Paper 3, Section II, C

Vector Calculus | Part IA, 2012

Consider the transformation of variables

x=1u,y=1v1uv.x=1-u, \quad y=\frac{1-v}{1-u v} .

Show that the interior of the unit square in the uvu v plane

{(u,v):0<u<1,0<v<1}\{(u, v): 0<u<1,0<v<1\}

is mapped to the interior of the unit square in the xyx y plane,

R={(x,y):0<x<1,0<y<1}.R=\{(x, y): 0<x<1,0<y<1\} .

[Hint: Consider the relation between vv and yy when u=αu=\alpha, for 0<α<10<\alpha<1 constant.]

Show that

(x,y)(u,v)=(1(1x)y)2x\frac{\partial(x, y)}{\partial(u, v)}=\frac{(1-(1-x) y)^{2}}{x}

Now let

u=1t1wt,v=1wu=\frac{1-t}{1-w t}, \quad v=1-w

By calculating

(x,y)(t,w)=(x,y)(u,v)(u,v)(t,w)\frac{\partial(x, y)}{\partial(t, w)}=\frac{\partial(x, y)}{\partial(u, v)} \frac{\partial(u, v)}{\partial(t, w)}

as a function of xx and yy, or otherwise, show that

Rx(1y)(1(1x)y)(1(1x2)y)2dxdy=1\int_{R} \frac{x(1-y)}{(1-(1-x) y)\left(1-\left(1-x^{2}\right) y\right)^{2}} d x d y=1

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