# Rewrite Indefinite Integral in Terms of Elliptic Coordinates

1. Sep 9, 2013

### wifi

Problem:

Rewrite the indefinite integral $\iint\limits_R\, (x+y) dx \ dy$ in terms of elliptic coordinates $(u,v)$, where $x=acosh(u)cos(v)$ and $y=asinh(u)sin(v)$.

Attempt at a Solution:

So would it be something like,

$\iint\limits_R\, (x+y) dx \ dy = \iint\limits_R\, [acosh(u)cos(v)+asinh(u)sin(v)] dx \ dy$. I need to calculate $dx$ and $dy$ in terms of $u$ and $v$, then substitute them in the integral, correct?

The only thing that comes to mind is the multivariable chain rule: $\frac{dx}{dt} = \frac{\partial x}{\partial u} \frac{du}{dt}+\frac{\partial y}{\partial v} \frac{dv}{dt}$ or $\frac{dy}{dt} = \frac{\partial y}{\partial u} \frac{du}{dt}+\frac{\partial y}{\partial v} \frac{dv}{dt}$

Any help is appreciated. Thanks in advance.

2. Sep 9, 2013

### wifi

I should note that the task is to merely rewrite the integral, not to evaluate it.

3. Sep 9, 2013

### vela

Staff Emeritus
Think Jacobian.

4. Sep 9, 2013

### wifi

I'm sorry, can you be more specific?

5. Sep 9, 2013

### vela

Staff Emeritus
6. Sep 9, 2013

### wifi

I think I got it, hold on.

7. Sep 9, 2013

### wifi

So I simply use $\iint\limits_R\ f(x,y) dx \ dy = \iint\limits_{R^*}f(x(u,v),y(u,v)) det(J) du \ dv$?

8. Sep 9, 2013

### vela

Staff Emeritus
Yup.

9. Sep 9, 2013

### wifi

Okay thanks! I'm gonna LaTeX my work so you can see.

10. Sep 9, 2013

### wifi

For the Jacobian I got $J = (a^2sinh^2(u)cos^2(v)-a^2cosh(u)sinh(u)cos(v)sin(v))du \ dv$. Is that correct?

11. Sep 9, 2013

### vela

Staff Emeritus
I don't get that.

12. Sep 9, 2013

### wifi

Darn. I'll try agian.

13. Sep 9, 2013

### wifi

I think I messed up a negative. Now I'm getting $J= (asinh(u)cos(v))^2+(acosh(u)sin(v))^2$

14. Sep 9, 2013

### vela

Staff Emeritus
I'm not at my computer right now, but it looks like what I got.

15. Sep 9, 2013

### wifi

Awesome! Thank you so much!