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(5) Transform + =0 into a function of r and

dx2 dy*

e, having given ar cos 0, y = r sin 0.

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into a function of r, 0, and p, having given

x = r cos 0, y = r sin 0 sin (, ≈ = r sin 0 cos p.

A slight artifice will enable us to do this with considerable facility. Assume pr sin 0,

=

y = p sin o,

p = r sin 0,

so that

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Taking first the two variables y and x, we find as in the preceding example

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In exactly the same way, the equations of condition being similar, we find

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Also, as in the first part of the last example,

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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.

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into one where u and v are the independent variables, x, y, u, v being connected by the equations

x + y = u, y = uv.

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Therefore dy dx = udud v,

and ffmy dy dx = ffum+n-1 (1 − v)TM-1 y"-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.

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into one where r and are the independent variables, having given

x = r cose, y = r sin 0,

[[ex2+y3 dx dy = − Sfer3r dr do.

(9) Having given

=

x = r cos 0, y = r sin 0 sin p, ≈r sin cos p,

transform the triple integral

fffV dx dy dz

into a function of r, 0, and p.

Using the same artifice as in Ex. 6, we find

ЛSSV dx dy dx = ƒƒƒ Vr2 dr sin 0 de dp.

This is a very important transformation, being that from rectangular to polar co-ordinates in space. If we suppose V = 1, ffdx dy dz is the expression for the volume of any solid referred to rectangular co-ordinates: and it becomes Jffr2 dr sine de do when referred to polar co-ordinates.

(10) Having given a function of x and y determined

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dz dz

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dx' dy

and do dy, we find

[sdæ dy {1 + (da)2 + (da) *}*

= ƒƒd0 dp sin 0 { a2b2 (cos 0)2 + (c sin 0)2 (a2 sin2 p + b2 cos3p)}1.

Ivory, Phil. Trans. 1809.

CHAPTER IV.

ELIMINATION OF CONSTANTS AND FUNCTIONS BY MEANS OF DIFFERENTIATION.

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To eliminate a, substitute its value given by (2) in (1);

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To eliminate both a and b, differentiate (2) again; then

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(4) Eliminate a and b from the equation

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