I'm trying to derive the formula for the volume of a sphere, using integration :
$\int_{0}^{\pi r}\pi r^{2}dc$
$\pi r^{2}$ is constant, so :
$\pi r^{2}\int_{0}^{\pi r}dc$
Integrating, I get only c, and plug $\pi r$ for c.
So I'm getting $\pi^{2}r^{3}$, instead of $\frac{4}{3}\pi r^{3}$.
My reasoning is that I start with a full disc ($\pi r^{2}$), then rotating the disc for half the circumference of the sphere ($\pi r$) to get the final volume.
What's wrong with my reasoning ?
You're wrong because you're basically trying to obtain the volume of the sphere via the following product:
$$(\pi r) (\pi r^2)$$
Let's see why it doesn't work, integral calculus apart.
If you had the area of the circle as $\pi r^2$, by multiplying the same by an arbitrary constant ($c$ in this case, which goes from zero to $\pi r$), would just transform the circle into a cylinder with height $r$, as the radius.
Now you multiplied it by $\pi$, so you have essentially the volume of the solid you obtained as $\pi^2 r^3$, that is a sort of expansion of the volume $\pi$ times.
Via Triple Integration
Back to your problem: with a triple integration you can handle it easy, provided you know about polar spherical coordinates.
Thence, as long as you use the correct volume element of $dr\,\, r \sin \theta \, dr\,\, r\,d\phi = r^2dr\,d\theta\,d\phi$ : $$ \int_{r=0}^R\int_{\theta = 0}^{\pi} \int_{\phi=0}^{2\pi} r^2\sin\theta\,dr\,d\theta\,d\phi = \\ 2\pi\int_{r=0}^R\int_{\theta = 0}^{\pi} r^2\sin\theta\,dr\,d\theta = \\ 2\pi\int_{r=0}^R \left[ -r^2\cos\theta \right]_{\theta = 0}^\pi dr = -2\pi\int_{r=0}^R \left[ \cos\pi - \cos 0\right]\,dr \\ = 4\pi \int_{r=0}^R r^2dr = \frac43\pi R^3 $$
Via Solid of revolution
Otherwise you should check for "volume of solid of rotations with integrals".
In that case you need to find a function $f(x)$ from the circumference and then apply the definition for the volume of a solid of revolution.
Start with the circumference $x^2 + y^2 = r^2$ to get $y = \sqrt{r^2 - x^2}$, which delines the arc in the first quadrant of the Cartesian plane. Since a rotation around the $Y$ axis of this arc (from $x = 0$ to $x = r$) would give you the volume of the semisphere, we need to multiply by a factor of $2$ for the whole volume.
Rotating the function around the $Y$ axis, with the factor $2$, we get: $$ V = 2\pi\int_0^{r} y^2\ \text{d}x = 2\pi \int_0^r r^2 - x^2\ \text{d}x = 2\pi\left(r^3 - \frac{r^3}{3}\right) = 4\pi\frac{r^3}{3}$$