in the Integral Calculus. 509 
and substituting, we get 
r= (1+ m*)P?; 
and, remembering that in polar coordinates 
‘e r? sin 8 
Pe i ee int o(%)*) 
/ (#) +r? sin? 8+ sin? 6 =) 
we get 
( a) + r*sin?6 + sin? Gy = (1+m?)r? sin? 6 
dg | do ; 
(<)'= sin? 04 mi? a) ‘} , 
b 
This, then, is the result of the transformation of the given 
equation to polar coordinates, the deduction of which by direct 
substitution would have been tedious and laborious, and the ex- 
pression thus reduced may be integrated by series. It can be 
easily proved that the quadrature of the surface represented by 
this equation is given by the simple formula 
y=(1+ m?)* V7? sin 6 dé dd, 
a result which indeed might have been anticipated from geome- 
trical considerations, 
or 
4, There are some differential equations whose integration 
may be facilitated by a partial employment of the above method. 
Thus if it be proposed to integrate the equation 
ydax —ady = (2? +-y?)*da, 
by partial transformation to polar coordinates we get 
rd0=rdz, or rdd=dz, 
or 
whence, at once, the integral 
wr 6 
tan 2 + ) =Cr, 
It is not difficult to reduce this solution to the form 
= u) -1 Ca—-1 
tan (2 =2 tan Cal)" 
5, Again, let it be proposed to integrate the equation 
y 
y(yda — xdy) = (2° + ye" + dz ; 
