Let $f: \mathbb{R}^3\to \mathbb{R}^3$ be given by $f(\rho, \phi, \theta) = (\rho\cos\theta \sin \phi, \rho \sin \theta \sin \phi, \rho \cos \phi).$ It is called the spherical coordinate transformation. Take $S$ to be the set $S = [1, 2] \times (0, \pi /2] \times [0, \pi /2]$.
(a) Calculate $D_f$ and $\det D_f$.
(b) Sketch the image under $f$ of the set $S$.
For (a), $D_f(\rho, \phi, \theta)=\begin{bmatrix}\cos \theta \sin \phi &\rho \cos \theta \cos \phi &-\rho\sin\theta\sin\phi \\ \sin\theta\sin\phi &\rho\sin\theta\cos\phi &\rho\cos\theta\sin\phi \\ \cos\phi & -\rho\sin\phi & 0\end{bmatrix}$ and so $\det D_f=\rho^2\cos^2\theta\sin^3\phi+\rho^2\cos^2\theta\cos^2\phi\sin\phi+\rho^2\sin^2\theta\sin^2\phi+\rho^2\sin^2\theta\sin\phi\cos^2\phi$
I do not know what to do in (b), could someone help me please? Thank you very much.

Using properties of the determinant, calculating over the last row, and then using some very basic trigonometry, \begin{align} \det D_f &=\begin{vmatrix}\cos \theta \sin \phi &\rho \cos \theta \cos \phi &-\rho\sin\theta\sin\phi \\ \sin\theta\sin\phi &\rho\sin\theta\cos\phi &\rho\cos\theta\sin\phi \\ \cos\phi & -\rho\sin\phi & 0\end{vmatrix}\\ \ \\ &=\rho^2\sin\phi\,\begin{vmatrix}\cos \theta \sin \phi & \cos \theta \cos \phi &- \sin\theta \\ \sin\theta\sin\phi & \sin\theta\cos\phi & \cos\theta \\ \cos\phi & - \sin\phi & 0\end{vmatrix}\\ \ \\ &=\rho^2\sin\phi\,[\cos\phi(\cos^2\theta\cos\phi+\sin^2\theta\cos\phi)+\sin\phi(\cos^2\theta\sin\phi+\sin^2\theta\sin\phi)]\\ \ \\ &=\rho^2\sin\phi\,[\cos^2\phi+\sin^2\phi]\\ \ \\ &=\rho^2\sin\phi. \end{align}
For part (b), start first with the set $[0,2]\times [0,2\pi]\times[0,\pi]$. Then move to $[1,2]\times [0,2\pi]\times[0,\pi]$, then $[1,2]\times [0,\pi/2]\times[0,\pi]$, and then $[1,2]\times [0,\pi/2]\times[0,\pi/2]$ and $[1,2]\times (0,\pi/2]\times[0,\pi/2]$. If you struggle with the first one, go refresh your spherical coordinates.