# Homework Help: 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

### mantgx

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

21. Jun 25, 2014

### Orodruin

Staff Emeritus
:thumbs:

I am assuming you got 36π by forgetting the r in the area element? (du dv = r dr dθ, i.e., |J|=r for the change to polar coordinates)

22. Jun 25, 2014

### Zondrina

Whoops, my bad. Didn't notice that while typing.

23. Jun 25, 2014

### mantgx

thank you guys 100x times

24. Jun 25, 2014

### Orodruin

Staff Emeritus
Copy and Paste, the source of 98% of all typos in published papers - also 73.4% of all statistics you will read are made up on the spot ;)

25. Jun 26, 2014

### ehild

What is the shape of the region bounded by the line $\frac{x^2}{4}+\frac{y^2}{9}=1$ ?

You can rewrite the square root in the integrand as $\sqrt{9x^2+4y^2}=6\sqrt{x^2/4+y^2/9}=6$, so you have the integral $\int _D{dxdy}$ which is the area of the shape bounded by the line $\frac{x^2}{2^2}+\frac{y^2}{9}=1$.

ehild