Question

Find the integral surface of (x-y)(p-q) = z(q+1) which contains the circle z=4; x²+y²=4².

Answer 1

Answer:-

(x-y)(p-q)=z(q+1)

xp-qx-yp+yq=z(q+1)

xp-qx-yp+yq-zq=z

(x-y)p+(-x+y-z)q=z

Corresponding symmetric equation of the given system is;

$\frac{dx}{x-y}=\frac{dy}{-x+y-z}=\frac{dz}{z}$

We need two independent first integral

Adding numerator and denominator we get,

$\frac{dx+dy+dz}{x-y-x+y-z+z}=\frac{d(x+y+z)}{0}$

Which means d(x+y+z)=0 so,

1^st independent integral is f₁=x+y+z

Next:

$\frac{dx-dy+dz}{x-y+x-y+z+z}=\frac{dz}{z}$

$\frac{d(x-y+z)}{x-y+z}=\frac{2dz}{z}$

$d(\frac{ln|x-y+z|}{z^{2}})=0$

and we get 2^nd independent integral

f₂=$\frac{x-y+z}{z^{2}}$

Taking t as a parameter and the given equation of x²+y²=4²,z=4. Can be put in parametric form as;

x=t

y=4+/-t

z=4

We then rewrite f₁ and f₂ as:

f₁=t+4+t+4(taking positive value of t in y)

f₁=2t+8

t=$\frac{f_1-8}{2}$

f₂=$\frac{t-(4+t)+4}{4^{2}}$

f₂=$\frac{2t}{4^{2}}$

Replace t in f₂

f₂=$\frac{f_1-8}{16}$

f₁-8-16f₂=0

Putting the values of f₁ and f₂ the desired integral surface is;

x+y+z-16($\frac{x-y+z}{z^{2}})$ -8=0