Why isn't the volume of a sphere $\frac{2\pi R^3}{3}$? I tried to get the volume of a sphere like this:
The cross section of a sphere is a circle and its area can be notated as $\pi r^2$
So the volume of the half of the sphere with radius $R$ is
$\int_{0}^R \pi r^2 dr$,
and it's equal to
$\frac{\pi R^3}{3}$.
So the whole volume of the sphere is
$\frac{2\pi R^3}{3}$.
I know how to get the volume of a sphere correctly, but can't understand what's wrong with the argument above.
 A: As your argument $r$ goes from $0$ to $R$, the shape you're computing is the volume of a cone. 
If you rewrite using the suggestive name $y$ for the height of a slice of the northern hemisphere, then you've computed
$$
\int_0^R \pi (R-y)^2 ~dy
$$
while you should have said "At height $y$, the radius of my disk will be $\sqrt{R^2 - y^2}$," and written
$$
\int_0^R \pi \left(\sqrt{R^2 - y^2}\right)^2 ~dy
$$
which would have given you the correct volume of a hemisphere. 
A: The area on height $r$ is not $\pi r^2$, but $\pi(R^2-r^2)$.
A: The integral you are calculating doesn't give the volume of sphere. Consider a small thin (thickness $dr$) spherical shell (radius $r$) inside sphere of radius $R$. The surface area of the spherical shell element is $4\pi r^2$. So the volume of the element shell would be $dV=4\pi ((r+dr)^2-r^2)= 4\pi r^2 dr$  The total volume can be found by integrating from $0$ to $R$. $V=\int_0^R 4\pi r^2 dr=(4/3)\pi r^2$
A: If you replace $r$ with $y$ just for better readability, then $\pi y^2$ would indeed be the cross-section of a sphere of radius $R$ you get by rotating the curve $y=\sqrt{R^2-x^2}$ (which is geometrically a semicircle centered at $(0,0)$ lying above the x-axis) around the $x$-axis in the xy-coordinate system.

But what is $y$ as you move along the $x$-axis? It's $\sqrt{R^2-x^2}$:
$$
V=\int_{-R}^{R}\pi y^2\,dx = \pi\int_{-R}^{R} (\sqrt{R^2-x^2})^2\,dx=
\pi\int_{-R}^{R} (R^2-x^2)\,dx
$$
Also notice that you don't have to go all the way from $-R$ to $R$ because the curve $y=\sqrt{R^2-x^2}$ is symmetric about the $y$-axis. Just integrate it from $0$ to $R$ and double the result:
$$
V=2\pi\int_{0}^{R} (R^2-x^2)\,dx=
2\pi\left[R^2 x-\frac{x^3}{3}\right]_{0}^{R}=
2\pi\left(R^2 \cdot R-\frac{R^3}{3}-0\right)=\\
2\pi\left(R^3-\frac{R^3}{3}\right)=
\frac{4}{3}\pi R^3\ cubic\ units.
$$
