Triple integral limits in spherical coordinates In the tripple integral to calculate the volume of a sphere 
why does setting the limits as follows not work?
$$ \int_{0}^{2\pi} \int_{0}^{\pi}
\int_{0}^{R} p^2 \sin{\phi}
\, dp\,d\theta\,d\phi $$
 A: Jacobian is p^2sin(ϕ). If you integrate from 0 to 2pi in ϕ you are going through a point in which the Jacobian is zero.
A: Taking into account my comment:
$$\int_0^{2\pi}\int_0^\pi\int_0^R \rho^2\sin\phi\, d\rho\,d\phi\,d\theta=2\pi\int_0^\pi\left.\frac13\rho^3\right|_0^R\sin\phi\,d\phi=$$
$$\frac{2\pi R^3}3\int_0^\pi\sin\phi\,d\phi=\frac{2\pi R^3}3(-\cos\phi|_0^\pi)=\frac{4\pi R^3}3$$
It looks like working fine to me...Perhaps, as Michael points out in the comments, your problem is because you had the wrong limits for the variables? It is $\;0\le \theta\le2\pi\;,\;\;0\le\phi\le\pi\;,\;\;0\le\rho\le R\;$
A: Let $R>0$ be the radius of the sphere $S=\{(x,y,z)\in\mathbb{R}^{3}\vert x^{2}+y^{2}+z^{2}=R^{2}\}$. You want to compute the volume of $B=\{(x,y,z)\in\mathbb{R}^{3}\vert x^{2}+y^{2}+z^{2}\le R^{2}\}$.
You can do it using cartesian coordinates, which requires more work and calculation. But you can also try to convert these coordinates into the so-called spherical coordinates. As you should know, if you switch to spherical coordinates, you get the factor $\rho^{2}\sin(\phi)$ called the "Jacobian".
The volume of a certain region of the space is obtained by taking the integral of $1$ over this region. This fact can be intuitively understood when looking for the area of a region in $\mathbb{R}^{2}$: a very simple example is the area of the rectangle $\mathcal{R}:=[x_{a},x_{b}]\times[y_{c},y_{d}]$ where $x_{a}<x_{b}$ and $y_{c}<y_{d}$. We know from basic geometry that its area is $(x_{b}-x_{a})(y_{d}-y_{c})$. But with integrals, it is obtained as follows:
\begin{align*}
\text{Area}(\mathcal{R})&=\int_{\mathcal{R}}1\text{d}\sigma\\
&=\int_{x_{a}}^{x_{b}}\int_{y_{c}}^{y_{d}}1\text{d}y\text{d}x\\
&=\int_{x_{a}}^{x_{b}}(y_{d}-y_{c})\text{d}x\\
&=(y_{d}-y_{c})\int_{x_{a}}^{x_{b}}1\text{d}x\\
&=(x_{b}-x_{a})(y_{d}-y_{c})
\end{align*}
This is true for any system of coordinates as long as we take the Jacobian (associated with the said system) into account. It is how we obtain:
$$\text{Vol}(B)=\color{magenta}{\int_{0}^{2\pi}}\color{limegreen}{\int_{0}^{\pi}}\color{blue}{\int_{0}^{R}}1\cdot\overbrace{\rho^{2}\sin(\phi)}^{\text{the Jacobian}}\color{blue}{\text{d}\rho}\color{limegreen}{\text{d}\phi}\color{magenta}{\text{d}\theta}$$
And the calculations are given in the other answers.

