Can't solve double integral

1. Jun 25, 2014

mantgx

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

∫∫D√(9x2+4y2) dx dy

D is the region: x2/4+y2/9=1

My understanding is that i have to integrate the function of a density to calculate the mass of plate which is ellipse. Problem is i cant and shouldnt be able to integrate this integral at my level, so am i missing some way of simplification?

Last edited: Jun 25, 2014
2. Jun 25, 2014

Orodruin

Staff Emeritus
Have you considered a change of variables (or two)?

3. Jun 25, 2014

mantgx

to polar cordinates?

4. Jun 25, 2014

Orodruin

Staff Emeritus
Try getting the coordinates to a form where making the change to polar coordinates make more sense first. Hint: An ellipse does not give the simplest boundary condition in polar coordinates.

5. Jun 25, 2014

Zondrina

Why not let $x = 2u$ and $y = 3v$.

Then switch to polar.

6. Jun 25, 2014

mantgx

i do not understand how this will help.

can please someone solve the problem for me im desperate

7. Jun 25, 2014

Orodruin

Staff Emeritus
Why don't you try doing it? It will be much more instructive than if we solve it for you.

8. Jun 25, 2014

Zondrina

There's a theorem you should be familiar with, it should be along the lines of:

$\int_R \int f(x,y) dxdy = \int_{R'} \int f(u,v) |J| dudv$.

Where $|J|$ is the Jacobian matrix of the transformation, $R$ and $R'$ are the regions. You have an integral, which looks like the left side of the above theorem. The theorem tells you that you can change the region and co-ordinates of a double integral, and get the same answer.

So making the substitution I noted in an earlier post, what happens to the region $D$? It gets mapped to some region $D'$ through an invertible map. What is the result of plugging the substitution into the integral on the right?

9. Jun 25, 2014

mantgx

∫∫R'√(36u^2+12v^2) J du dv

?

10. Jun 25, 2014

Zondrina

No, the integral becomes $36 \int \int_{D'} \sqrt{u^2 + v^2} |J| dudv$. Where of course $dx = 2du$ and $dy = 3dv$.

The region $D → D' := u^2 + v^2 = 1$. Notice this is the equation of a circle with radius 1.

The Jacobian is the matrix of partial derivatives of the transformation. Can you compute it?

For reference: http://en.wikipedia.org/wiki/Jacobian_matrix_and_determinant

11. Jun 25, 2014

Orodruin

Staff Emeritus
What is $4\times 3^2$ and what is the expression for $J$?

If you are not familiar with jacobians, try the following instead: what is $du$ if $u = x/2$? (It is ok to use this type of reasoning as long as you are making parameter transformations of the type considered here, i.e., u = f(x) and v = g(y))

12. Jun 25, 2014

Orodruin

Staff Emeritus
$\sqrt{36} = 6$ ...

13. Jun 25, 2014

mantgx

ok, i understand a little now.

J=6 ?

14. Jun 25, 2014

mantgx

is the final solution 36pi?

15. Jun 25, 2014

Orodruin

Staff Emeritus
Yes, J=6 (or rather |J|=6) in the first variable change.

What is the area element $du\, dv$ expressed in polar coordinates?

16. Jun 25, 2014

mantgx

24pi

17. Jun 25, 2014

Zondrina

Don't forget $dx$ and $dy$.

18. Jun 25, 2014

Orodruin

Staff Emeritus
I would have agreed if you did not still have the jacobian in your expression.

19. Jun 25, 2014

Orodruin

Staff Emeritus
What are your integration limits in polar coordinates? (The answer is the same as I get by doing it in my head, which I do not trust at 1am, but I just want to make sure you've got it right)

20. Jun 25, 2014

r=(0,1)
fi=(0,2pi)