# Double integral help

1. Oct 29, 2008

can someone please give me some help on this integral. it should be solvable analytically.

$$\int^{1}_{-1}\int^{2\pi}_{0}\frac{1}{2+x+\sqrt{1-x^2}cos(y)+\sqrt{1-x^2}sin(y)}dxdy$$

2. Oct 30, 2008

### Staff: Mentor

This one might be easier to work in polar coordinates:
x = r*cos($$\theta$$)
y = r*sin($$\theta$$)
dxdy = rdrd$$\theta$$

You'd want to change the limits of integration, also, to represent the rectangle R={(x,y): 0 <= x <= 2 $$\pi$$, -1 <= y <= 1} in its polar form.

I haven't worked this problem, but that's what I would try.

3. Oct 30, 2008

i don't think the integral is in cartesian coordinates. x ranges from (-1,1) and y from (0,2pi). i'm not sure what this coordinate system is called, but it's different. so ur suggestion wouldn't work.

4. Oct 30, 2008

### Staff: Mentor

It's ***definitely*** in Cartesian coordinates, as evidenced by dx and dy. The fact that y ranges from 0 to 2pi has nothing to do with whether this is or isn't in Cartesian coordinates.

5. Oct 30, 2008

sorry, i did a variable change because i tried to save time by just using x and y instead of greek symbols mu and gamma.

i hope that clears it up that they are not cartesian coordinates.

6. Oct 30, 2008

### Staff: Mentor

No, it does not.

You can start with a double integral over a closed and bounded region R and get this integral:
$$\int \int$$ f(x, y) dA
The Cartesian iterated integral has the form $$\int \int$$ f(x, y) dx dy.
The polar iterated integral has the form $$\int \int$$ f(r, $$\theta$$) r dr d$$\theta$$.

It doesn't matter if you use mu and gamma instead of x and y, it's still Cartesian.

7. Oct 30, 2008

### gabbagabbahey

Do you really have to solve this by hand, analytically? If you are allowed to use Mathematica, I would definitely consider it....anyways, I haven't solved this by hand but if you choose to do so, you will have to do the y-integral first (unless you can find a useful change of variables)....Mathematica's solution for the y-integral involves Cos(y/2) and Sin(y/2) terms, so using the double angle formula might help you.

I may be able to come up with something more useful if you show me the original problem; there may be a way to avoid this particular integral altogether.

8. Oct 30, 2008

### gabbagabbahey

There is no way to determine the nature of the Co-ordinate system from the given integral...for example, what if your $f(r,\theta)=g(r,\theta)/r$? then the integral looks like $\int g(r,\theta)dr d \theta$ and a simple renaming of r to x and theta to y could make it look like the given integral.

That being said, I would guess based on the form of the integrand that if this integral represents some sort of physical application; it is given in curvilinear coordinates. In either case, it is unimportant since the integrand is a scalar for this problem.

9. Oct 30, 2008

### Staff: Mentor

I stand corrected...

10. Oct 31, 2008

the original integral was $$\int_{4\pi}\frac{1}{1+A.\Omega}d\Omega$$ where A is a vector and $$\Omega$$ is the solid angle.

11. Oct 31, 2008

### gabbagabbahey

What is the vector $\vec{A}$? And what surface is $$\Omega$$ the solid A angle of (sphere, cone...etc)?

12. Oct 31, 2008

to make it simple, vector A = (1,2,3) in the cartesian coordinates and it is the solid angle over a sphere. i'm pretty sure there's a way to simplify this. i've tried multiplying top and bottom by 1+A.Omega and then separating them into two integrals. One is an even function and one is an odd function. So the odd function drops to zero and I am left with:

$$2.\int_{A.\Omega}\frac{1}{1-(A.\Omega)^2}d\Omega$$

i'm not sure if i made it worst!

Last edited: Oct 31, 2008
13. Oct 31, 2008

### gabbagabbahey

Usually, the solid angle for a sphere is just the scalar quantity $\Omega=4 \pi$ and $d \Omega= sin(\theta) d \theta d\phi$ in the usual spherical coordinates... Perhaps you'd better explain to me what expressions you are using, and how you are taking he dot product of a vector with a scalar????

14. Oct 31, 2008

oh ok. $$\Omega = (\mu,\sqrt{1-\mu^2}cos(\gamma),\sqrt{1-\mu^2}sin(\gamma))$$ in the Cartesian coordinate.
hmmm...I've never seen the solid angle expressed as a vector before.. what are $\mu$ and $\gamma$ and what expression are you using for $d \Omega$?