Area of Plane Region Bounded by Curve: Find Solution

[bluskies]

Homework Statement

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

Homework Equations

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

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.

 

[SammyS]

Homework Statement

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

Homework Equations

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

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.

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

 

[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?

 

[SammyS]

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?

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?

 

[SammyS]

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

 

[SammyS]

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

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)​

 

[SammyS]

Definitely, use polar coordinates.