Transformation rule

**vincisonfire** · Nov 4th 2008, 08:14 AM

I have to integrate $\int \int \limits_{D} \frac{xy}{x^2-y^2}\cdot dA$ where D is (x, y) such that x > 0 , −1 < y < 2 , 4 < {x^2 + y^2} < 16 , 1 < {x^2 − y^2} < 9 . The lesser sign are lesser or equal but I don't know how to do them.
I use change of coordinates $u = x^2 + y^2$ and $v = x^2 - y^2$
I tried to divise the domain in "curved rectangles" only because in this case it is easy to evalute but I can't figure how to get rid of the straight boundaries. Can someone tell me about another transformation, some deduction leading to answer of anything helpful?
I know integral becomes $\int \int \limits_{R} \frac{1}{8 \cdot v}\cdot dA$ R : 4 0 , y > 0 , 4 < {x^2 + y^2} < 16 , 1 < {x^2 − y^2} < 9 but then I'm stuck.

**shawsend** · Nov 4th 2008, 09:52 AM

This looks like the region to me. Is that indeed the region? Looks pretty messy.

**vincisonfire** · Nov 4th 2008, 01:26 PM

This is the good region.

**shawsend** · Nov 4th 2008, 04:28 PM

Must be an easier way. When I use the change of coordinates you specified:

$u=x^2+y^2$

$v=x^2-y^2$

and transform the region to the u-v plane, I get the indicated region in the second plot below (the diagonal line emanating from the origin is an artifact of the plotting code) :

The domain in the u-v plane now looks a little easier to integrate over, but seems to me you'd still have to split it up into two integrals not to mention you'd have to figure out what each line is and adjust the integration limits accordingly. Also I can't say I'm 100% confident I've mapped it correctly as I used some pretty hack Mathematica code to make the mapping. Here it is if you're interested. We could of course easily change the code to mappings we do know are correct to verify the code if that was necessary.

Code:

rplot = RegionPlot[x >= 0 && y >= -1 && 
    4 <= x^2 + y^2 <= 16 && 
    1 <= x^2 - y^2 <= 9 && y <= 2, 
   {x, 0, 5}, {y, -2, 2}, PlotPoints -> 75, 
   AxesLabel -> {Style["x", 20], 
     Style["y", 20]}, Frame -> None, 
   Axes -> True]
transform = rplot /. GraphicsComplex[pnts_, 
      data__] :> GraphicsComplex[
      ({#1[[1]]^2 + #1[[2]]^2, #1[[1]]^2 - 
          #1[[2]]^2} & ) /@ pnts, data]; 
newplot = Show[transform, PlotRange -> All, 
   AxesLabel -> {Style["u", 20], 
     Style["v", 20]}, Frame -> None, 
   Axes -> True]
GraphicsArray[{{rplot, newplot}}]

**shawsend** · Nov 5th 2008, 12:48 AM

Know what, I'm struggling with this. For starters I believe my illustration of the mapping is correct. Bet a dollar anyway. Let me just assume that that boundary is as it seems: Top diagonal is v=u and the bottom diagonal is v=u-8. That would need to be verified of course. Then:

$\mathop\int\limits_{D}\frac{xy}{x^2+y^2}dA=\pm\fra c{1}{8}\left(\int_4^9\int_1^u \frac{1}{v}dvdu+\int_9^{16}\int_{u-8}^9 \frac{1}{v}dvdu\right)=\pm\frac{1}{4}-\ln\left(\frac{81}{16}\right)$

With the $\pm$ indicating that I've not yet determined how (u,v) traces the boundary of the new domain. Of course if I had to turn this in I would definitely calculate the integral numerically to verify that the answer is correct.

**vincisonfire** · Nov 5th 2008, 03:05 AM

Pay attention the denominator is $x^2-y^2$ .
I understood the mapping but it is the diagonals I had difficulty with. I you could tell me how to find there expression in the changed system of coordinates.
Now I think I understand how to do it. You just set y =2 and that implies u = x^2+4 and v = x^2 -4 and u = v +8.
Anyway thanks a lot.
PS This seem complicated to calculate with matlab. And remember in math reasoning is important not calculations.

**shawsend** · Nov 5th 2008, 04:32 AM

Originally Posted by vincisonfire

Pay attention the denominator is $x^2-y^2$ .

Ok. And I agree with the reasoning part but a good numerical analysis is a good practical confirmation of that reasoning.

However I think my calculations above are wrong because the original integral can be split up (horizontally) into three integrals:

$\mathop\int\limits_{D}\frac{xy}{x^2-y^2}dA=$

$\int_{-1}^{\sqrt{3/2}}\int_{\sqrt{4-y^2}}^{\sqrt{9+y^2}} fdxdy+ \int_{\sqrt{3/2}}^{\sqrt{7/2}}\int_{\sqrt{1+y^2}}^{\sqrt{9+y^2}} fdxdy+ \int_{\sqrt{7/2}}^{2}\int_{\sqrt{1+y^2}}^{\sqrt{16-y^2}} fdxdy\approx1.592$

and $\pm\left(1/4+\ln(81/16)\right)\approx \pm 1.872$

. . . I just don't understand big business . . .

