jacobian

**kittycat** · October 1st 2007, 12:45 PM

use the transformation T(u,v) = < u, u+v> to evaluate the iterated integral
the integral from 1 to 2 the integral from x+3 to x+6 of 1/(sqrt(xy-x^2)) dy dx.

please help if you know how to solve this question. thank you.

**Jhevon** · October 1st 2007, 06:45 PM

Originally Posted by kittycat

use the transformation T(u,v) = < u, u+v> to evaluate the iterated integral
the integral from 1 to 2 the integral from x+3 to x+6 of 1/(sqrt(xy-x^2)) dy dx.

please help if you know how to solve this question. thank you.

I will attach the graphs later if i have time. i will show you how to set up the integral.

Recall how we perform change of variables with Jacobians:

Assuming all the conditions are fulfilled (look up these conditions in your text), we calculate the integral under the $C^1$ transformation T as follows:

$\iint_{R}f(x,y)~dA = \iint_{S}f(x(u,v),y(u,v)) \cdot \left| \frac {\partial (x,y)}{ \partial (u,v)} \right|~dudv$

where $\frac {\partial (x,y)}{ \partial (u,v)} = \left| \begin{array}{cc} \frac {\partial x}{\partial u} & \frac {\partial x}{\partial v} \\ & \\ \frac {\partial y}{\partial u} & \frac {\partial y}{\partial v} \end{array}\right|$

we are given:

$\boxed {x = u}$ and $y = u + v \implies \boxed{v = y - x}$

So, we have: $\frac {\partial (x,y)}{\partial (u,v)} = \left| \begin{array}{cc} 1 & 0 \\ 1 & 1 \end{array}\right| = 1$

Now, graph the region over which the original integral is being done, it will be a parallelogram. label the lines as i describe so you can follow, we have to transform each line under the directions of T. Label the line y = x + 3 as $f_1$ , the line x = 2 as $f_2$ , the line y = x + 6 as $f_3$ , and the line x = 1 as $f_4$ . note the intervals on which these lines are defined, as we have to change them under the transformation T as well. ( $f_1: 1 \le x \le 2$ , $f_2: 5 \le y \le 8$ , $f_3: 1 \le x \le 2$ , and $f_4: 4 \le y \le 7$ )

Now, let's get this show on the road.

For $f_1:$

$\begin{array}{c|c} y = x + 3 & x = u \\ \Rightarrow u + v = u + 3 & 1 \le x \le 2 \\ \Rightarrow \boxed {v = 3} & \boxed{1 \le u \le 2} \end{array}$

For $f_2:$

$\begin{array}{c|c} x = 2 & y = 2 + v \\ \Rightarrow \boxed{u = 2} & v = y - 2 \\ & 5 \le y \le 8 \\ & \boxed{3 \le v \le 6} \end{array}$

For $f_3:$

$\begin{array}{c|c} y = x + 6 & x = u \\ \Rightarrow u + v = u + 6 & 1 \le x \le 2 \\ \Rightarrow \boxed {v = 6} & \boxed{1 \le u \le 2} \end{array}$

For $f_4:$

$\begin{array}{c|c} x = 1 & y = 1 + v \\ \Rightarrow \boxed{u = 1} & v = y - 1 \\ & 4 \le y \le 7 \\ & \boxed{3 \le v \le 6} \end{array}$

Now we are ready. draw all those graphs between the given intervals on a new pair of axis. Let the vertical axis be $v$ and the horizontal axis be $u$ . you will notice the new region is a rectangle, which is a much simpler region to integrate over. not only that, but the integral gets a lot easier as well:

We know: $\iint_R f(x,y)~dA = \iint_S f(x(u,v),y(u,v)) \cdot \left| \frac {\partial (x,y)}{\partial (u,v)} \right|~dvdu$

$= \int_{1}^{2} \int_{3}^{6} \frac 1{\sqrt{u^2 + uv - u^2}} |1|~dvdu$

$= \int_{1}^{2} \int_{3}^{6} (uv)^{-\frac 12}~dvdu$

Now continue

Check my computations, i only got 2 hours of sleep last night, so i am really tired and prone to silly mistakes right now

EDIT: I attached the diagrams of the regions

**kittycat** · October 2nd 2007, 05:57 AM

Hi Jhevon,

Many thanks for your help.

I should give you a hug for that.

I have the final answer (2sqrt(6) -2sqrt(3)) *(2sqrt(2) -2) or approximately 1.189. Am I right?

**Jhevon** · October 2nd 2007, 01:34 PM

Originally Posted by kittycat

Hi Jhevon,

Many thanks for your help.

I should give you a hug for that.

I have the final answer (2sqrt(6) -2sqrt(3)) *(2sqrt(2) -2) or approximately 1.189. Am I right?

that's what i got. hopefully we're not both wrong

i take it you followed all the steps ok?

**Krizalid** · October 24th 2007, 11:10 AM

Just to confirm, I have the same answer

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