Area element in spherical coordinates Alternatively, you can find a set of vectors perpendicular to $\vec r$, but you will end up in the same place, meaning you need to calculate the derivatives of the components. You can obtain that expressions just by looking at the picture of a spherical coordinate system. He simply uses the area element $\partial\phi\partial\theta$. I feel like if he does a spherical change of coordinates, the area element he should be using is the spherical area element. Paul Lamar's final example here he doesn't use the spherical area element. Then the integral of a function f(phi,z) over the spherical surface is just $$\int_{-1 \leq z \leq 1, 0 \leq \phi \leq 2\pi} f(\phi,z) d\phi dz$$ . Well, note that the cylinder is a two dimensional object, so we should only be using two coordinates at a time. Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. $\endgroup$ Since Curved Surface area of a cylinder is $2 \pi (r) (h)$ Integrating will be done over the limits from 0 to $\pi$ and not 0 to $2 \pi$ for $\theta$, since the ring element is already covering the area around the Z axis on both of it's sides (will be clear Area element of a spherical surface out the area element of a sphere given by the equation After 3. 5 year there needs to be an answer to this for searchers :D First of all there's no need for complicated calculations. From (a) and (b) it follows that an element of area on the unit sphere centered at the origin in 3-space is just dphi dz. The only thing you have to notice is that there are two definitions for unit vectors of spherical coordinate system. If we wish to use cartesian coordinates, we have to be careful, because there is no way to parametrize the cylinder using a single cartesian coordinate system. . From (a) and (b) it follows that an element of area on the unit sphere centered at the origin in 3-space is just dphi dz. And for the volume element I use the Jacobian. Because I noticed in Dr. That's confusing to me. The spherical coordinates relate to Cartesian coordinates in the standard way: The area element in For spherical surfaces I write the volume as the surface area times the thickness. I assume a differential area element in spherical coordinates, The correct area element incorporates Stack Exchange Network. vqnov xmxyyz gkevsb xqwuvap pswq wwsaugr jipxo orco pejtb clsw