An Integral Error I was studying the derivations for the volume and surface area of a sphere . 
One  derivation , for the volume of the sphere is the disk-method .
A circle of radius $r$ is considered , centred at origin . The equation of the circle is $x^2+y^2=r^2$ .
This implies that $y^2=r^2-x^2$ .
To find the volume of the sphere  , $\int _{-r} ^r \pi y^2 \cdot dy $ = $ \int _{-r} ^r \pi (r^2-x^2) \cdot dx $ is calculated . This yields $ \frac{4}{3} \pi r^3 $ , the correct answer .
I wondered if a similar method could be used to calculate the surface area of the sphere . So I calculated $\int ^r _{-r} 2 \pi y \cdot dy $ = $\int ^r _{-r} 2 \pi \sqrt{r^2-x^2} \cdot dx $ . Theoretically , to me , this meant summing up the circumferences , which must yield the surface area . 
However , the definite integral evaluates to $\pi^2 r^2 $ , which is not the correct answer. I would like to know how my approach is wrong , and if at all the area can be calculated by summing up the circumferences (without the use of $\theta$ please ) .
Note:- I am aware of other proofs for the surface area , and am not looking for them :) .
 A: The way you're doing it is not the way it's done. In order to find the surface area of a circle, you need to explore the topic of surfaces of revolution. There, you will learn that the formula for surfaces of revolution for two-dimensional objects (for things in the Cartesian plane, in other words) looks like this:
$$
S=2\pi\int_{a}^{b}f(x)\sqrt{1+\left(\frac{dy}{dx}\right)^2}\,dx
$$
In that formula, $f(x)$ is your curve and $\frac{dy}{dx}$ is the first derivative of that curve. In your case, the curve is going to be given by the function $f(x)=\sqrt{r^2-x^2}$ whose derivative is:
$$
\frac{dy}{dx}=\frac{d}{dx}\left(\sqrt{r^2-x^2}\right)=-\frac{x}{\sqrt{r^2-x^2}}
$$
And then you just plug all that information into the formula and integrate (notice the result of the integration should be multiplied by $2$ since integration from $0$ to $r$ will only give us the surface area of a hemisphere):
$$
S=2\cdot2\pi\int_{0}^{r}\sqrt{r^2-x^2}\sqrt{1+\left(-\frac{x}{\sqrt{r^2-x^2}}\right)^2}\,dx=\\
4\pi\int_{0}^{r}\sqrt{r^2-x^2}\frac{\sqrt{r^2-x^2+x^2}}{\sqrt{r^2-x^2}}\,dx=
4\pi r\int_{0}^{r}\,dx=4\pi rx\bigg|_{0}^{r}=4\pi r(r-0)=4\pi r^2.
$$
A: You actually need to calculate $2\int_{y\in [0,\,r]}2\pi y ds$ with $s$ satisfying $s^{\prime 2}=1+y^{\prime 2},\,f^\prime:=\frac{df}{dx}$. This is because two circumferences separated by a distance $dy$ border a ribbon with a sloped edge of length $ds$.
