# Area of plane region

1. Dec 24, 2011

### bluskies

1. The problem statement, all variables and given/known data

Find the area of the plane region bounded by the curve
$$(x^2+y^2)^3 = x^4+y^4$$

2. Relevant equations
The change of variables formula:
$$\int\int_R F(x,y)dxdy = \int\int_S G(u,v)\left| \frac{∂(x,y)}{∂(u,v)}\right| dudv$$

3. The attempt at a solution

I recognize this as a change of variables problem, and in general I understand how to do change of variables, but for this one I cannot figure out what to use as the new variables u = u(x,y) and v = v(x,y). Previously I could tell what I needed by inspection, but I cannot tell with this problem. Is there some method for directly computing what u and v must be?

I thought of using ${u = x^2 + y^2}$ and ${v = x^4 + y^4}$, but I can't figure out how to use this to solve for x and y, so I reached a dead end.

Any help would be appreciated.

2. Dec 24, 2011

### SammyS

Staff Emeritus
Try $\displaystyle u=x^2+y^2\ \ \text{ and }\ \ v=2xy\,.$

3. Dec 24, 2011

### bluskies

Thank you for your help! Using the above change of variables, I solved for x and y in terms of u and v (to compute the Jacobian $\frac{∂(x,y)}{∂(u,v)}$):

$v = 2xy \Rightarrow y = \frac{v}{2x}$

$u = x^2+y^2=x^2 + \frac{v^2}{4x^2} \Rightarrow 4x^4-4ux^2+v^2=0$

Let $z=x^2$. Then

$4z^2-4uz+v^2=0 \Rightarrow z = \frac{1}{2}\left( u \pm \sqrt{u^2-v^2} \right)$

$\Rightarrow x = \sqrt{ \frac{1}{2}\left( u \pm \sqrt{u^2-v^2} \right) }$

$\Rightarrow y = \frac{v}{2\sqrt{ \frac{1}{2}\left( u \pm \sqrt{u^2-v^2} \right) }}$

The curve becomes

$u^3 = x^4+y^4 = (x+y)^4 - 2xy(2y^2+3xy+2x^2) = (u+v)^2-4xy(x+y)^2+2x^2y^2 = (u+v)^2 - 2v(u+v)+\frac{1}{2}v^2 = u^2-\frac{1}{2}v^2$

$\Rightarrow u^3 = u^2-\frac{1}{2}v^2$

This seems overly complicated, like I made a mistake somewhere, but I cannot find it - does it look correct? Also, when I graphed the new curve, I got the below graph, so I think I didn't do something right. Isn't it supposed to be a nice curve?

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4. Dec 24, 2011

### SammyS

Staff Emeritus
I think that's all right. I got the same graph.

I transformed the expression by noticing that $\displaystyle u^2=(x^2+y^2)^2=x^4+2x^2y^2+y^4\,.$

So subtract 2v2 from that to get x4+y4 .

Of course, (x2+y2)3 was obvious.

I haven't looked at the resulting integral, but solving for v is fairly easy:
$v=\pm\sqrt{2(u^2-u^3)}=\pm|u|\sqrt{2}\sqrt{1-u}\,.$​

Your graph is consistent with this from the point of view that the loop corresponds to 0 ≤ u ≤ 1. $\sqrt{1-u}$ is real there.

Have you worked out the Jacobian yet?

5. Dec 24, 2011

### SammyS

Staff Emeritus
After looking at the graph of u^3 < u^2-v^2, I also have my doubts about this transformation.

6. Dec 25, 2011

### SammyS

Staff Emeritus
Before I go on about my previous suggestion for a translation, have you tried polar coordinates? ...

After messing around with a few sets of ordered pairs and a few other details, like graphing the region in the xy-plane, I like this transformation again.

Here's WolframAlpha's graph of the region in the xy-plane, along with the graphs of two circles, one with radius 1 the other with radius = 1/√(2) :

Here's WolframAlpha's graph of the region in the xy-plane, along with the graph of y = ±x .

Since x2 + y2 ≥ 0, u ≥ 0, for the region of interest.

The point (x,y)=(a, ±a) → (u,v)=(2a2, ±2a2), where the signs are correlated.
Specifically, the point (x,y)=(1/2, ±1/2) → (u,v)=(1/2, ±1/2) .​

The point (x,y)=(±a,0) → (u,v)=(a2,0) also the point (x,y)=(0, ±a) → (u,v)=(a2,0) .
Specifically, the points (x,y)=(±1,0) and (x,y)=(0, ±1) all go to (u,v)=(1,0)​

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7. Dec 25, 2011

### SammyS

Staff Emeritus
Definitely, use polar coordinates.