**shawsend** · Nov 5th 2008, 05:26 AM

I'm really amazed that I can plot both the domain and the region of the plot extending over (and under) this domain. Just looking at the scale of the plot below we can estimate the integral will be rather small, say in the order of no more than 10. The red surface is the region of the function to be integrated over the blue domain. Pretty cool I think.

For the record, the reason I think my calculations aren't working is that the transformation:

$u=x^2+y^2$
$v=x^2-y^2$

is not one-to-one over the domain in question: For example consider two points on the arc edge of the circle $x^2+y^2=4$ . That's the left border of the domain.

$v[2\cos[a],2\sin[a]]=v[2\cos(-a),2\sin(-a))$

for example when $a=\pi/12$

For this reason, I do not believe this integral can be solved using these transformations unless it can be broken up into regions which are one-to-one.

**vincisonfire** · Nov 5th 2008, 06:20 AM

I get the same answer working it without transformation but I don't get the same for the other one. Maybe we both did mistakes.

**shawsend** · Nov 5th 2008, 06:37 AM

Hey vin, the plot below shows where the mapping is not one-to-one I think (the blue part). I'm wondering if I integrate over the whole domain once then add to it, integrating over just the blue part? I'm curious if it works that way when the mapping is not one-to-one? You know this is exactly what real math is: real problems involve sometimes getting it wrong along the way to getting it right and that is where the learning comes from

**vincisonfire** · Nov 5th 2008, 07:59 AM

Doesn't work this way. At least for me.

**shawsend** · Nov 5th 2008, 08:02 AM

Ok, I think I'm making some progress: Suppose we use the transformation where it is actually one-to-one. That would be the case if we clip the domain $D$ to the portion above the x-axis. In this case, the actual integral is:

$ \int_{0}^{\sqrt{3/2}}\int_{\sqrt{4-y^2}}^{\sqrt{9+y^2}} fdxdy+ \int_{\sqrt{3/2}}^{\sqrt{7/2}}\int_{\sqrt{1+y^2}}^{\sqrt{9+y^2}} fdxdy+ \int_{\sqrt{7/2}}^{2}\int_{\sqrt{1+y^2}}^{\sqrt{16-y^2}} fdxdy\approx 1.87186 $

and in terms of the transformed coordinates:

$ \mathop\int\limits_{D}\frac{xy}{x^2+y^2}dA=\pm\fra c{1}{8}\left(\int_4^9\int_1^u \frac{1}{v}dvdu+\int_9^{16}\int_{u-8}^9 \frac{1}{v}dvdu\right)\approx\pm 1.87186$

Hey vin, I think we have it. The discrepancy may be coming from the direction of path over both components of the transformation: we may have to subtract the second component because the direction over the domain is changing. I'll check this hypothesis later.

**shawsend** · Nov 6th 2008, 01:37 AM

The transformation: $T:\{u=x^2+y^2, v=x^2-y^2\}$ needs to be restricted to domains over which it is one-to-one and onto. That was the cause of all the problems. In the case of $x\geq 0$ it's obvious the x-axis separates domains over which the transformation is one-to-one and thus we'd have to partition the domain $D=D_1+D_2$ along the x-axis. In the case of the inverses: $x=\sqrt{(u+v)/2}, y=\sqrt{(u-v)/2}$ , we can see that these are onto $D$ in a region between the diagonals $v=u$ and $v=-u$ . Noting this and how the point $(x,y)$ travels around the border of the respective images, it's relatively a simple matter to express the integral in the new coordinates:

$\mathop\int\limits_{D}\frac{xy}{x^2-y^2}dA=-1/8\left(\hspace{3pt}\mathop\int\limits_{D_1}\frac{1 }{v}dA+\mathop\int\limits_{D_2}\frac{1}{v}dA\right )\$

where the regions are indicated in the plots below. Note in the case of $D_1$ the orientation is reversed so we'll need a negative sign leading to:

$\int\limits_{D}\frac{xy}{x^2-y^2}dA=-1/8\left[-\left\{\int_4^9\int_1^u \frac{1}{v}dvdu+\int_9^{16}\int_{u-8}^9 \frac{1}{v}dvdu\right\}\right.$
$\hspace{100pt}+\left. \left\{\int_4^9\int_{u-2}^u \frac{1}{v}dvdu +\int_9^{11}\int_{u-2}^9 \frac{1}{v}dvdu\right\}\right] $

$=-1/8\left[-(2+8\ln(9/8)+8\ln(9)-\ln(256))+2+\ln(81/64)\right]$

$\approx 1.59241$

which agrees with the numerical estimate of the integral in the variables x,y. Alright, I'm done.

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