<s>[INST] Ratio between the area of a region and its transformation by a change of variables. The square $T:\{a<x<a+h, b<y<b+h\}, \text{with } a>0,b>0,h>0$ is tranformed into the region $S$ of the plane $uv$ by the following change of variables:
$$
u=\frac{y^2}{x}
$$
$$
v=x^{1/2}y^{1/2}
$$
Find $\frac{area(S)}{area(T)}$.
This is what I have thought:
Using the formula for the change of variables, I have that
$$
area(S)=\iint_Sdudv=\iint_T \left|\frac{\partial(u,v)}{\partial(x,y)}\right|dxdy=
$$
$$
=\int_a^{a+h}\int_b^{b+h}\frac{3}{2}x^{-\frac{3}{2}}y^{\frac{3}{2}}dydx=-\frac{6}{5}(\frac{1}{\sqrt{a+h}}-\frac{1}{\sqrt{a}})((b+h)^{5/2}-b^{5/2})
$$
Therefore, $\frac{area(S)}{area(T)}=\frac{1}{h^2}(-\frac{6}{5}(\frac{1}{\sqrt{a+h}}-\frac{1}{\sqrt{a}})((b+h)^{5/2}-b^{5/2}))$.
However, the solution in my book is:
$$\frac{area(S)}{area(T)}=\frac{6}{5}\frac{b^2+(b+h)^2+b(b+h)+(2b+h)\sqrt{b(b+h)}}{(\sqrt{a+h}+\sqrt{a})(\sqrt{b+h}+\sqrt{b})\sqrt{a(a+h)}}$$
I have tried to do some algebraic transformations to my solution but I do not arrive to the given solution. What am I doing wrong?
Thanks for your help.
 [/INST] Your solution is the same as the solution in your book.
You have
$$\frac{area(S)}{area(T)}=\frac{1}{h^2}\bigg(-\frac{6}{5}\bigg(\frac{1}{\sqrt{a+h}}-\frac{1}{\sqrt{a}}\bigg)\bigg((b+h)^{5/2}-b^{5/2}\bigg)\bigg)$$
Firstly, we get
$$\begin{align}\frac{1}{\sqrt{a+h}}-\frac{1}{\sqrt{a}}&=\frac{\sqrt a-\sqrt{a+h}}{\sqrt a\sqrt{a+h}}
\\\\&=\frac{(\sqrt a-\sqrt{a+h})(\sqrt a+\sqrt{a+h})}{\sqrt{a(a+h)}(\sqrt a+\sqrt{a+h})}
\\\\&=\frac{-h}{\sqrt{a(a+h)}(\sqrt a+\sqrt{a+h})}\end{align}$$
Secondly, letting $x=\sqrt{b+h},y=\sqrt b$, we get
$$\begin{align}&(b+h)^{5/2}-b^{5/2}
\\\\&=x^5-y^5
\\\\&=(x-y)(x^4 + x^3 y + x^2 y^2 + x y^3 + y^4)
\\\\&=(\sqrt{b+h}-\sqrt b)\bigg((b+h)^2+(b+h)\sqrt{b(b+h)}+(b+h)b+b\sqrt{b(b+h)}+b^2\bigg)
\\\\&=(\sqrt{b+h}-\sqrt b)\bigg((b+h)^2+(b+h)\sqrt{b(b+h)}+(b+h)b+b\sqrt{b(b+h)}+b^2\bigg)\times\frac{\sqrt{b+h}+\sqrt b}{\sqrt{b+h}+\sqrt b}
\\\\&=\frac{h}{\sqrt{b+h}+\sqrt b}\bigg(b^2+(b+h)^2+b(b+h)+(2b+h)\sqrt{b(b+h)}\bigg)\end{align}$$
So, your solution can be written as
$$\frac{area(S)}{area(T)}=\frac{1}{h^2}\bigg(-\frac 65\bigg)\bigg(\frac{-h}{\sqrt{a(a+h)}(\sqrt a+\sqrt{a+h})}\bigg)\times\frac{h}{\sqrt{b+h}+\sqrt b}\bigg(b^2+(b+h)^2+b(b+h)+(2b+h)\sqrt{b(b+h)}\bigg)$$
$$=\frac{6}{5}\cdot\frac{b^2+(b+h)^2+b(b+h)+(2b+h)\sqrt{b(b+h)}}{(\sqrt{a+h}+\sqrt{a})(\sqrt{b+h}+\sqrt{b})\sqrt{a(a+h)}}$$
which is the solution in your book.
</s>