we find when 2 + y2+ x2 = r2, φ dr2 d2 V = 0. (5) Transform + = 0 into a function of r and dx2 dy 0, having given ar cos 0, y = r sin 0. into a function of r, 0, and p, having given x = r cos 0, y = r sin 0 sin 4, x = r sin 0 cos p. A slight artifice will enable us to do this with considerable facility. Assume pr sin 0, so that x = p cos &, y = p sin o, p = r sin 0, x = r cos 0. Taking first the two variables y and x, we find as in the preceding example In exactly the same way, the equations of condition being similar, we find Also, as in the first part of the last example, By substituting for p its value, and making some obvious reductions, this becomes This important equation is the basis of the Mathematical Theories of Attraction and Electricity. The artifice here used is given by Mr A. Smith in the Cambridge Mathematical Journal, Vol. 1. p. 122. into one where u and v are the independent variables, x, y, u, v being connected by the equations x + y = u, y = UV. Therefore dy dx = udud v, and ffx-1y-1 dy dx = ffum+n-1 (1 − v)m−1 y2−1 du dv. This transformation is given by Jacobi in Crelle's Journal, Vol. XI. p. 307: it is of great use in the investigation of the values of definite integrals. into one where r and are the independent variables, having given x = r cos 0, y = r sin 0, [Sex2+ y2 dx dy = - Sfer3r dr do. Using the same artifice as in Ex. 6, we find [[[V dx dy dx = [[[Vr2 dr sin ✪ de dø. This is a very important transformation, being that from rectangular to polar co-ordinates in space. If we suppose V = 1, fffdx dy do is the expression for the volume of any solid referred to rectangular co-ordinates: and it becomes fffr2 dr sine de dp when referred to polar co-ordinates. |