how to find surface area of a sphere

could any one tell me how to calculate surfaces area of a sphere using elementary mathematical knowledge? I am in Undergraduate second year doing calculus 2. I know its $4\pi r^2$ if the sphere is of radius $r$, I also want to know what is the area of unit square on a sphere.

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How do you think we should proceed? –  hjpotter92 Mar 20 '13 at 6:10
What is a unit square on a sphere? –  ronno Nov 8 '13 at 4:41

You can even do it by using techniques from first-year calculus, and without using polar coordinates. Rotate the half-circle $y=\sqrt{r^2-x^2}$, from $x=-r$ to $x=r$, about the $x$-axis. Better, rotate the quarter circle, $0$ to $r$ about the $x$-axis, and then double the answer. So by the usual formula for the surface area of a solid of revolution. we want $$\int_0^r 2\pi x\sqrt{1+\left(\frac{dy}{dx}\right)^2}\,dx.$$

Find $\frac{dy}{dx}$. We get $-\frac{x}{\sqrt{r^2-x^2}}$. Square this, add $1$, bring to a common denominator, take the square root. So now we need $$\int_0^r 2\pi r\frac{x\,dx}{\sqrt{r^2-x^2}}.$$ The integration is straightforward. Either let $u=r^2-x^2$, or recognize directly that $-\sqrt{r^2-x^2}$ is an antiderivative of $\frac{x}{\sqrt{r^2-x^2}}$. This is easy to see, since earlier we differentiated $\sqrt{r^2-x^2}$.

Finally, plug in. We end up with $2\pi r^2$. And multiply by $2$, since we were calculating the surface area of the half-sphere.

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Note that the parametrization of sphere is given by $$r(u,v)=(r\sin u\cos v,r\sin u\sin v,r\cos u)$$ where $0\leq u\leq \pi$ and $0\leq v\leq2\pi.$ So the surface area is given by the formula $$\int_0^{2\pi}\int_0^{\pi}||r_u\times r_v|| du\hspace{1mm} dv$$ Now calculate the integral to get the formula.

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what is $r_u$, $r_v$? –  bantus Mar 20 '13 at 6:27
@bantus: $r_u, r_v$ are the partial derivatives w.r.t. $u$ and $v$, see here:en.wikipedia.org/wiki/Surface_area . –  pritam Mar 20 '13 at 15:19

Hint

Take the sphere to be formed by the rotation of semicircle(why?) about x-axis $x = r \cos \theta$, $y = r \sin \theta$ where $\theta \in [0, \pi]$.

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could you tell me what will be the expression for surface integral? is it this $\int_{\theta=0}^{\pi}\int_{r=0}^{2\pi}$? I dont know what will be inside the the integral. –  bantus Mar 20 '13 at 6:59
I won't go into a proof that the surface of the sphere $S^2_r$ of radius $r$ is $4\pi r^2$.
Your last question asks for the area of a unit square on $S^2_r$. The notion "unit square" can mean two things: A "square" $Q\subset S^2_r$ of unit area, or a "square" $Q\subset S^2_r$ of unit side length. I guess you mean the latter.
Such a square $Q$ is the union of $8$ congruent rectangular spherical triangles $T$. Denote by $T'$ the corresponding triangle on the unit sphere $S^2$ obtained by scaling. One leg of $T'$ has length $a:={1\over 2 r}$, and the angle opposite to $a$ is $\alpha:={\pi\over 4}$. The third angle $\beta$ is related to $a$ and $\alpha$ via the following formula from spherical trigonometry: $$\cos\alpha=\cos a\sin\beta\ .$$ It follows that $$\beta=\arcsin{1\over\sqrt{2}\cos(1/(2r))}\ .$$ The area of $T'$ then computes to $${\rm area}(T')=\left(\alpha+\beta+{\pi\over2}\right)-\pi=\beta-{\pi\over4}\ ,$$ and finally we obtain the area of $Q$ as $${\rm area}(Q)=8r^2\>{\rm area}(T')\ .$$