Therefore in your situation it remains to compute the vector product ${\bf x}_\phi\times {\bf x}_\theta$ Latitude is either geocentric latitude, measured at the Earth's center and designated variously by , q, , c, g or geodetic latitude, measured by the observer's local vertical, and commonly designated . It is also convenient, in many contexts, to allow negative radial distances, with the convention that In polar coordinates: \[\int\limits_{0}^{\infty}\int\limits_{0}^{2\pi} A^2 e^{-2ar^2}r\;d\theta dr=A^2\int\limits_{0}^{\infty}e^{-2ar^2}r\;dr\int\limits_{0}^{2\pi}\;d\theta =A^2\times\dfrac{1}{4a}\times2\pi=1 \nonumber\]. The square-root factor comes from the property of the determinant that allows a constant to be pulled out from a column: The following equations (Iyanaga 1977) assume that the colatitude is the inclination from the z (polar) axis (ambiguous since x, y, and z are mutually normal), as in the physics convention discussed. Perhaps this is what you were looking for ? ) the orbitals of the atom). When , , and are all very small, the volume of this little . Lets see how we can normalize orbitals using triple integrals in spherical coordinates. The line element for an infinitesimal displacement from (r, , ) to (r + dr, + d, + d) is. I am trying to find out the area element of a sphere given by the equation: r 2 = x 2 + y 2 + z 2 The sphere is centered around the origin of the Cartesian basis vectors ( e x, e y, e z). The differential surface area elements can be derived by selecting a surface of constant coordinate {Fan in Cartesian coordinates for example} and then varying the other two coordinates to tIace out a small . In order to calculate the area of a sphere we cover its surface with small RECTANGLES and sum up their total area. From these orthogonal displacements we infer that da = (ds)(sd) = sdsd is the area element in polar coordinates. Using the same arguments we used for polar coordinates in the plane, we will see that the differential of volume in spherical coordinates is not \(dV=dr\,d\theta\,d\phi\). In three dimensions, this vector can be expressed in terms of the coordinate values as \(\vec{r}=x\hat{i}+y\hat{j}+z\hat{k}\), where \(\hat{i}=(1,0,0)\), \(\hat{j}=(0,1,0)\) and \(\hat{z}=(0,0,1)\) are the so-called unit vectors. The geometrical derivation of the volume is a little bit more complicated, but from Figure \(\PageIndex{4}\) you should be able to see that \(dV\) depends on \(r\) and \(\theta\), but not on \(\phi\). Three dimensional modeling of loudspeaker output patterns can be used to predict their performance. We also mentioned that spherical coordinates are the obvious choice when writing this and other equations for systems such as atoms, which are symmetric around a point. These relationships are not hard to derive if one considers the triangles shown in Figure \(\PageIndex{4}\): In any coordinate system it is useful to define a differential area and a differential volume element. (8.5) in Boas' Sec. [2] The polar angle is often replaced by the elevation angle measured from the reference plane towards the positive Z axis, so that the elevation angle of zero is at the horizon; the depression angle is the negative of the elevation angle. for any r, , and . I'm able to derive through scale factors, ie $\delta(s)^2=h_1^2\delta(\theta)^2+h_2^2\delta(\phi)^2$ (note $\delta(r)=0$), that: The precise standard meanings of latitude, longitude and altitude are currently defined by the World Geodetic System (WGS), and take into account the flattening of the Earth at the poles (about 21km or 13 miles) and many other details. ( The straightforward way to do this is just the Jacobian. This page titled 10.2: Area and Volume Elements is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by Marcia Levitus via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request. There is an intuitive explanation for that. In spherical coordinates, all space means \(0\leq r\leq \infty\), \(0\leq \phi\leq 2\pi\) and \(0\leq \theta\leq \pi\). There is yet another way to look at it using the notion of the solid angle. The relationship between the cartesian coordinates and the spherical coordinates can be summarized as: (25.4.5) x = r sin cos . As the spherical coordinate system is only one of many three-dimensional coordinate systems, there exist equations for converting coordinates between the spherical coordinate system and others. 180 The spherical coordinate system generalizes the two-dimensional polar coordinate system. This will make more sense in a minute. $$ In cartesian coordinates, all space means \(-\infty