# Double integral problem

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?

## Answers and Replies

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Dick
Homework Helper
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?

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?
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.

Dick
Homework Helper
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)|?

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)|?
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?

Dick