Double Integral Help - Solve Analytically

[madeinmsia]

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

 

[Mark44]

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.

 

[madeinmsia]

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.

 

[Mark44]

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.

 

[madeinmsia]

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.

 

[Mark44]

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.

 

[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.

 

[gabbagabbahey]

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.

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.

 

[Mark44]

I stand corrected...

 

[madeinmsia]

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.

 

[gabbagabbahey]

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.

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

 

[madeinmsia]

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!

 

[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?

 

[madeinmsia]

oh ok. \Omega = (\mu,\sqrt{1-\mu^2}cos(\gamma),\sqrt{1-\mu^2}sin(\gamma)) in the Cartesian coordinate.

 

[gabbagabbahey]

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?