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The expresion for my n may be deduced from that opening by putting - 0 for 0. We then get

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

My di dv , 1 dv .1 d'.

dpi + d 232 = dps + pot do + dri Also, as in the first part of the last example,

1 d V 1 d V cot Od V

p de 7 dr ' pol do' Adding these three expressions,

[ dv dᏤ dᏙ

dy + d + dx
dy 1 đY 1 ly 9 do cot a dy.

d mo? + y de?+ pdge+, dr + pi do = By substituting for p its value, and making some obvious reductions, this becomes

d? (rV). 1 d V d line

dy2 * sin 0 do + d. cos A (sins & d coc A) = 0. 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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(7) Transform the double integral

Slæm-yn-dy dx into one where u and v are the independent variables, x, y, u, v being connected by the equations

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Here dy dx = (damento dim) du dv.

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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.

(8) Transform the double integral

Sfer*+ y* dx dy into one where r and 0 are the independent variables, having

given

X = r cos 0, y = r sin 0,

Sle=+y*dx dy = - 55er dr do. (9) Having given

X = r cos 0, y = y sin o sind, 3 = r sin cos , transform the triple integral

SSI V dx dy da into a function of r, 0, and p. Using the same artifice as in Ex. 6, we find

SSS V dx dy dx = SSS V dr sin 0 de do. This is a very important transformation, being that from rectangular to polar co-ordinates in space. If we suppose

V = 1, SSfdx dy dx is the expression for the volume of any solid referred to rectangular co-ordinates: and it becomes SS fpad dr sin do do when referred to polar co-ordinates.

(10) Having given z a function of x and y determined by the equation

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into a function of O and when

x = a sin 0 cos , y = b sin @ sin ,

and consequently x= c cos 0. In this case

da = a cos 6 cos , dx=- a sin & sin 0,

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CHAPTER IV.

ELIMINATION OF CONSTANTS AND FUNCTIONS BY MEANS OF

DIFFERENTIATION.

Ex. (1) y? = ax + b.........(1).
To eliminate b, differentiate, when we have

24 g = a........(2).
To eliminate a, substitute its value given by (2) in (1);

then g* = 2 ay yet be

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

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r

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the result is () - w + m = 0. (4) Eliminate a and 6 from the equation

y - a m? bx = 0; the result is any

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