Question

 Write the mathematical expressions for dv  in spherical and cylindrical coordinates.

Answer 1

First, for the cylindrical coordinates we have

$x=r \cos\theta \\ y=r\sin\theta \\ z=z$

Then, we use the jacobian to perform the change of variables $(x,y,z)\to(r,\theta,z)$ and for this particular case:

$dV_{cylindrical}=J^* {dr}{d\theta}{dz}, \\ \text{where J* is the jacobian:} \\ J^*=\text{absolute value of }det \begin{vmatrix} x_{r} & x_{\theta} & x_{z} \\ y_{r} & y_{\theta} & y_{z} \\ z_{r} & z_{\theta} & z_{z} \end{vmatrix} \\ \text{we compute the partial} \\ \text{derivatives to find:} \\ J^*= \begin{vmatrix} \cos \theta & -r \sin \theta & 0 \\ \sin \theta & r \cos \theta & 0 \\ 0 & 0 & 1 \end{vmatrix} \\ J^*= (1)\begin{vmatrix} \cos \theta & -r \sin \theta \\ \sin \theta & r \cos \theta \end{vmatrix} \\ \text{ } \\\implies J^*=r(\cos^2\theta+\sin^2\theta)=r$

This conclusion leads us to: $dV_{cylindrical}=r{dr}{d\theta}{dz}$.

Now, for the spherical coordinates, we have

$x=\rho \sin\phi\cos\theta \\ y=\rho \sin\phi\sin\theta \\ z=\rho\cos\phi$

and the volume differential (for the new change of variables $(x,y,z)\to(\rho,\phi,\theta)$) is defined as

$dV_{spherical}=J {d\rho}{d\phi}{d\theta}, \\ \text{where J is the jacobian:} \\ J=\text{absolute value of }det \begin{vmatrix} x_{\rho} & x_{\phi} & x_{\theta} \\ y_{\rho} & y_{\phi} & y_{\theta} \\ z_{\rho} & z_{\phi} & z_{\theta} \end{vmatrix} \\ \text{we compute the partial} \\ \text{derivatives to find:} \\ J= \begin{vmatrix} \sin \phi \cos \theta & \rho \cos \phi \cos \theta & - \rho \sin \phi \sin \theta \\ \sin \phi \sin \theta & \rho \cos \phi \sin \theta & \rho \sin \phi \cos \theta \\ \cos \phi & - \rho \sin \phi & 0 \end{vmatrix} \\ \text{ } \\ \implies J= \cos\phi \begin{vmatrix} \rho \cos \phi \cos \theta & - \rho \sin \phi \sin \theta \\ \rho \cos \phi \sin \theta &\rho \sin \phi \cos \theta \end{vmatrix}+\rho \sin\phi \begin{vmatrix} \sin \phi \cos \theta & - \rho \sin \phi \sin \theta \\ \sin \phi \sin \theta & \rho \sin \phi \cos \theta \end{vmatrix}$

Then we continue to reduce and solve the determinants to find J:

$J=\cos \phi (\rho^2\sin\phi\cos\phi\cos^2\theta+\rho^2\sin\phi\cos\phi\sin^2\theta)+... \\ ...+(\rho\sin\theta)( \rho\sin^2\phi\cos^2\theta+\rho\sin^2\phi\sin^2\theta ) \\ \text{once we reduce and use trigonometric identities}, \\ \text{ we find } J=\rho^2 \sin \phi$

Thus we finally have: $dV_{spherical}=\rho^2\sin\theta {d\rho}{d\phi}{d\theta}$

Reference:

• Department of Mathematics. (1996). Triple Integrals in Cylindrical and Spherical Coordinates. Triple integrals in cylindrical and spherical coordinates. Retrieved October 3, 2021, from http://sites.science.oregonstate.edu/math/home/programs/undergrad/CalculusQuestStudyGuides/vcalc/255cs/255cs.html