# Double integral problem

1. Mar 12, 2008

### naggy

I'm supposed to prove that

$$\int\int_{S}^{}\ f(ax + by + c) \, dA \ =2 \int_{-1}^{1} \sqrt{1 - u^2} f(u\sqrt{a^2 + b^2} + c) \, du$$

Where S is the disk x^2 + y^2 <= 1. It is also given that a^2 + b^2 is not zero

I can´t use polar coordinates and I can´t see how you simplify the surface S in any other way. What is the change of variable and why?

2. Mar 12, 2008

### Dick

It should be pretty clear from the formula what the change of variables should be, right? u*sqrt(a^2+b^2)+c=ax+by+c. So in vector terms u=(a,b).(x,y)/|(a,b)|. u is the dot product of (x,y) with the normalization of the vector (a,b). It's just a rotation of the unit disk so that f is a function only of u. Does that help?

3. Mar 13, 2008

### naggy

I thought about that, it's sort of obvious, but then I can´t see what the other substitution is, that is

u = u(x,y)
v = v(x,y)

Now I know what to substitute for u, but not v.

4. Mar 13, 2008

### Dick

Set v to be the dot product of (x,y) with a unit vector perpendicular to (a,b)/|(a,b)|. How about (-b,a)/|(a,b)|?

5. Mar 13, 2008

### naggy

WOw this works. I don´t really understand why. What do you mean that it is a rotation of the unit disk? What does that even mean? Normalization of the vector (a,b)?

How did you know the other parameter should be perpendicular to the first one?

I think you are parametrizing the functions with vectors, but can you treat the variables x and y as regular vectors like i,j and k?

6. Mar 13, 2008

### Dick

The function f is constant along lines that are parallel to the line ax+by=0 (lines ax+by=C for C a constant). If you don't want to work with vectors and dot products then you can just make a choice of (u,v) by rotating (x,y) so that f is constant along one of those directions. A rotation looks like u=x*cos(t)-y*sin(t), v=x*sin(t)+y*cos(t), right? So a clever choice of angle t is the direction of the line ax+by=0. This means tan(t)=-a/b. If that's tan(t) then sin(t)=b/sqrt(a^2+b^2), cos(t)=-a/sqrt(a^2+b^2). If you start working this stuff out, you'll see you are lead back to the same transformation we've been talking about. I may have some details wrong, like getting u and v backwards, but you get the idea, right?