_3(r)=\frac{4}{3}π r^3, S_2(r)=4π r^2$. To recover all formulas, we note that $$ 1=\int_{\mathbb R^n}e^{-π\vert x\vert ^2} dx=\int_0^{+\infty}e^{-π r^2} r^{n-1} dr S_{n-1}(1), \quad\text{so that } $$ $\quad$ $$\boxed{ S_{n-1}(R)=R^{n-1}\frac{2\pi^{n/2}}{\Gamma(n/2)},\qquad V_{n}(R)=R^n\frac{\pi^{n/2}}{\Gamma(1+\frac n2)}.} $$ We need only $\Gamma(1/2)=π^{1/2},\ \text{for $x>0$:}\Gamma (x+1)=x\Gamma (x),\ \text{for $n\in \mathbb N: \Gamma(n+1)=n!$} $

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Let us take the circle $$x^2+y^2=1$$ If we rotate this circle about the x-axis(or y-axis),we get a sphere of unit radius.

To compute the volume:

Let us take a small element dx with height y.Volume of the sphere formed is given by $$\int_0^1 \pi y^2~dx$$ But this is only the volume obtained for rotating only a quarter of a circle(since we took limit from (0,1)).However when we rotate the quarter circle along the x-axis we get a hemisphere(not "a quarter of a sphere"). Hence to get the volume of the sphere is $$2\int_0^1 \pi y^2~dx$$

A similar logic can be applied in computing the surface area. You will have to keep in mind that to find the surface area you will have to take a small arc length dS at a height of y.And rotate this length about the x-axis(or y-axis) to get a small circular element of width dS.

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