# Thread: trouble with change of variables

1. ## trouble with change of variables

i'm having a small problem with changing variables.
u = x and v = xy, i am having trouble getting y in terms of v.

if my region of integration is needed here it is: from x = 1 to x= 2 and xy = 1 to xy = 2

2. Originally Posted by ihaveamathproblem
i'm having a small problem with changing variables.
u = x and v = xy, i am having trouble getting y in terms of v.

if my region of integration is needed here it is: from x = 1 to x= 2 and xy = 1 to xy = 2
In the uv-plane you will have the lines u=x=1 and u=x=2 and v=xy=1 v=xy=2

You will get the square with corners (1,1) (1,2) (2,1) and (2,2)

y will be in terms of u and v

$y=\frac{v}{x}=\frac{v}{u}$

The jacobian matrix would be

$\frac{\partial{(x,y)}}{\partial{(u,v)}}=
\begin{vmatrix}
1 && 0 \\
-\frac{v}{u^2} && \frac{1}{u} \\
\end{vmatrix}
$

3. Originally Posted by ihaveamathproblem
i'm having a small problem with changing variables.
u = x and v = xy, i am having trouble getting y in terms of v.

if my region of integration is needed here it is: from x = 1 to x= 2 and xy = 1 to xy = 2
The transformation is $\bold{G}(u,v) = (u,v/u)$. If $f$ is the function you are integrating then $\int_R f = \int_{\bold{G}^{-1} (R) } f\circ \bold{G}$. Where $R = \{ (x,y)\in \mathbb{R}^2 | 1/x \leq y\leq 2/x \mbox{ and }1\leq x\leq 2\}$. And $\bold{G}^{-1}(R) = \{ (u,v) \in \mathbb{R}^2 | \bold{G}(u,v) \in R\}$. Thus, for a point $(u,v)$ to we be in $\bold{G}^{-1}(R)$ we require that $(u,v/u) \in R$, thus, $1\leq u\leq 2 \mbox{ and }1/u \leq v/u\leq 2/u \implies 1\leq u\leq 2 \mbox{ and }1\leq v\leq 2$. Thus, the this new set you are integrating over is simply the rectangle $[1,2]\times [1,2]$.