Solving \int\int_D (x^4-y^4) dxdy with Optimal Change of Variables

[Icebreaker]

Which change of variable should I use to find:

$$\int\int_D (x^4-y^4) dxdy$$

Where D is in the first quadrant with"

$$1 \leq x^2-y^2 \leq 3, 2\leq xy \leq 3$$

 

[Physics Monkey]

Try $$u = x^2 - y^2$$ and $$v = xy$$ since this makes the integration region simple.

 

[Icebreaker]

It does, but finding the Jacobian and solving for x and y are almost impossible. Or so I tried.

 

[Physics Monkey]

Yeah, I just gave it a try and it looks that way doesn't it. Not impossible, but certainly a mess on the face of it.

Edit: Ok, a better choice is $$u = x^2 + y^2$$ and $$v = x^2 - y^2$$.

 

[arildno]

It is quite simple:
In order to change variables, you have been taught to calculate the quantities $\frac{\partial{x}}{\partial{u}},\frac{\partial{x}}{\partial{v}},\frac{\partial{y}}{\partial{u}},\frac{\partial{y}}{\partial{v}}$ then form a matrix, and then calculate the determinant of this matrix, right?

But to go from the (x,y) description to the (u,v) description should be the INVERSE of going from the (u,v) description to the (x,y) description!.
Agreed?

Thus, the matrix you seek should be the inverse of the matrix you get by calculating the quantities $\frac{\partial{u}}{\partial{x}},\frac{\partial{u}}{\partial{y}},\frac{\partial{v}}{\partial{x}},\frac{\partial{x}}{\partial{y}}$
In particular, its determinant is the reciprocal of the other matrix' determinant.

does that help you?

 

[Icebreaker]

Yes, the Jacobian I have. But how do I get the expression x^4-y^4 from the previous substitution?

 

[Physics Monkey]

Are you talking about $$u = x^2 + y^2$$ and $$v = x^2 - y^2$$? If you are then try multiplying u and v.

 

[Icebreaker]

No I meant u = x^2-y^2 and v = xy.

 

[arildno]

Yes, the Jacobian I have. But how do I get the expression x^4-y^4 from the previous substitution?

Well, if you have calculated correctly, your integrand should now be:
$$\frac{x^{4}-y^{4}}{2(x^{2}+y^{2})}=\frac{(x^{2}-y^{2})(x^{2}+y^{2})}{2(x^{2}+y^{2})}=\frac{x^{2}-y^{2}}{2}=\frac{v}{2}$$

 

[Physics Monkey]

Nice, both ways work pretty well.

 

[Icebreaker]

Ah, of course! Man I should slap myself silly for not seeing that.

 

[arildno]

Ah, of course! Man I should slap myself silly for not seeing that.

I'm not in a habit of slapping others, so I can't help you out on that account.