The volume of sphere using integrals In spherical coordinate system I have the volume element $$dV=r^{2}\sin(\theta)\ d\theta\ d\varphi\ dr$$
I want to calculate the volume for the radius equal to $R$. I calculate the integral:
$$\int_{0}^{R} \int_{0}^{2\pi} \int_{0}^{\pi}  r^{2}\sin(\theta)\ d\theta\ d\varphi\ dr = \left [-\frac{1}{3}r^{3}\cos(\theta)  \right ]_{0,0,0}^{r=R,\varphi=2\pi,\theta=\pi}=\frac{2}{3}\pi R^{3}$$
What did I do wrong?
 A: It often helps to write out all of the computation in excruciating detail.  It is harder to make errors that way, and easier to spot them.
\begin{align}
\int_{0}^{R}\int_{0}^{2\pi}\int_{0}^{\pi} r^2\sin(\theta)\, \mathrm{d}\theta\,\mathrm{d}\varphi\,\mathrm{d}r
&= \int_{0}^{R} \int_{0}^{2\pi} \Big[r^2 (-\cos(\theta))\Big]_{\theta=0}^{\pi}\,\mathrm{d}\varphi\, \mathrm{d}r \\
&= \int_{0}^{R} \int_{0}^{2\pi} r^2\left(-\cos(\pi) + \cos(0)\right)\,\mathrm{d}\varphi\, \mathrm{d}r \tag{$\ast$} \\
&= \int_{0}^{R} \int_{0}^{2\pi} 2r^2\,\mathrm{d}\varphi\, \mathrm{d}r\\
&= \int_{0}^{R} 4\pi r^2\,\mathrm{d}r \\
&= \left[ \frac{4}{3}R^3 \right]_{r=0}^{R} \\
&= \frac{4}{3} \pi R^3.
\end{align}
In this case, as others have pointed out, it appears that your error was somewhere around the step I labeled with ($\ast$).
A: Sphere of radius $ r $ can be generated by revolving the upper semicircular disk enclosed between the $ x-$ axis and 
$$ x^2+y^2 = r^2 $$
about the $x-$axis. Since the upper half of this circle is the graph of 
$$ y = f(x)= \sqrt{r^2 - x^2},$$
it follows that the volume of the sphere is
$$ V = \int_{a}^{b}\pi[f(x)]^2dx = \int_{-r}^{r}\pi(r^2 -x^2)dx = \pi\left[r^2x -\frac{x^3}{3}\right] = \frac{4}{3}\pi r^3.$$
A: When you evaluate your integral it should be $$-\frac{1}{3}r^3[\cos(\pi)-\cos(0)].$$
A: $$(-\cos (\pi))-(-\cos (0))=2$$
the volume is
$$\frac {R^3}{3} .2\pi.2=\frac {4\pi R^3}{3} $$
